JEE Functions Questions

The function $$f : \mathbb{R} \to \mathbb{R}$$ satisfies the additive rule
$$f(x+y)=f(x)+f(y)\quad\forall\,x,y\in\mathbb{R}.$$ For such a function:

• $$f(0)=f(0+0)=f(0)+f(0)\;\Rightarrow\;f(0)=0$$.
• For any integer $$n$$, $$f(nx)=nf(x)$$ (apply the rule $$n$$ times).
• For any rational $$r=\dfrac{p}{q}$$, $$f(rx)=rf(x)$$ (use the integer result on $$q x$$ and divide).

Given $$f\!\left(-\dfrac35\right)=12$$, we first find $$f\!\left(\dfrac35\right)$$:
$$f\!\left(\dfrac35\right)=-f\!\left(-\dfrac35\right)=-12.$$ Now choose $$x=\dfrac35$$ and a rational multiplier $$r=\dfrac{1/4}{3/5}=\dfrac{5}{12}$$ so that $$r\,x=\dfrac14$$. Using the rational-multiple rule,

$$f\!\left(\dfrac14\right)=r\,f(x)=\dfrac{5}{12}\,(-12)=-5.$$

The function $$g : \mathbb{R} \to (0,\infty)$$ satisfies the multiplicative rule
$$g(x+y)=g(x)\,g(y)\quad\forall\,x,y\in\mathbb{R}.$$ For such a function:

• $$g(0)=g(0+0)=g(0)^2\;,\;g(0)\gt 0\;\Rightarrow\;g(0)=1.$$
• For any integer $$n$$, $$g(nx)=g(x)^n$$ (apply the rule $$n$$ times).
• Consequently, for rational $$r=\dfrac{p}{q}$$, $$g(rx)=g(x)^r$$ (use the integer result and take roots).

Given $$g\!\left(-\dfrac13\right)=2$$, we obtain $$g\!\left(\dfrac13\right)=\dfrac1{g(-1/3)}=\dfrac12.$$

Write $$-2=6\!\left(-\dfrac13\right)$$. Using the integer-multiple rule,

$$g(-2)=g\!\left(-\dfrac13\right)^{\,6}=2^{6}=64.$$

Now evaluate the required expression:

$$\bigl(f(1/4)+g(-2)-8\bigr)\,g(0)=\bigl((-5)+64-8\bigr)\,(1)=51.$

Hence the value is 51.

We want to determine $$\sqrt{3\alpha + 1}$$ where $$\underbrace{f \circ f \circ \cdots \circ f}_{10}(x) = \frac{2^{10}x}{\sqrt{1+9\alpha x^2}}$$ and $$f(x) = \frac{2x}{\sqrt{1+9x^2}}\,. $$

To begin, we compute the second iterate by substituting $$f(x)=\frac{2x}{\sqrt{1+9x^2}}$$ into itself, yielding $$f(f(x)) = \frac{2 \cdot \frac{2x}{\sqrt{1+9x^2}}}{\sqrt{1 + 9\bigl(\frac{2x}{\sqrt{1+9x^2}}\bigr)^2}} = \frac{\frac{4x}{\sqrt{1+9x^2}}}{\sqrt{1+\frac{36x^2}{1+9x^2}}} = \frac{4x}{\sqrt{(1+9x^2)+36x^2}} = \frac{4x}{\sqrt{1+45x^2}}\,. $$ Hence $$f^{(2)}(x)=\frac{2^2x}{\sqrt{1+9\cdot 5x^2}}\,. $$

Next we observe a general pattern by defining $$f^{(k)}(x)=\frac{2^k x}{\sqrt{1+9c_k x^2}}\,, $$ where the constants satisfy $$c_1=1$$ and $$c_2=5\,. $$ Applying one more iteration gives $$f^{(k+1)}(x)=f\!\Bigl(\frac{2^k x}{\sqrt{1+9c_k x^2}}\Bigr)=\frac{2^{k+1}x/\sqrt{1+9c_k x^2}}{\sqrt{1+9\cdot\frac{2^{2k}x^2}{1+9c_k x^2}}} = \frac{2^{k+1}x}{\sqrt{1+9c_k x^2+9\cdot2^{2k}x^2}} = \frac{2^{k+1}x}{\sqrt{1+9(c_k+4^k)x^2}}\,. $$ Therefore the recurrence for the coefficients is $$c_{k+1}=c_k+4^k\,. $$

Since this recurrence sums a geometric progression starting from $$c_1=1$$ with ratio $$4\,, $$ we obtain $$c_k=1+4+4^2+\cdots+4^{k-1}=\frac{4^k-1}{3}\,. $$ Checking small values confirms that $$c_1=(4-1)/3=1$$ and $$c_2=(16-1)/3=5\,. $$ Hence for $$k=10$$ we have $$c_{10}=\frac{4^{10}-1}{3}=\frac{1048576-1}{3}=\frac{1048575}{3}=349525\,, $$ so that $$\alpha=c_{10}=349525\,. $$

Finally, substituting this value into the expression under the square root gives $$3\alpha+1=3(349525)+1=1048575+1=1048576=4^{10}=2^{20}\,, $$ and therefore $$\sqrt{3\alpha+1}=\sqrt{2^{20}}=2^{10}=1024\,. $$

Thus, the desired value is 1024.

We need to find natural number pairs $$(x, y)$$ that satisfy $$2x + 3y = 23$$. Since $$3y$$ must be odd (because $$23$$ is odd and $$2x$$ is even), $$y$$ must be odd.

• If $$y=1: 2x + 3 = 23 \implies 2x = 20 \implies x = 10$$. Pair: $$(10, 1)$$

• If $$y=3: 2x + 9 = 23 \implies 2x = 14 \implies x = 7$$. Pair: $$(7, 3)$$

• If $$y=5: 2x + 15 = 23 \implies 2x = 8 \implies x = 4$$. Pair: $$(4, 5)$$

• If $$y=7: 2x + 21 = 23 \implies 2x = 2 \implies x = 1$$. Pair: $$(1, 7)$$

• If $$y=9: 2x + 27 = 23$$ (No natural number solution).

So, $$A = \{(10, 1), (7, 3), (4, 5), (1, 7)\}$$. The number of elements $$n(A) = 4$$.

Finding the elements of Set $$B$$

$$B$$ consists of the $$x$$-coordinates from $$A$$.

$$B = \{10, 7, 4, 1\}$$. The number of elements $$n(B) = 4$$.

 Calculating One-One Functions

A one-one function (injection) from a set of size $$n$$ to a set of size $$n$$ is simply a permutation of the elements.

$$\text{Number of functions} = n! = 4! = 4 \times 3 \times 2 \times 1 = 24$$

$$a = [a] + \{a\}$$. Let $$[a] = n$$ and $$\{a\} = f$$, where $$f \in [0, 1)$$.

Equation: $$|2(n+f) - 1| = 3n + 2f$$.

Case 1: $$2a - 1 \ge 0 \implies a \ge 1/2$$

$$2n + 2f - 1 = 3n + 2f \implies n = -1$$.

Since $$a = n+f = -1+f$$, the max value is $$<0$$, which contradicts $$a \ge 1/2$$. No solution.

Case 2: $$2a - 1 < 0 \implies a < 1/2$$

$$-(2n + 2f - 1) = 3n + 2f \implies 1 - 2n - 2f = 3n + 2f \implies 4f = 1 - 5n$$.

Since $$0 \le f < 1$$, then $$0 \le \frac{1-5n}{4} < 1$$.

• $$1-5n < 4 \implies -3 < 5n \implies n > -0.6$$

• $$1-5n \ge 0 \implies 5n \le 1 \implies n \le 0.2$$

The only integer $$n$$ in this range is $$n = 0$$.

If $$n = 0$$, $$4f = 1 \implies f = 1/4$$.

$$a = 0 + 1/4 = 1/4$$.

Sum of $$a \in S$$ is $$1/4$$.

$$72 \times (1/4) = 18$$.

We analyze the relation $$R = \{(a,b) : \gcd(a,b) = 1, 2a \neq b, a, b \in \mathbb{Z}\}$$.

Reflexive?

For reflexivity, we need $$(a, a) \in R$$ for all $$a \in \mathbb{Z}$$. This requires $$\gcd(a, a) = 1$$ and $$2a \neq a$$.

$$\gcd(a, a) = |a|$$, which equals 1 only when $$a = \pm 1$$. For $$a = 2$$: $$\gcd(2,2) = 2 \neq 1$$. So $$(2, 2) \notin R$$.

Therefore, R is NOT reflexive.

Symmetric?

For symmetry, if $$(a, b) \in R$$, we need $$(b, a) \in R$$. We need $$\gcd(b, a) = 1$$ (same as $$\gcd(a, b) = 1$$, so this holds) AND $$2b \neq a$$.

Consider $$(1, 2)$$: $$\gcd(1, 2) = 1$$ and $$2(1) = 2 = b$$, so $$(1, 2) \notin R$$ (fails $$2a \neq b$$).

Consider $$(2, 1)$$: $$\gcd(2, 1) = 1$$ and $$2(2) = 4 \neq 1$$, so $$(2, 1) \in R$$.

But we need the reverse: $$(1, 2)$$. We showed $$(1, 2) \notin R$$.

So $$(2, 1) \in R$$ but $$(1, 2) \notin R$$. Therefore, R is NOT symmetric.

Transitive?

For transitivity, if $$(a, b) \in R$$ and $$(b, c) \in R$$, we need $$(a, c) \in R$$.

Consider $$(3, 2)$$: $$\gcd(3, 2) = 1$$, $$2(3) = 6 \neq 2$$. So $$(3, 2) \in R$$.

Consider $$(2, 3)$$: $$\gcd(2, 3) = 1$$, $$2(2) = 4 \neq 3$$. So $$(2, 3) \in R$$.

Now check $$(3, 3)$$: $$\gcd(3, 3) = 3 \neq 1$$. So $$(3, 3) \notin R$$.

We have $$(3, 2) \in R$$ and $$(2, 3) \in R$$ but $$(3, 3) \notin R$$. Therefore, R is NOT transitive.

The correct answer is Option 4: neither symmetric nor transitive.

We begin by determining which of the two statements is true.

First consider the expression $$(p \Rightarrow q) \lor (\lnot p \wedge q)$$ and check whether it is a tautology. Recall that $$p \Rightarrow q \equiv \lnot p \lor q$$. Hence

since $$\lnot p \wedge q$$ is absorbed by $$\lnot p \lor q$$. Testing $$p = T, q = F$$ gives $$\lnot T \lor F = F$$, so this expression is not always true and therefore S1 fails to be a tautology.

Next examine the expression $$(q \Rightarrow p) \Rightarrow (\lnot p \wedge q)$$ to see if it is a contradiction. Noting that $$q \Rightarrow p \equiv \lnot q \lor p$$, we find

and testing $$p = F, q = T$$ yields $$\lnot F \wedge T = T$$. Since the expression can be true, S2 is not a contradiction.

For completeness, the following truth table verifies the values of S1 and S2 under all assignments of $$p$$ and $$q$$:

Since S1 is not a tautology (it fails when $$p = T, q = F$$) and S2 is not a contradiction (it holds when $$p = F, q = T$$), neither statement is true.

The correct answer is Option A: Neither (S1) nor (S2) is True.

Step 1: Understand the Logical Equivalence Rules

The standard logical identity for an implication with a disjunction (OR) in its antecedent is given by:

$$(p \vee q) \Rightarrow r \equiv (p \Rightarrow r) \wedge (q \Rightarrow r)$$

Step 2: Analyze Statement S1

$$S1: ((p \vee q) \Rightarrow r) \Leftrightarrow (p \Rightarrow r)$$

Let us construct a counterexample using truth values to see if this is a tautology:

- Let $$p = \text{False}$$

- Let $$q = \text{True}$$

- Let $$r = \text{False}$$

Evaluate LHS: $$(p \vee q) \Rightarrow r$$

- $$p \vee q = \text{False} \vee \text{True} = \text{True}$$

- $$\text{True} \Rightarrow \text{False} = \text{False}$$

Evaluate RHS: $$p \Rightarrow r$$

- $$\text{False} \Rightarrow \text{False} = \text{True}$$

Since $$\text{LHS} = \text{False}$$ and $$\text{RHS} = \text{True}$$, their bi-conditional equivalence ($$\Leftrightarrow$$) evaluates to $$\text{False}$$.

Therefore, $$S1$$ is not a tautology.

Step 3: Analyze Statement S2

$$S2: ((p \vee q) \Rightarrow r) \Leftrightarrow ((p \Rightarrow r) \vee (q \Rightarrow r))$$

Using our known logical identity from Step 1, the true distribution rule uses an AND ($$\wedge$$) operator, not an OR ($$\vee$$) operator. Let us test with another counterexample:

- Let $$p = \text{True}$$

- Let $$q = \text{False}$$

- Let $$r = \text{False}$$

Evaluate LHS: $$(p \vee q) \Rightarrow r$$

- $$p \vee q = \text{True} \vee \text{False} = \text{True}$$

- $$\text{True} \Rightarrow \text{False} = \text{False}$$

Evaluate RHS: $$(p \Rightarrow r) \vee (q \Rightarrow r)$$

- $$p \Rightarrow r = \text{True} \Rightarrow \text{False} = \text{False}$$

- $$q \Rightarrow r = \text{False} \Rightarrow \text{False} = \text{True}$$

- $$\text{False} \vee \text{True} = \text{True}$$

Since $$\text{LHS} = \text{False}$$ and $$\text{RHS} = \text{True}$$, their bi-conditional equivalence ($$\Leftrightarrow$$) evaluates to $$\text{False}$$.

Therefore, $$S2$$ is also not a tautology.

Conclusion:

Neither (S1) nor (S2) is a tautology.

To determine the truth values of $$p$$, $$q$$, and $$r$$ that make the logical expression false, let's analyze the implication.

The given expression is:

$$\{(p \lor q) \land (\neg p \lor r)\} \to (\neg q \lor r)$$

Let the premise be $$A = (p \lor q) \land (\neg p \lor r)$$and the conclusion be $$B = \neg q \lor r$$.

The expression is of the form $$A \to B$$.

An implication $$A \to B$$ is false if and only if the premise $$A$$ is True and the conclusion $$B$$ is False.

Step 1: Make the ($$B$$) False

We need $$B = \neg q \lor r$$ to be False.

For a logical OR ($$\lor$$) to be false, both of its components must be false.

$$\neg q$$ must be False $$ \implies $$q is True

$$r$$ must be False $$ \implies$$ r is False

So, we have established that $$q = True $$ and $$r = False$$.

Step 2: Ensure the premise ($$A$$) is True

We need $$A = (p \lor q) \land (\neg p \lor r)$$ to be True.

Substitute the known values ($$q = \text{True}$$, $$r = \text{False}$$) into the premise:

$$p \lor q = p \lor \text{True} = \text{True}$$(since $$q$$ is True, the OR statement is True regardless of $$p$$)

$$\neg p \lor r = \neg p \lor \text{False} = \neg p$$

For the premise $$A$$ to be True, both parts of the AND ($$\land$$) statement must be True.

$$\text{True} \land \neg p$$ must be True.

This means $$\neg p$$ must be True, which implies p is False .

Conclusion

The combination of truth values that makes the given logical expression false is:

$$p = \text{False}$$

$$q = \text{True}$$

$$r = \text{False}$$

We need to find the converse of $$((\sim p) \wedge q) \Rightarrow r$$.

Recall the definition of converse.

The converse of $$P \Rightarrow Q$$ is $$Q \Rightarrow P$$.

Apply to the given statement.

Here $$P = (\sim p) \wedge q$$ and $$Q = r$$.

The converse is: $$r \Rightarrow ((\sim p) \wedge q)$$.

Check the options.

Option C states: $$(\sim r) \Rightarrow ((\sim p) \wedge q)$$. This is the inverse, not the converse.

Option D states: $$(p \vee (\sim q)) \Rightarrow (\sim r)$$. This is the contrapositive of the converse.

let me re-examine. The converse of $$(\sim p \wedge q) \Rightarrow r$$ is $$r \Rightarrow (\sim p \wedge q)$$.

Looking at the options more carefully:

Option C: $$(\sim r) \Rightarrow ((\sim p) \wedge q)$$ — this is the inverse (negating both sides of original).

None of the options directly states $$r \Rightarrow (\sim p \wedge q)$$. Let me re-read the options.

The answer given is Option D: $$(p \vee (\sim q)) \Rightarrow (\sim r)$$.

This is actually the contrapositive of the converse. The converse is $$r \Rightarrow (\sim p \wedge q)$$. Its contrapositive is $$\sim(\sim p \wedge q) \Rightarrow \sim r$$, i.e., $$(p \vee \sim q) \Rightarrow \sim r$$.

Since a statement and its contrapositive are logically equivalent, Option D is equivalent to the converse.

The correct answer is Option D.

We need to simplify $$\sim(p \wedge (p \to \sim q))$$.

First, recall that $$p \to \sim q \equiv \sim p \vee \sim q$$.

So: $$p \wedge (p \to \sim q) \equiv p \wedge (\sim p \vee \sim q)$$

Using distribution: $$= (p \wedge \sim p) \vee (p \wedge \sim q)$$

Since $$p \wedge \sim p$$ is always false (contradiction):

$$= F \vee (p \wedge \sim q) = p \wedge \sim q$$

Therefore: $$\sim(p \wedge (p \to \sim q)) = \sim(p \wedge \sim q)$$

By De Morgan's law: $$= \sim p \vee q$$

$$= (\sim p) \vee q$$

The correct answer is Option 3: $$(\sim p) \vee q$$.

Given: $$f : \{1, 2, 3, 4\} \to \{a \in \mathbb{Z} : |a| \leq 8\}$$ satisfying $$f(n) + \frac{1}{n}f(n+1) = 1$$ for all $$n \in \{1, 2, 3\}$$.

From the recurrence relation, we express all values in terms of $$f(1)$$:

For $$n = 1$$: $$f(1) + f(2) = 1 \implies f(2) = 1 - f(1)$$

For $$n = 2$$: $$f(2) + \frac{1}{2}f(3) = 1 \implies f(3) = 2(1 - f(2)) = 2(1 - (1 - f(1))) = 2f(1)$$

For $$n = 3$$: $$f(3) + \frac{1}{3}f(4) = 1 \implies f(4) = 3(1 - f(3)) = 3(1 - 2f(1)) = 3 - 6f(1)$$

All function values must be integers with $$|f(i)| \leq 8$$. Since $$f(1)$$ is an integer, we need:

• $$|f(1)| \leq 8$$
• $$|1 - f(1)| \leq 8 \implies -7 \leq f(1) \leq 9$$
• $$|2f(1)| \leq 8 \implies -4 \leq f(1) \leq 4$$
• $$|3 - 6f(1)| \leq 8 \implies -\frac{5}{6} \leq f(1) \leq \frac{11}{6}$$

The most restrictive constraint gives $$f(1) \in \{0, 1\}$$ (integers in $$[-0.83, 1.83]$$).

The two valid functions are:

• $$f(1) = 0$$: $$f = (0, 1, 0, 3)$$
• $$f(1) = 1$$: $$f = (1, 0, 2, -3)$$

The number of such functions is 2.

Therefore, the correct answer is Option D: $$\mathbf{2}$$.

Let $$A = \{1, 3, 4, 6, 9\}$$ and $$B = \{2, 4, 5, 8, 10\}$$. The relation $$R$$ on $$A \times B$$ is defined as:

$$ R = \{((a_1, b_1), (a_2, b_2)) : a_1 \leq b_2 \text{ and } b_1 \leq a_2\} $$

Elements of $$A \times B$$ are ordered pairs $$(a, b)$$ with $$a \in A, b \in B$$. The relation $$R$$ relates two such pairs $$((a_1, b_1), (a_2, b_2))$$ when both conditions $$a_1 \leq b_2$$ and $$b_1 \leq a_2$$ hold.

Since $$a_1$$ and $$b_2$$ come from different pairs, and $$b_1$$ and $$a_2$$ come from different pairs, the two conditions are independent.

Count pairs $$(a_1, b_2)$$ with $$a_1 \in A, b_2 \in B, a_1 \leq b_2$$:

$$a_1 = 1$$: $$b_2 \in \{2,4,5,8,10\}$$ → 5 pairs

$$a_1 = 3$$: $$b_2 \in \{4,5,8,10\}$$ → 4 pairs

$$a_1 = 4$$: $$b_2 \in \{4,5,8,10\}$$ → 4 pairs

$$a_1 = 6$$: $$b_2 \in \{8,10\}$$ → 2 pairs

$$a_1 = 9$$: $$b_2 \in \{10\}$$ → 1 pair

Total: $$5 + 4 + 4 + 2 + 1 = 16$$

Count pairs $$(b_1, a_2)$$ with $$b_1 \in B, a_2 \in A, b_1 \leq a_2$$:

$$b_1 = 2$$: $$a_2 \in \{3,4,6,9\}$$ → 4 pairs

$$b_1 = 4$$: $$a_2 \in \{4,6,9\}$$ → 3 pairs

$$b_1 = 5$$: $$a_2 \in \{6,9\}$$ → 2 pairs

$$b_1 = 8$$: $$a_2 \in \{9\}$$ → 1 pair

$$b_1 = 10$$: $$a_2 \in \{\}$$ → 0 pairs

Total: $$4 + 3 + 2 + 1 + 0 = 10$$

$$ |R| = 16 \times 10 = 160 $$

The number of elements in $$R$$ is Option A: 160.

$$f : \mathbb{R} - \{0, 1\} \to \mathbb{R}$$ such that $$f(x) + f\left(\frac{1}{1-x}\right) = 1 + x$$. Find $$f(2)$$.

Substituting $$x = 2$$ into the given equation yields $$f(2) + f\left(\frac{1}{1-2}\right) = 1 + 2$$, which simplifies to $$f(2) + f(-1) = 3 \quad (1)$$.

Next, letting $$x = -1$$ gives $$f(-1) + f\left(\frac{1}{1-(-1)}\right) = 1 + (-1)$$, so $$f(-1) + f\left(\frac{1}{2}\right) = 0 \quad (2)$$.

Finally, putting $$x = \frac{1}{2}$$ leads to $$f\left(\frac{1}{2}\right) + f\left(\frac{1}{1-\frac{1}{2}}\right) = 1 + \frac{1}{2}$$, hence $$f\left(\frac{1}{2}\right) + f(2) = \frac{3}{2} \quad (3)$$.

From (1) we have $$f(-1) = 3 - f(2)$$. Substituting into (2) gives $$(3 - f(2)) + f\left(\frac{1}{2}\right) = 0$$, so $$f\left(\frac{1}{2}\right) = f(2) - 3 \quad (4)$$. Substituting (4) into (3) yields $$(f(2) - 3) + f(2) = \frac{3}{2}$$, hence $$2f(2) = \frac{3}{2} + 3 = \frac{9}{2}$$ and therefore $$f(2) = \frac{9}{4}$$.

Answer: Option B $$\left(\frac{9}{4}\right)$$

The function is:

$$f(x) = \frac{\log_{(x+1)}(x-2)}{e^{2\log_e x} - (2x+3)}$$

We need to evaluate the constraints for both the numerator and the denominator independently, and then find their intersection.

1. Constraints for the Numerator

For the logarithmic function $$\log_{(x+1)}(x-2)$$ to be defined, the following conditions must be met:

• The argument must be strictly positive:

  $$x - 2 > 0 \implies x > 2$$

• The base must be strictly positive:

  $$x + 1 > 0 \implies x > -1$$

• The base cannot equal 1:

  $$x + 1 \neq 1 \implies x \neq 0$$

Intersecting these three conditions for the numerator, we get $$x > 2$$.

2. Constraints for the Denominator

The denominator must not equal zero, and any functions within it must also be defined.

• The argument of the logarithm in the exponent must be strictly positive:

  $$x > 0$$

• The denominator cannot be zero:

  $$e^{2\log_e x} - (2x+3) \neq 0$$

Using the logarithm power rule, $$2\log_e x = \log_e(x^2)$$. For $$x > 0$$, the expression simplifies using the property $$e^{\log_e(a)} = a$$:

$$e^{\log_e(x^2)} = x^2$$

Substitute this back into the denominator inequality:

$$x^2 - (2x + 3) \neq 0$$
$$x^2 - 2x - 3 \neq 0$$

Factoring the quadratic equation gives:

$$(x - 3)(x + 1) \neq 0$$

This means $$x \neq 3$$ and $$x \neq -1$$.

3. Final Domain

To find the domain of $$f(x)$$, we take the intersection of the valid sets for both the numerator and the denominator:

• From the numerator: $$x > 2$$
• From the denominator: $$x > 0$$, $$x \neq 3$$, and $$x \neq -1$$

Combining these constraints, $$x$$ must be strictly greater than $$2$$, but it cannot equal $$3$$.

Therefore, the domain of $$f(x)$$ is:

$$x \in (2, 3) \cup (3, \infty)$$ or $$(2, \infty) \setminus \{3\}$$.

We need to find the negation of $$(p \wedge (\neg q)) \vee (\neg p)$$.

Simplify the given expression.

Using the distributive law, we expand:

$$(p \wedge (\neg q)) \vee (\neg p) = (\neg p \vee p) \wedge (\neg p \vee \neg q)$$

Since $$\neg p \vee p$$ is a tautology (always true):

$$= T \wedge (\neg p \vee \neg q) = \neg p \vee \neg q$$

Apply De Morgan's Law.

By De Morgan's Law, $$\neg p \vee \neg q = \neg(p \wedge q)$$. So:

$$(p \wedge (\neg q)) \vee (\neg p) \equiv \neg(p \wedge q)$$

Find the negation.

The negation of $$\neg(p \wedge q)$$ is:

$$\neg[\neg(p \wedge q)] = p \wedge q$$

Verification using truth table:

When $$p = T, q = T$$: Original = $$(T \wedge F) \vee F = F$$. Negation = $$T$$. And $$p \wedge q = T$$. ✔

When $$p = T, q = F$$: Original = $$(T \wedge T) \vee F = T$$. Negation = $$F$$. And $$p \wedge q = F$$. ✔

When $$p = F, q = T$$: Original = $$(F \wedge F) \vee T = T$$. Negation = $$F$$. And $$p \wedge q = F$$. ✔

When $$p = F, q = F$$: Original = $$(F \wedge T) \vee T = T$$. Negation = $$F$$. And $$p \wedge q = F$$. ✔

The negation matches $$p \wedge q$$ in all cases.

The answer is Option B: $$p \wedge q$$.

The function is $$f(x) = \dfrac{[x]}{1 + x^2}$$, where $$[x]$$ denotes the greatest integer less than or equal to $$x$$, and the domain is $$[2, 6)$$. We evaluate $$f$$ on each unit interval where $$[x]$$ is constant.

On $$[2, 3)$$: $$[x] = 2$$, so $$f(x) = \dfrac{2}{1+x^2}$$. Since $$1+x^2$$ increases from $$5$$ to $$10$$, $$f$$ decreases from $$\tfrac{2}{5}$$ to just above $$\tfrac{2}{10} = \tfrac{1}{5}$$. On $$[3, 4)$$: $$f(x) = \dfrac{3}{1+x^2}$$, ranging from $$\tfrac{3}{10}$$ down to just above $$\tfrac{3}{17}$$. On $$[4, 5)$$: $$f(x) = \dfrac{4}{1+x^2}$$, from $$\tfrac{4}{17}$$ to just above $$\tfrac{4}{26} = \tfrac{2}{13}$$. On $$[5, 6)$$: $$f(x) = \dfrac{5}{1+x^2}$$, from $$\tfrac{5}{26}$$ to just above $$\tfrac{5}{37}$$.

Combining all intervals, the minimum value approaches $$\tfrac{5}{37}$$ (from above) and the maximum value is $$\tfrac{2}{5}$$ (attained at $$x = 2$$). The range is $$\left(\dfrac{5}{37},\;\dfrac{2}{5}\right]$$, which matches $$\boxed{\left\{\dfrac{5}{37},\;\dfrac{2}{5}\right\}}$$, i.e., option (D).

Find the domain of $$f(x) = \ln(4x^2+11x+6) + \sin^{-1}(4x+3) + \cos^{-1}\left(\dfrac{10x+6}{3}\right)$$, then compute $$36|\alpha+\beta|$$.

Domain constraints.

(i) $$4x^2+11x+6 > 0$$: Factor: $$(4x+3)(x+2) > 0$$. Solutions: $$x < -2$$ or $$x > -\dfrac{3}{4}$$.

(ii) $$-1 \leq 4x+3 \leq 1$$: $$-4 \leq 4x \leq -2$$, so $$-1 \leq x \leq -\dfrac{1}{2}$$.

(iii) $$-1 \leq \dfrac{10x+6}{3} \leq 1$$: $$-3 \leq 10x+6 \leq 3$$, so $$-\dfrac{9}{10} \leq x \leq -\dfrac{3}{10}$$.

From (ii): $$[-1, -1/2]$$

From (iii): $$[-9/10, -3/10]$$

Intersection of (ii) and (iii): $$[-9/10, -1/2]$$

Now intersect with (i): $$x < -2$$ or $$x > -3/4$$.

$$[-9/10, -1/2]$$ intersected with ($$x < -2$$ or $$x > -3/4$$):

$$-9/10 = -0.9$$ and $$-3/4 = -0.75$$. So $$x > -3/4$$ intersected with $$[-9/10, -1/2]$$ gives $$(-3/4, -1/2]$$.

Also check $$x < -2$$: no overlap with $$[-9/10, -1/2]$$.

So domain = $$(-3/4, -1/2] = (\alpha, \beta]$$.

$$\alpha = -3/4, \beta = -1/2$$

$$\alpha + \beta = -3/4 + (-1/2) = -5/4$$

$$36|\alpha + \beta| = 36 \times 5/4 = 45$$

The correct answer is Option D: 45.

Given: $$f(x) = \frac{1}{\sqrt{\lceil x \rceil - x}}$$ where $$\lceil x \rceil$$ denotes the smallest integer greater than or equal to $$x$$ (ceiling function).

Domain $$A$$: We need $$\lceil x \rceil - x > 0$$, which holds for all non-integer $$x$$.

For integer $$x$$: $$\lceil x \rceil = x$$, so $$\lceil x \rceil - x = 0$$. Not in domain.

$$A = \mathbb{R} \setminus \mathbb{Z}$$

Range $$B$$: For non-integer $$x$$ with $$\lfloor x \rfloor = n$$, the fractional part $$\{x\} \in (0, 1)$$, and $$\lceil x \rceil - x = 1 - \{x\}$$.

As $$\{x\}$$ ranges over $$(0, 1)$$: $$1 - \{x\} \in (0, 1)$$, so $$\frac{1}{\sqrt{1 - \{x\}}} \in (1, \infty)$$.

$$B = (1, \infty)$$

Checking (S1): $$A \cap B = (\mathbb{R} \setminus \mathbb{Z}) \cap (1, \infty) = (1, \infty) \setminus \{2, 3, 4, \ldots\} = (1, \infty) \setminus \mathbb{N}$$.

The statement says $$A \cap B = (1, \infty) - \mathbb{N}$$. This is correct since $$\mathbb{N} \cap (1, \infty) = \{2, 3, 4, \ldots\}$$. (S1) is TRUE.

Checking (S2): $$A \cup B = (\mathbb{R} \setminus \mathbb{Z}) \cup (1, \infty)$$.

This union contains non-integers less than or equal to 1 (like $$0.5$$, $$-0.5$$, etc.), so $$A \cup B \neq (1, \infty)$$.

For example, $$0.5 \in A$$ but $$0.5 \notin (1, \infty)$$. (S2) is FALSE.

Only (S1) is true.

The correct answer is Option B.

We need to solve $$x^2 - 4x + [x] + 3 = x[x]$$, where $$[x]$$ is the greatest integer function.

We start by rearranging the equation.

$$(x^2 - 4x + 3) = x[x] - [x] = [x](x - 1)$$

$$(x - 1)(x - 3) = [x](x - 1)$$

Next, we analyze cases based on the factor $$(x - 1)$$.

Case 1: $$x = 1$$.

The left-hand side becomes 0 and the right-hand side is $$[1](1 - 1) = 0$$, so $$x = 1$$ is a solution.

Case 2: $$x \neq 1$$.

Dividing both sides of $$(x - 1)(x - 3) = [x](x - 1)$$ by $$(x - 1)$$ gives

$$x - 3 = [x]$$

Since $$[x]$$ is an integer, $$x - 3$$ must be an integer, implying $$x$$ is an integer.

If $$x$$ is an integer, then $$[x] = x$$, so the equation becomes $$x - 3 = x$$, which leads to $$-3 = 0$$, a contradiction.

Alternatively, if $$x$$ were not an integer but $$x - 3$$ were an integer, then $$x$$ would be an integer, which is also a contradiction. Hence no solutions arise for $$x \neq 1$$.

Therefore, the only solution to the equation is $$x = 1$$.

Hence, the correct answer is Option 4, indicating a unique solution in $$(-\infty, \infty)$$.

$$f(x) = \sqrt{3-x} + \sqrt{2+x}$$. Domain: $$-2 \leq x \leq 3$$.

Square: $$f(x)^2 = (3-x) + (2+x) + 2\sqrt{(3-x)(2+x)} = 5 + 2\sqrt{(3-x)(2+x)}$$.

Let $$g(x) = (3-x)(2+x) = -x^2 + x + 6$$. This is a downward parabola with max at $$x = 1/2$$.

$$g(1/2) = (5/2)(5/2) = 25/4$$.

$$g(-2) = 5 \cdot 0 = 0$$ and $$g(3) = 0 \cdot 5 = 0$$.

So $$g(x) \in [0, 25/4]$$, and $$\sqrt{g(x)} \in [0, 5/2]$$.

$$f(x)^2 \in [5, 5 + 5] = [5, 10]$$.

$$f(x) \in [\sqrt{5}, \sqrt{10}]$$.

To solve for sets A and B, we will break them down individually and then determine their relationship.

The set is defined as $$A\ =\ \left\{x\in\ R\ :\ \left|x+3\right|+\left|x+4\right|\le3\right\}$$

We evaluate this absolute value inequality by checking three critical regions determined by the roots x = -4 and x = -3.

Case 1: x < -4

Both terms inside the absolute values are negative:

$$-(x + 3) - (x + 4) \le 3 \implies -2x - 7 \le 3 \implies -2x \le 10 \implies x \ge -5$$

In this region, the solution is [-5, -4).

Case 2: -4 <= x <= -3

The first term is negative, the second is positive:

$$-(x + 3) + (x + 4) \le 3 \implies 1 \le 3$$

This is always true. So, the entire interval [-4, -3] is included.

Case 3: x > -3

Both terms are positive:

$$(x + 3) + (x + 4) \le 3 \implies 2x + 7 \le 3 \implies 2x \le -4 \implies x \le -2$$

In this region, the solution is (-3, -2].

Combining the cases:

$$A = [-5, -2]$$

Now Solve for Set(B):

The set is defined as 

First, let's simplify the summation. It is an infinite geometric series with first term $$a\ =\ \frac{3^{x-3}}{10}$$ and common ratio $$q\ =\ \frac{1}{10}$$:

$$\sum_{r=1}^{\infty} \frac{3^{x-3}}{10^r} = 3^{x-3} \left( \frac{1/10}{1 - 1/10} \right) = 3^{x-3} \left( \frac{1/10}{9/10} \right) = \frac{3^{x-3}}{9} = \frac{3^{x-3}}{3^2} = 3^{x-5}$$

Now, substitute this back into the inequality:

$$3^x.3^{x-5}<3^{-3x}$$

Since the base 3 > 1, we can compare the exponents directly:

$$3^{2x-5}<3^{-3x}$$

So, B = (-$$\infty\ $$, 1).

So Finally,

We have:

• $A = [-5, -2]
• $B = (-$$\infty\ $$, 1)

Checking the relationship between the two sets:

Since every value in [-5, -2] is strictly less than 1, all elements of A are contained within B.

Final Result:

$$A \subset B$$

$$B\ =\ \left\{x\ \in\ R\ :\ 3^x\Sigma\ \frac{3^{x-3}}{10^r}<3^{-3x}\right\}$$

Let

$$[x]=n,$$

where $$n$$ is an integer.

Since the square root is in the denominator, we require

$$n^2-3n-10>0.$$

Factorising,

$$(n-5)(n+2)>0.$$

Therefore,

$$n<-2\quad\text{or}\quad n>5.$$

Now,

$$[x]\le-3$$

implies

$$x<-2,$$

and

$$[x]\ge6$$

implies

$$x\ge6.$$

Hence, the domain of the function is

$$(-\infty,-2)\cup[6,\infty).$$

Therefore,

$$\boxed{(-\infty,-2)\cup[6,\infty)}$$

Total onto functions from $$R = \{a,b,c,d,e\}$$ to $$S = \{1,2,3,4\}$$:

Using inclusion-exclusion: $$4^5 - \binom{4}{1} \cdot 3^5 + \binom{4}{2} \cdot 2^5 - \binom{4}{3} \cdot 1^5$$

$$= 1024 - 4(243) + 6(32) - 4(1) = 1024 - 972 + 192 - 4 = 240$$

Now subtract onto functions where $$f(a) = 1$$. When $$f(a) = 1$$, we need $$f : \{b,c,d,e\} \to \{1,2,3,4\}$$ to be onto (since every element of $$S$$ must be hit).

But wait — element 1 is already hit by $$f(a) = 1$$. So $$\{b,c,d,e\}$$ must cover $$\{2,3,4\}$$ (all of them), and may also map to 1.

Number of such functions = total functions from 4 elements to 4 elements that cover $$\{2,3,4\}$$:

$$= 4^4 - \binom{3}{1} \cdot 3^4 + \binom{3}{2} \cdot 2^4 - \binom{3}{3} \cdot 1^4$$

$$= 256 - 3(81) + 3(16) - 1 = 256 - 243 + 48 - 1 = 60$$

Onto functions with $$f(a) \neq 1$$ = $$240 - 60 = 180$$

Given $$f(x) = ax - 3$$, $$g(x) = x^b + c$$ with $$a, b, c \in \mathbb{N}$$, and $$(f \circ g)^{-1}(x) = \left(\dfrac{x-7}{2}\right)^{1/3}$$.

Finding $$a, b, c$$:

$$f(g(x)) = a(x^b + c) - 3 = ax^b + ac - 3$$

If $$(f \circ g)^{-1}(x) = \left(\frac{x-7}{2}\right)^{1/3}$$, then:

$$f \circ g(x) = 2x^3 + 7$$

Comparing coefficients: $$a = 2$$, $$b = 3$$, $$ac - 3 = 7 \implies c = 5$$.

Computing $$(f \circ g)(ac)$$:

$$ac = 2 \times 5 = 10$$

$$(f \circ g)(10) = 2(10)^3 + 7 = 2000 + 7 = 2007$$

Computing $$(g \circ f)(b)$$:

$$f(3) = 2(3) - 3 = 3$$

$$g(3) = 3^3 + 5 = 32$$

Final answer:

$$(f \circ g)(ac) + (g \circ f)(b) = 2007 + 32 = 2039$$

The answer is $$2039$$.

Let $$A = \{1,2,3,4,5\}$$, $$B = \{1,2,3,4,5,6\}$$. Find the number of functions $$f: A \to B$$ with $$f(1) + f(2) = f(4) - 1$$.

We need $$f(1) + f(2) = f(4) - 1$$, where $$f(1), f(2), f(4) \in \{1,2,3,4,5,6\}$$.

Let $$f(4) = k$$, so $$f(1) + f(2) = k - 1$$.

Since $$f(1), f(2) \geq 1$$, we need $$k - 1 \geq 2$$, i.e., $$k \geq 3$$.

Since $$f(1), f(2) \leq 6$$, we need $$k - 1 \leq 12$$, always satisfied.

$$k = 3$$: $$f(1) + f(2) = 2$$. Only $$(1,1)$$ → 1 pair

$$k = 4$$: $$f(1) + f(2) = 3$$. Pairs: $$(1,2), (2,1)$$ → 2 pairs

$$k = 5$$: $$f(1) + f(2) = 4$$. Pairs: $$(1,3), (2,2), (3,1)$$ → 3 pairs

$$k = 6$$: $$f(1) + f(2) = 5$$. Pairs: $$(1,4), (2,3), (3,2), (4,1)$$ → 4 pairs

Total constrained combinations: $$1 + 2 + 3 + 4 = 10$$

$$f(3)$$ and $$f(5)$$ are unconstrained, each with 6 choices.

Free choices: $$6 \times 6 = 36$$

$$ \text{Total} = 10 \times 36 = 360 $$

The number of functions is 360.

We have $$A = \{1,2,3,\dots,10\}$$, $$B = \{0,1,2,3,4\}$$, and we seek the number of ordered pairs $$(a,b)\in A\times A$$ satisfying $$2(a-b)^2 + 3(a-b)\in B\,. $$ To analyze this, set $$d = a - b$$ and define $$f(d) = 2d^2 + 3d\,. $$ We look for integer values of $$d$$ such that $$f(d)\in B\,. $$

When $$d = 0\,,\;f(0) = 0\in B$$, which accounts for the 10 diagonal pairs $$(a,a)\,,\;a=1,2,\dots,10\,. $$ For $$d = 1\,,\;f(1) = 5\notin B$$, and for $$d = -1\,,\;f(-1) = 2 - 3 = -1\notin B\,. $$ When $$d = 2\,,\;f(2) = 8 + 6 = 14\notin B$$, but for $$d = -2\,,\;f(-2) = 8 - 6 = 2\in B$$. The condition $$a - b = -2$$ gives $$b = a + 2$$, yielding the pairs $$(1,3), (2,4), \dots, (8,10)$$, which are 8 in number. All other values of $$d$$ produce $$f(d)$$ outside of $$B\,. $$

In total there are $$10 + 8 = 18$$ ordered pairs satisfying the condition. The answer is 18.

We need to find the number of functions $$f : A \to A$$ such that $$f(m \cdot n) = f(m) \cdot f(n)$$ for every $$m, n \in A$$ with $$m \cdot n \in A$$, where $$A = \{1, 2, 3, 5, 8, 9\}$$.

We begin by identifying all pairs $$(m, n)$$ with $$m, n \in A$$ and $$m \cdot n \in A$$. The relevant products within $$A$$ are $$1 \times k = k$$ for all $$k \in A$$ and $$3 \times 3 = 9 \in A$$. All other products (for example, $$2 \times 3 = 6$$, $$2 \times 5 = 10$$, $$3 \times 5 = 15$$, etc.) fall outside $$A$$.

From $$f(1 \cdot k) = f(1) \cdot f(k)$$ for all $$k \in A$$ we obtain $$f(k) = f(1) \cdot f(k)$$. Since $$f(k) \ge 1$$ for all $$k$$ (because $$f : A \to A$$ and $$0 \notin A$$), we must have $$f(1) = 1$$.

Next, considering $$f(3 \times 3) = f(3)^2$$, we get $$f(9) = f(3)^2$$ with both $$f(3)$$ and $$f(9)$$ lying in $$A = \{1, 2, 3, 5, 8, 9\}$$. Checking all possibilities for $$f(3)$$ shows: if $$f(3)=1$$ then $$f(9)=1$$ (valid); if $$f(3)=2$$ then $$f(9)=4 \notin A$$ (invalid); if $$f(3)=3$$ then $$f(9)=9$$ (valid); if $$f(3)=5$$ then $$f(9)=25 \notin A$$ (invalid); if $$f(3)=8$$ then $$f(9)=64 \notin A$$ (invalid); and if $$f(3)=9$$ then $$f(9)=81 \notin A$$ (invalid). Therefore $$f(3) \in \{1, 3\}$$, giving 2 choices, each determining $$f(9)$$.

The values $$f(2)$$, $$f(5)$$, and $$f(8)$$ have no multiplicative constraints, since no product involving these elements remains in $$A$$. Hence each of these can be chosen freely as any of the 6 elements of $$A$$.

Multiplying the number of choices yields $$\text{Total} = 1 \times 2 \times 6 \times 6 \times 6 = 432$$ because $$f(1)$$ is fixed, $$f(3)$$ has 2 choices (determining $$f(9)$$), and each of $$f(2)$$, $$f(5)$$, and $$f(8)$$ has 6 choices.

Thus the correct answer is \boxed{432}.

To find the value of $$a + b$$, we can compute the first few iterations of the composite function to identify the underlying pattern.

The first iteration is given as:

$$f^1(x) = \frac{3x+2}{2x+3}$$

---

Step 1: Compute the second iteration $$f^2(x)$$

$$f^2(x) = f^1(f^1(x)) = \frac{3\left(\frac{3x+2}{2x+3}\right) + 2}{2\left(\frac{3x+2}{2x+3}\right) + 3}$$

Multiplying the numerator and denominator by $$(2x+3)$$ to clear the fractions:

$$f^2(x) = \frac{3(3x+2) + 2(2x+3)}{2(3x+2) + 3(2x+3)} = \frac{9x + 6 + 4x + 6}{6x + 4 + 6x + 9} = \frac{13x + 12}{12x + 13}$$

Notice that the sum of the coefficients in the numerator of $$f^2(x)$$ is $$13 + 12 = 25 = 5^2$$.

---

Step 2: Compute the third iteration $$f^3(x)$$

$$f^3(x) = f^1(f^2(x)) = \frac{3\left(\frac{13x+12}{12x+13}\right) + 2}{2\left(\frac{13x+12}{12x+13}\right) + 3}$$

Multiplying the numerator and denominator by $$(12x+13)$$:

$$f^3(x) = \frac{3(13x+12) + 2(12x+13)}{2(13x+12) + 3(12x+13)} = \frac{39x + 36 + 24x + 26}{26x + 24 + 36x + 39} = \frac{63x + 62}{62x + 63}$$

Notice that the sum of the coefficients in the numerator of $$f^3(x)$$ is $$63 + 62 = 125 = 5^3$$.

---

Step 3: Generalize the pattern for $$f^n(x)$$

From the observed results, any iteration $$f^n(x)$$ can be represented in the symmetric form:

$$f^n(x) = \frac{a_n x + b_n}{b_n x + a_n}$$

Furthermore, the sum of the coefficients in the numerator satisfies the following exponential relation:

$$a_n + b_n = 5^n$$

Additionally, by tracking the difference of the coefficients ($$a_1 - b_1 = 3 - 2 = 1$$, $$a_2 - b_2 = 13 - 12 = 1$$, etc.), we have $$a_n - b_n = 1$$ for all $$n$$. Since their difference is always 1, $$a_n$$ and $$b_n$$ are consecutive integers, which guarantees that $$\gcd(a_n, b_n) = 1$$ for any iteration.

---

Step 4: Calculate $$a + b$$ for $$f^5(x)$$

For $$n = 5$$, the coefficients are $$a$$ and $$b$$ such that $$f^5(x) = \frac{ax+b}{bx+a}$$ with $$\gcd(a,b) = 1$$. Following our general relation:

$$a + b = 5^5$$

Therefore, the value of $$a + b$$ is equal to 3125.

Let $$y = \log_5 x$$. Then $$x = 5^{y}$$ (because logarithm base 5 is the inverse of the exponential base 5).

Rewrite the given equation using $$y$$:
$$x^{\bigl(16(\log_5 x)^3 - 68\log_5 x\bigr)} = 5^{-16}$$ becomes
$$(5^{y})^{\,16y^{3} \;-\; 68y} = 5^{-16}.$$

Using the law $$\bigl(a^{m}\bigr)^{n}=a^{mn}$$, the left side simplifies to
$$5^{\,y\,(16y^{3} - 68y)} = 5^{\,16y^{4} - 68y^{2}}.$$

Because the bases are equal (both are 5), equate the exponents:
$$16y^{4} - 68y^{2} = -16.$$

Divide every term by 2 to make the numbers smaller:
$$8y^{4} - 34y^{2} + 8 = 0 \qquad -(1)$$

Put $$z = y^{2} \;(\text{note: } z \ge 0)$$ to convert the quartic to a quadratic:

Equation $$(1)$$ becomes
$$8z^{2} - 34z + 8 = 0.$$

Solve this quadratic using the quadratic formula $$z = \dfrac{-b \pm \sqrt{b^{2}-4ac}}{2a}$$ with $$a=8,\, b=-34,\, c=8$$:

Discriminant $$\Delta = (-34)^{2} - 4\cdot 8 \cdot 8 = 1156 - 256 = 900.$$
$$\sqrt{\Delta} = 30.$$

Therefore
$$z = \dfrac{34 \pm 30}{16}.$$

Case 1: $$z = \dfrac{34 + 30}{16} = \dfrac{64}{16} = 4.$$
Case 2: $$z = \dfrac{34 - 30}{16} = \dfrac{4}{16} = \dfrac14.$$

Recall $$z = y^{2}$$, so

From Case 1: $$y^{2}=4 \quad\Longrightarrow\quad y = \pm 2.$$
From Case 2: $$y^{2}=\dfrac14 \quad\Longrightarrow\quad y = \pm \dfrac12.$$

Convert back to $$x$$ using $$x = 5^{y}$$:

For $$y = 2: \; x = 5^{2} = 25.$$
For $$y = -2: \; x = 5^{-2} = \dfrac1{25}.$$
For $$y = \dfrac12: \; x = 5^{1/2} = \sqrt{5}.$$
For $$y = -\dfrac12: \; x = 5^{-1/2} = \dfrac1{\sqrt{5}}.$$

All four values are positive real numbers, so they all satisfy the original equation.

The required product is
$$25 \times \dfrac1{25} \times \sqrt{5} \times \dfrac1{\sqrt{5}} = 1.$$

Hence, the product of all positive real solutions is 1.

We are given the statements:

$$p$$: Ramesh listens to music

$$q$$: Ramesh is out of his village

$$r$$: It is Sunday

$$s$$: It is Saturday

The statement to express is: "Ramesh listens to music only if he is in his village and it is Sunday or Saturday."

The phrase "P only if Q" translates to $$P \Rightarrow Q$$. So "Ramesh listens to music only if (condition)" becomes $$p \Rightarrow \text{(condition)}$$.

Now the condition is "he is in his village and it is Sunday or Saturday." Being "in his village" is the negation of $$q$$ (since $$q$$ means "out of his village"), so this is $$\sim q$$. "It is Sunday or Saturday" is $$r \vee s$$.

So the condition is $$(\sim q) \wedge (r \vee s)$$.

Therefore the full statement is $$p \Rightarrow ((\sim q) \wedge (r \vee s))$$.

Hence, the correct answer is Option 4.

We are given the formulas $$S_1 : ((\sim p) \vee q) \vee ((\sim p) \vee r)$$ and $$S_2 : p \to (q \vee r)$$.

To simplify $$S_1$$, we use the associative and idempotent laws of disjunction: $$S_1 = (\sim p \vee q) \vee (\sim p \vee r) = \sim p \vee q \vee \sim p \vee r = \sim p \vee q \vee r.$$

Next, the implication in $$S_2$$ is equivalent to a disjunction, namely $$S_2 = \sim p \vee (q \vee r) = \sim p \vee q \vee r.$$

Since both $$S_1$$ and $$S_2$$ simplify to the same expression $$\sim p \vee q \vee r$$, they are logically equivalent: $$S_1 \equiv S_2$$.

We now evaluate each option:

Option A: “If $$S_2$$ is True, then $$S_1$$ is True.” This holds because $$S_1 \equiv S_2$$.

Option B: “If $$S_2$$ is False, then $$S_1$$ is False.” This also holds by equivalence.

Option C: “If $$S_2$$ is False, then $$S_1$$ is True.” This would require different truth values for $$S_1$$ and $$S_2$$, contradicting their equivalence, so it is not true.

Option D: “If $$S_1$$ is False, then $$S_2$$ is False.” This holds by equivalence.

Therefore, the statement that is NOT true is Option C.

We need to determine what the statement $$(\sim(p \Leftrightarrow \sim q)) \wedge q$$ is equivalent to.

$$p \Leftrightarrow \sim q$$ is true when p and $$\sim q$$ have the same truth value, i.e., when p and q have opposite truth values.

Therefore, $$\sim(p \Leftrightarrow \sim q)$$ is true when p and q have the same truth value, which means $$\sim(p \Leftrightarrow \sim q) \equiv (p \Leftrightarrow q)$$.

The statement becomes $$(p \Leftrightarrow q) \wedge q$$.

$$p = T, q = T$$: $$(T \Leftrightarrow T) \wedge T = T \wedge T = T$$

$$p = T, q = F$$: $$(T \Leftrightarrow F) \wedge F = F \wedge F = F$$

$$p = F, q = T$$: $$(F \Leftrightarrow T) \wedge T = F \wedge T = F$$

$$p = F, q = F$$: $$(F \Leftrightarrow F) \wedge F = T \wedge F = F$$

The statement is true only when $$p = T$$ and $$q = T$$, which is $$p \wedge q$$.

Option D: $$(p \Rightarrow q) \wedge p$$

$$p = T, q = T$$: $$T \wedge T = T$$

$$p = T, q = F$$: $$F \wedge T = F$$

$$p = F, q = T$$: $$T \wedge F = F$$

$$p = F, q = F$$: $$T \wedge F = F$$

This is also equivalent to $$p \wedge q$$. ✓

Therefore, the correct answer is Option D: equivalent to $$(p \Rightarrow q) \wedge p$$.

We are given the statements: $$A$$: Rishi is a judge, $$B$$: Rishi is honest, and $$C$$: Rishi is not arrogant.

The statement "if Rishi is a judge and he is not arrogant, then he is honest" can be written as: $$(A \wedge C) \to B$$

Recall that the negation of an implication $$p \to q$$ is $$p \wedge (\sim q)$$, so $$\sim[(A \wedge C) \to B] = (A \wedge C) \wedge (\sim B)$$.

This can be rewritten as: $$(\sim B) \wedge (A \wedge C)$$

Therefore, the negation is Option B: $$(\sim B) \wedge (A \wedge C)$$.

For an implication

$$A\to B$$

to be FALSE,

$$A=\text{TRUE},\qquad B=\text{FALSE}$$

Hence,

$$p=\text{TRUE}$$

and

$$\sim p\lor q=\text{FALSE}$$

Now, an OR statement is FALSE only when both parts are FALSE.

Therefore,

$$\sim p=\text{FALSE},\qquad q=\text{FALSE}$$

Thus,

$$p=\text{TRUE},\qquad q=\text{FALSE}$$

Now,

$$P_1=\sim(p\to\sim q)$$

Since

$$p=\text{TRUE},\qquad \sim q=\text{TRUE}$$

we get

$$p\to\sim q=\text{TRUE}$$

Hence,

$$P_1=\sim(\text{TRUE})=\text{FALSE}$$

Also,

$$P_2=(p\land\sim q)\land(\sim p\lor q)$$

$$=(\text{TRUE}\land\text{TRUE})\land(\text{FALSE}\lor\text{FALSE})$$

$$=\text{TRUE}\land\text{FALSE}$$

$$=\text{FALSE}$$

Therefore, both $$P_1$$ and $$P_2$$ are FALSE.

We need to find $$\Delta \in \{\wedge, \vee, \Rightarrow, \Leftrightarrow\}$$ such that $$(p \wedge q) \;\Delta\; ((p \vee q) \Rightarrow q)$$ is a tautology.

Evaluate both components for all truth value combinations:

Let $$X = p \wedge q$$ and $$Y = (p \vee q) \Rightarrow q$$.

Case 1: $$p = T, q = T$$: $$X = T$$, $$p \vee q = T$$, $$Y = T \Rightarrow T = T$$. So $$T \;\Delta\; T$$.

Case 2: $$p = T, q = F$$: $$X = F$$, $$p \vee q = T$$, $$Y = T \Rightarrow F = F$$. So $$F \;\Delta\; F$$.

Case 3: $$p = F, q = T$$: $$X = F$$, $$p \vee q = T$$, $$Y = T \Rightarrow T = T$$. So $$F \;\Delta\; T$$.

Case 4: $$p = F, q = F$$: $$X = F$$, $$p \vee q = F$$, $$Y = F \Rightarrow F = T$$. So $$F \;\Delta\; T$$.

For a tautology, all four cases must give True:

From Case 1: $$T \;\Delta\; T = T$$ — true for all four connectives.

From Case 2: $$F \;\Delta\; F = T$$ — checking: $$\wedge$$: F, $$\vee$$: F, $$\Rightarrow$$: T, $$\Leftrightarrow$$: T. Only $$\Rightarrow$$ and $$\Leftrightarrow$$ work.

From Cases 3 and 4: $$F \;\Delta\; T = T$$ — checking: $$\Rightarrow$$: T, $$\Leftrightarrow$$: F. Only $$\Rightarrow$$ works.

Only $$\Delta = \Rightarrow$$ satisfies all cases.

The answer is Option C: $$\Rightarrow$$.

We need to find $$\Delta, \nabla \in \{\wedge, \vee\}$$ such that $$p \nabla q \to ((p \Delta q) \nabla r)$$ is a tautology, then determine what $$(p \nabla q) \Delta r$$ is logically equivalent to.

Since $$\nabla$$ and $$\Delta$$ can each be $$\vee$$ or $$\wedge$$, there are four possibilities. First, if $$\nabla = \vee$$ and $$\Delta = \vee$$, then $$p \vee q \to (p \vee q) \vee r$$. This is always true since $$(p \vee q) \vee r$$ contains $$p \vee q$$, so it is a tautology. Next, if $$\nabla = \vee$$ and $$\Delta = \wedge$$, then $$p \vee q \to (p \wedge q) \vee r$$. Taking $$p = T, q = F, r = F$$ yields $$T \to F$$, which is false, so this case is not a tautology. Now if $$\nabla = \wedge$$ and $$\Delta = \vee$$, then $$p \wedge q \to (p \vee q) \wedge r$$. Taking $$p = T, q = T, r = F$$ gives $$T \to F$$, which is false. Similarly, if $$\nabla = \wedge$$ and $$\Delta = \wedge$$, then $$p \wedge q \to (p \wedge q) \wedge r$$, and choosing the same values yields $$T \to F$$ again. Therefore the only tautology arises when $$\nabla = \vee$$ and $$\Delta = \vee$$.

Substituting these back shows that $$(p \nabla q) \Delta r$$ becomes $$(p \vee q) \vee r$$, which simplifies to $$p \vee q \vee r$$.

Among the answer choices, Option A: $$(p \Delta r) \vee q = (p \vee r) \vee q = p \vee q \vee r$$ matches this expression. The correct answer is Option A.

The correct answer is Option A.

First, we need to find $$r \in \{p, q, \sim p, \sim q\}$$ such that $$r \vee (\sim p) \Rightarrow (p \wedge q) \vee r$$ is a tautology.

Next, recall that an implication $$P \Rightarrow Q$$ is a tautology when $$Q$$ is true whenever $$P$$ is true. Therefore, we need: whenever $$r \vee (\sim p)$$ is true, $$(p \wedge q) \vee r$$ must also be true.

Now, test $$r = \sim p$$. The statement becomes:

$$ (\sim p) \vee (\sim p) \Rightarrow (p \wedge q) \vee (\sim p) $$

Substituting, this simplifies to:

$$ \sim p \Rightarrow (p \wedge q) \vee (\sim p) $$

Since the implication must hold in all cases, consider the following:

Case 1: $$p$$ is True (so $$\sim p$$ is False). The implication $$F \Rightarrow \text{anything}$$ is True.

Case 2: $$p$$ is False (so $$\sim p$$ is True). The RHS becomes $$(F \wedge q) \vee T = T$$. So $$T \Rightarrow T$$ is True.

Since the implication is true in all cases, it is a tautology. ✓

Next, verify that the other options fail.

For $$r = p$$: Let $$p = F, q = F$$. Then LHS: $$F \vee T = T$$, RHS: $$(F \wedge F) \vee F = F$$. So $$T \Rightarrow F$$ is False. ✗

For $$r = q$$: Let $$p = F, q = F$$. Then LHS: $$F \vee T = T$$, RHS: $$(F \wedge F) \vee F = F$$. So $$T \Rightarrow F$$ is False. ✗

For $$r = \sim q$$: Let $$p = T, q = T$$. Then LHS: $$F \vee F = F$$, so implication is True. Let $$p = F, q = T$$. Then LHS: $$F \vee T = T$$, RHS: $$(F \wedge T) \vee F = F$$. So $$T \Rightarrow F$$ is False. ✗

Therefore, the answer is Option C: $$\sim p$$.

We need to find the expression equivalent to $$(\sim(p \wedge q)) \vee q$$.

Using De Morgan's law: $$\sim(p \wedge q) = \sim p \vee \sim q$$. Substituting gives $$(\sim(p \wedge q)) \vee q = (\sim p \vee \sim q) \vee q = \sim p \vee (\sim q \vee q) = \sim p \vee T = T$$ (since $$\sim q \vee q$$ is always true), so the original expression is a tautology.

Next, we examine each option:

Option A: $$q \to (p \wedge q) = \sim q \vee (p \wedge q) = (\sim q \vee p) \wedge (\sim q \vee q) = (\sim q \vee p) \wedge T = \sim q \vee p$$. This is NOT always true (fails when $$p = F, q = T$$), so it is not a tautology.

Option B: $$p \to q = \sim p \vee q$$. This is NOT always true (fails when $$p = T, q = F$$), so it is not a tautology.

Option C: $$p \to (p \to q) = p \to (\sim p \vee q) = \sim p \vee (\sim p \vee q) = \sim p \vee q$$. This is the same as Option B and thus not a tautology.

Option D: $$p \to (p \vee q) = \sim p \vee (p \vee q) = (\sim p \vee p) \vee q = T \vee q = T$$. This is always true — a tautology.

Since the given expression simplifies to a tautology, and Option D is also a tautology, they are logically equivalent.

The correct answer is Option D.

We need to determine which statement is equivalent to $$(p \wedge q) \Rightarrow (p \wedge r)$$.

We begin by rewriting the implication using the identity $$A \Rightarrow B \equiv \neg A \vee B$$. We have $$(p \wedge q) \Rightarrow (p \wedge r) \equiv \neg(p \wedge q) \vee (p \wedge r) \equiv (\neg p \vee \neg q) \vee (p \wedge r)$$.

Now let us examine Option D: $$(p \wedge q) \Rightarrow r \equiv \neg(p \wedge q) \vee r \equiv (\neg p \vee \neg q) \vee r$$.

We claim these two expressions are logically equivalent. To verify, we check all cases based on $$p$$.

Case 1: If $$p$$ is false, then $$\neg p$$ is true, so both $$(\neg p \vee \neg q) \vee (p \wedge r)$$ and $$(\neg p \vee \neg q) \vee r$$ evaluate to true regardless of $$q$$ and $$r$$.

Case 2: If $$p$$ is true and $$q$$ is false, then $$\neg q$$ is true, so both expressions again evaluate to true.

Case 3: If $$p$$ is true and $$q$$ is true, then $$\neg p \vee \neg q = \text{false}$$. The first expression becomes $$(p \wedge r) = r$$ (since $$p$$ is true), and the second becomes simply $$r$$. Hence both equal $$r$$.

In every case the two expressions agree, confirming that $$(p \wedge q) \Rightarrow (p \wedge r) \equiv (p \wedge q) \Rightarrow r$$.

Hence, the correct answer is Option D.

We need to determine which of the given statements is a tautology (always true regardless of truth values of $$ p $$ and $$ q $$).

Recall that $$ A \Rightarrow B $$ is equivalent to $$ \sim A \lor B $$.

$$q \Rightarrow (\sim p \lor q) \equiv \sim q \lor (\sim p \lor q) \equiv \sim p \lor (\sim q \lor q) \equiv \sim p \lor T \equiv T$$

Since $$ \sim q \lor q $$ is always true (law of excluded middle), the entire expression simplifies to True. This is a tautology.

Option A: $$ (\sim p \lor q) \Rightarrow p \equiv \sim(\sim p \lor q) \lor p \equiv (p \land \sim q) \lor p \equiv p $$. This is not a tautology (false when $$ p = F $$).

Option B: $$ p \Rightarrow (\sim p \lor q) \equiv \sim p \lor (\sim p \lor q) \equiv \sim p \lor q $$. This is not a tautology (false when $$ p = T, q = F $$).

Option C: $$ (\sim p \lor q) \Rightarrow q \equiv \sim(\sim p \lor q) \lor q \equiv (p \land \sim q) \lor q \equiv (p \lor q) \land (\sim q \lor q) \equiv p \lor q $$. This is not a tautology (false when $$ p = F, q = F $$).

The tautology is $$ q \Rightarrow (\sim p \lor q) $$, which corresponds to Option D.

The statement $$(P \wedge (\sim R)) \to ((\sim R) \wedge Q)$$ is given to be False. An implication $$p \to q$$ is false only when $$p$$ is True and $$q$$ is False, so we require $$P \wedge (\sim R) = T$$ and $$(\sim R) \wedge Q = F$$. From $$P \wedge (\sim R) = T$$ we deduce $$P = T$$ and $$\sim R = T$$, hence $$R = F$$. Since $$(\sim R) \wedge Q = F$$ and $$\sim R = T$$, it follows that $$Q = F$$. Therefore $$P = T,\ Q = F,\ R = F$$.

Substituting these truth values into each option, we find:

Option A: $$P \vee Q \to \sim R = (T \vee F) \to T = T \to T = T$$

Option B: $$R \vee Q \to \sim P = (F \vee F) \to F = F \to F = T$$

Option C: $$\sim(P \vee Q) \to \sim R = \sim(T) \to T = F \to T = T$$

Option D: $$\sim(R \vee Q) \to \sim P = \sim(F) \to F = T \to F = F$$

The statement with truth value False is $$\boxed{\sim(R \vee Q) \to \sim P}$$, so the answer is Option D.

We need to find the ordered pair $$(*, \odot)$$ from $$\{\wedge, \vee\}$$ such that $$(p * q) \odot (p \odot \sim q)$$ is a tautology.

Step 1: Test each option systematically.

Option B: $$(*, \odot) = (\vee, \vee)$$

The expression becomes: $$(p \vee q) \vee (p \vee \sim q)$$

$$= p \vee q \vee p \vee \sim q$$

$$= p \vee (q \vee \sim q)$$

$$= p \vee T$$

$$= T$$

This is a tautology.

Step 2: Verify that other options are not tautologies.

Option A: $$(*, \odot) = (\vee, \wedge)$$

$$(p \vee q) \wedge (p \wedge \sim q)$$

When $$p = F, q = T$$: $$(F \vee T) \wedge (F \wedge F) = T \wedge F = F$$. Not a tautology.

Option C: $$(*, \odot) = (\wedge, \wedge)$$

$$(p \wedge q) \wedge (p \wedge \sim q)$$

$$= p \wedge (q \wedge \sim q) = p \wedge F = F$$. Not a tautology (always false).

Option D: $$(*, \odot) = (\wedge, \vee)$$

$$(p \wedge q) \vee (p \vee \sim q)$$

When $$p = F, q = T$$: $$(F \wedge T) \vee (F \vee F) = F \vee F = F$$. Not a tautology.

Only Option B gives a tautology.

The correct answer is Option B: $$(\vee, \vee)$$

We need to find how many choices for $$\Delta \in \{\wedge, \vee, \Rightarrow, \Leftrightarrow\}$$ make $$(p \Delta q) \Rightarrow ((p \Delta \sim q) \vee ((\sim p) \Delta q))$$ a tautology.

Recall that $$P \Rightarrow Q$$ is false only when $$P$$ is true and $$Q$$ is false. So the expression is a tautology if whenever $$(p \Delta q)$$ is true, at least one of $$(p \Delta \sim q)$$ or $$(\sim p \Delta q)$$ is also true.

Case 1: $$\Delta = \wedge$$ (AND)

$$(p \wedge q) \Rightarrow ((p \wedge \sim q) \vee (\sim p \wedge q))$$

When $$p = T, q = T$$: LHS = T. RHS = $$(T \wedge F) \vee (F \wedge T) = F \vee F = F$$.

So the implication is F. Not a tautology.

Case 2: $$\Delta = \vee$$ (OR)

$$(p \vee q) \Rightarrow ((p \vee \sim q) \vee (\sim p \vee q))$$

RHS: $$(p \vee \sim q) \vee (\sim p \vee q) = (p \vee \sim p) \vee (q \vee \sim q) = T \vee T = T$$.

Since RHS is always T, the implication is always T. Tautology.

Case 3: $$\Delta = \Rightarrow$$ (Implication)

$$(p \Rightarrow q) \Rightarrow ((p \Rightarrow \sim q) \vee (\sim p \Rightarrow q))$$

$$(\sim p \Rightarrow q)$$ is equivalent to $$(p \vee q)$$. This is false only when $$p = F, q = F$$.

When $$p = F, q = F$$: LHS = $$(F \Rightarrow F) = T$$. RHS = $$(F \Rightarrow T) \vee (T \Rightarrow F) = T \vee F = T$$.

When $$p = T, q = T$$: LHS = T. RHS = $$(T \Rightarrow F) \vee (F \Rightarrow T) = F \vee T = T$$.

When $$p = T, q = F$$: LHS = F. So implication is T.

When $$p = F, q = T$$: LHS = T. RHS = $$(F \Rightarrow F) \vee (T \Rightarrow T) = T \vee T = T$$.

All cases give T. Tautology.

Case 4: $$\Delta = \Leftrightarrow$$ (Biconditional)

$$(p \Leftrightarrow q) \Rightarrow ((p \Leftrightarrow \sim q) \vee (\sim p \Leftrightarrow q))$$

Note: $$(p \Leftrightarrow \sim q) = \sim(p \Leftrightarrow q)$$ and $$(\sim p \Leftrightarrow q) = \sim(p \Leftrightarrow q)$$.

So RHS = $$\sim(p \Leftrightarrow q) \vee \sim(p \Leftrightarrow q) = \sim(p \Leftrightarrow q)$$.

The expression becomes: $$(p \Leftrightarrow q) \Rightarrow \sim(p \Leftrightarrow q)$$.

When $$p = T, q = T$$: $$(T) \Rightarrow (F) = F$$. Not a tautology.

Therefore, exactly 2 choices ($$\vee$$ and $$\Rightarrow$$) make the expression a tautology.

The answer is Option B: 2.

We need to determine which statement is NOT equivalent to $$(p \Rightarrow q) \vee (p \Rightarrow r)$$.

We first simplify the given statement. We know $$p \Rightarrow q \equiv \sim p \vee q$$ and $$p \Rightarrow r \equiv \sim p \vee r$$. So $$(p \Rightarrow q) \vee (p \Rightarrow r) \equiv (\sim p \vee q) \vee (\sim p \vee r) \equiv \sim p \vee q \vee r$$. This is equivalent to $$p \Rightarrow (q \vee r)$$.

Now we check each option:

Option A: $$(p \wedge \sim r) \Rightarrow q \equiv \sim(p \wedge \sim r) \vee q \equiv \sim p \vee r \vee q$$. This equals $$\sim p \vee q \vee r$$. Equivalent.

Option B: $$\sim q \Rightarrow (\sim r \vee p) \equiv q \vee \sim r \vee p$$. This is $$p \vee q \vee \sim r$$, which is NOT the same as $$\sim p \vee q \vee r$$. For example, when $$p = T, q = F, r = T$$: our expression gives $$F \vee F \vee T = T$$, but Option B gives $$T \vee F \vee F = T$$. Try $$p = T, q = F, r = F$$: our expression gives $$F \vee F \vee F = F$$, but Option B gives $$T \vee F \vee T = T$$. So they differ. NOT equivalent.

Option C: $$p \Rightarrow (q \vee r) \equiv \sim p \vee q \vee r$$. Equivalent (same as our simplified form).

Option D: $$(p \wedge \sim q) \Rightarrow r \equiv \sim(p \wedge \sim q) \vee r \equiv \sim p \vee q \vee r$$. Equivalent.

Hence, the correct answer is Option B: $$\sim q \Rightarrow (\sim r \vee p)$$.

We need to identify which of the given statements is a tautology (always true regardless of truth values of $$p$$ and $$q$$).

Analyze Option A: $$((\sim q) \wedge p) \wedge q$$

This simplifies to $$p \wedge q \wedge (\sim q)$$, which contains $$q \wedge (\sim q) = F$$.

So this is always False (contradiction). Not a tautology.

Analyze Option B: $$((\sim q) \wedge p) \wedge (p \wedge (\sim p))$$

The term $$p \wedge (\sim p) = F$$.

So the entire expression is $$(\text{anything}) \wedge F = F$$.

This is always False (contradiction). Not a tautology.

Analyze Option C: $$((\sim q) \wedge p) \vee (p \vee (\sim p))$$

The term $$p \vee (\sim p) = T$$ (this is a tautology by itself).

So the entire expression is $$(\text{anything}) \vee T = T$$.

This is always True. This is a tautology!

Analyze Option D: $$(p \wedge q) \wedge (\sim(p \wedge q))$$

Let $$r = p \wedge q$$. Then this becomes $$r \wedge (\sim r) = F$$.

This is always False (contradiction). Not a tautology.

Therefore, the tautology is Option C: $$((\sim q) \wedge p) \vee (p \vee (\sim p))$$.

The correct answer is Option C.

We need to find the negation of the Boolean expression $$p \leftrightarrow (q \rightarrow p)$$.

$$q \rightarrow p \equiv \sim q \vee p$$

$$p \leftrightarrow (\sim q \vee p) \equiv [p \rightarrow (\sim q \vee p)] \wedge [(\sim q \vee p) \rightarrow p]$$

The first part: $$p \rightarrow (\sim q \vee p) \equiv \sim p \vee \sim q \vee p \equiv T$$ (tautology, since $$\sim p \vee p = T$$)

So: $$p \leftrightarrow (\sim q \vee p) \equiv T \wedge [(\sim q \vee p) \rightarrow p] \equiv (\sim q \vee p) \rightarrow p$$

$$(\sim q \vee p) \rightarrow p \equiv \sim(\sim q \vee p) \vee p \equiv (q \wedge \sim p) \vee p$$

$$\equiv (q \vee p) \wedge (\sim p \vee p) \equiv (q \vee p) \wedge T \equiv q \vee p \equiv p \vee q$$

So $$p \leftrightarrow (q \rightarrow p) \equiv p \vee q$$.

$$\sim(p \vee q) \equiv \sim p \wedge \sim q$$

The negation is T only when both $$p$$ and $$q$$ are F, which is $$\sim p \wedge \sim q$$. $$\checkmark$$

Therefore, the correct answer is Option D: $$\sim p \wedge \sim q$$.

We need to find the negation of $$(p \vee q) \Rightarrow ((\sim r) \vee p)$$. The negation of $$A \Rightarrow B$$ is $$A \wedge (\sim B)$$. Here $$A = (p \vee q)$$ and $$B = ((\sim r) \vee p)$$, so its negation is $$(p \vee q) \wedge \sim((\sim r) \vee p)$$.

By De Morgan's law, $$\sim((\sim r) \vee p) = r \wedge (\sim p)$$, therefore the expression becomes $$(p \vee q) \wedge r \wedge (\sim p)$$. Since $$\sim p$$ holds, $$(p \vee q) \wedge (\sim p)$$ simplifies to $$q \wedge (\sim p)$$, giving $$(\sim p) \wedge q \wedge r$$.

Therefore, the answer is Option C: $$(\sim p) \wedge q \wedge r$$.

We are given: $$P$$: Ramu is intelligent, $$Q$$: Ramu is rich, $$R$$: Ramu is not honest. So "Ramu is honest" is $$\sim R$$.

The statement "Ramu is intelligent and honest if and only if Ramu is not rich" translates to $$(P \wedge \sim R) \leftrightarrow (\sim Q)$$.

The negation of a biconditional is: $$\sim(A \leftrightarrow B) = (A \wedge \sim B) \vee (\sim A \wedge B)$$.

Here $$A = P \wedge (\sim R)$$ and $$B = \sim Q$$.

The first part: $$A \wedge \sim B = (P \wedge (\sim R)) \wedge Q$$.

The second part: $$\sim A \wedge B = \sim(P \wedge (\sim R)) \wedge (\sim Q) = ((\sim P) \vee R) \wedge (\sim Q)$$.

The full negation is $$((P \wedge (\sim R)) \wedge Q) \vee (((\sim P) \vee R) \wedge (\sim Q))$$.

The answer is Option D: $$((P \wedge (\sim R)) \wedge Q) \vee ((\sim Q) \wedge ((\sim P) \vee R))$$.

The function $$f : \mathbb{N} \to \mathbb{N}$$ is defined as:

$$f(n) = \begin{cases} 2n, & n = 2, 4, 6, 8, \ldots \text{ (even numbers)} \\ n-1, & n = 3, 7, 11, 15, \ldots \text{ (} n \equiv 3 \pmod{4}\text{)} \\ \frac{n+1}{2}, & n = 1, 5, 9, 13, \ldots \text{ (} n \equiv 1 \pmod{4}\text{)} \end{cases}$$

For even inputs $$n = 2,4,6,\ldots$$, we have $$f(n)=2n$$, which yields $$4,8,12,16,\ldots$$—all multiples of 4, i.e.\ $$\{n\in\mathbb{N}:n\equiv0\pmod4\}$$. If $$n\equiv3\pmod4$$ (so $$n=3,7,11,\ldots$$), then $$f(n)=n-1$$ gives $$2,6,10,14,\ldots$$—the set $$\{n\in\mathbb{N}:n\equiv2\pmod4\}$$. Finally, if $$n\equiv1\pmod4$$ (so $$n=1,5,9,\ldots$$), then $$f(n)=\frac{n+1}{2}$$ produces $$1,3,5,7,\ldots$$—all odd natural numbers.

Thus the range of $$f$$ is $$\{1,3,5,7,\ldots\}\cup\{2,6,10,14,\ldots\}\cup\{4,8,12,16,\ldots\}=\{\text{odd numbers}\}\cup\{\text{numbers}\equiv2\pmod4\}\cup\{\text{multiples of }4\}=\mathbb{N},$$ so $$f$$ is onto.

The three output sets—odd numbers, numbers congruent to 2 mod 4, and multiples of 4—are mutually disjoint, and in each case the rule is strictly increasing (hence injective). Therefore $$f$$ is one-one as well.

Therefore, the correct answer is Option A: One-one and onto.

We have $$f: \mathbb{R} \to \mathbb{R}$$ continuous, with $$f(3x) - f(x) = x$$, $$f(8) = 7$$.

Let $$f(x) = ax + b$$. Then:

$$f(3x) - f(x) = 3ax + b - ax - b = 2ax = x$$

$$\implies a = \frac{1}{2}$$

So $$f(x) = \frac{x}{2} + b$$.

$$f(8) = \frac{8}{2} + b = 4 + b = 7 \implies b = 3$$

$$f(x) = \frac{x}{2} + 3$$

$$f(14) = \frac{14}{2} + 3 = 7 + 3 = 10$$

Verification: $$f(3 \times 14) - f(14) = f(42) - f(14) = (21 + 3) - (7 + 3) = 24 - 10 = 14$$ ✓

Therefore, the correct answer is Option B: 10.

Given statement,

$$\sim q\land p\Rightarrow \sim p\lor q$$

Now,

$$\sim p\lor q\equiv p\Rightarrow q$$

and

$$\sim q\land p\equiv \sim(p\Rightarrow q)$$

Therefore, the statement becomes

$$\sim(p\Rightarrow q)\Rightarrow (p\Rightarrow q)$$

Let

$$A=(p\Rightarrow q)$$

Then the statement is

$$\sim A\Rightarrow A$$

The negation of

$$\sim A\Rightarrow A$$

is

$$\sim A\land \sim A$$

$$=\sim A$$

Substituting back,

$$=\sim(p\Rightarrow q)$$

Hence, the required negation is $$\boxed{\sim(p\Rightarrow q)}$$.

We need to count the total number of functions $$ f: \{1, 2, 3, 4\} \to \{1, 2, 3, 4, 5, 6\} $$ such that $$ f(1) + f(2) = f(3) $$.

We need $$ f(1) + f(2) = f(3) $$, where $$ f(1), f(2) \in \{1, 2, 3, 4, 5, 6\} $$ and $$ f(3) \in \{1, 2, 3, 4, 5, 6\} $$.

So we need $$ 2 \le f(1) + f(2) \le 6 $$ (since both are at least 1, and the sum must be at most 6).

Count pairs $$ (f(1), f(2)) $$ for each value of $$ f(3) $$:

• $$ f(3) = 2 $$: $$ f(1) + f(2) = 2 $$: only $$ (1,1) $$ — 1 pair
• $$ f(3) = 3 $$: $$ f(1) + f(2) = 3 $$: $$ (1,2), (2,1) $$ — 2 pairs
• $$ f(3) = 4 $$: $$ f(1) + f(2) = 4 $$: $$ (1,3), (2,2), (3,1) $$ — 3 pairs
• $$ f(3) = 5 $$: $$ f(1) + f(2) = 5 $$: $$ (1,4), (2,3), (3,2), (4,1) $$ — 4 pairs
• $$ f(3) = 6 $$: $$ f(1) + f(2) = 6 $$: $$ (1,5), (2,4), (3,3), (4,2), (5,1) $$ — 5 pairs

Total valid triples = $$ 1 + 2 + 3 + 4 + 5 = 15 $$.

$$ f(4) $$ can be any value in $$ \{1, 2, 3, 4, 5, 6\} $$, so there are 6 choices.

$$\text{Total} = 15 \times 6 = 90$$

The total number of functions is 90, which corresponds to Option B.

We need to determine whether $$f \circ g$$ is one-one and/or onto, where $$f : \mathbb{R} \to \mathbb{R}$$ is defined as $$f(x) = x - 1$$ and $$g : \mathbb{R} \setminus \{1, -1\} \to \mathbb{R}$$ is defined as $$g(x) = \frac{x^2}{x^2 - 1}$$.

First, we find the composite function. The domain of $$f \circ g$$ is the domain of $$g$$, which is $$\mathbb{R} \setminus \{-1, 1\}$$, and the codomain is $$\mathbb{R}$$. Substituting gives $$f(g(x)) = g(x) - 1 = \frac{x^2}{x^2 - 1} - 1 = \frac{x^2 - (x^2 - 1)}{x^2 - 1} = \frac{1}{x^2 - 1}.$$

Next, we check if $$f \circ g$$ is one-one (injective) by testing whether $$f(g(x_1)) = f(g(x_2))$$ implies $$x_1 = x_2$$. Since $$f(g(x)) = \frac{1}{x^2 - 1}$$ depends only on $$x^2$$, we have $$f(g(x)) = f(g(-x))$$ for all $$x$$ in the domain. For example, take $$x = 2$$ and $$x = -2$$: $$f(g(2)) = \frac{1}{4 - 1} = \frac{1}{3}, \quad f(g(-2)) = \frac{1}{4 - 1} = \frac{1}{3}.$$ Since $$2 \neq -2$$ but $$f(g(2)) = f(g(-2))$$, the function is not one-one.

Now, we check if $$f \circ g$$ is onto (surjective). We need to determine whether every real number $$y$$ is in the range of $$h(x) = \frac{1}{x^2 - 1}$$. Setting $$y = \frac{1}{x^2 - 1}$$ gives $$x^2 - 1 = \frac{1}{y}, \quad x^2 = 1 + \frac{1}{y}.$$ For a real solution to exist, we need $$y \neq 0$$ and $$1 + \frac{1}{y} \geq 0$$, i.e., $$\frac{y + 1}{y} \geq 0.$$ Therefore, such values of $$y$$ satisfy $$y > 0$$ or $$y \leq -1$$ (and additionally $$x \neq \pm 1$$, which means $$x^2 \neq 1$$, i.e., $$\frac{1}{y} \neq 0$$, which is always true).

Case-by-case analysis of values NOT in the range:

Case 1: $$y = 0$$. Since $$\frac{1}{x^2 - 1} = 0$$ has no solution, $$y = 0$$ is not in the range.

Case 2: $$-1 < y < 0$$. Here $$\frac{y + 1}{y} < 0$$, so $$x^2 < 0$$, which gives no real solution. These values are not in the range. For instance, $$y = -\frac{1}{2}$$ gives $$x^2 = 1 - 2 = -1 < 0$$, which has no real solution.

Since not every real number is achieved (the interval $$(-1, 0]$$ is not in the range), the function is not onto.

Therefore, $$f \circ g$$ is neither one-one nor onto.

The answer is Option D: Neither one-one nor onto.

We need to find $$\alpha$$ such that $$\sum_{k=1}^{10}f(\alpha + k)=\frac{512}{3}(2^{20}-1)\,.$$

First, observe that $$f(x+y)=2f(x)f(y)$$ with $$f(1)=2$$. Substituting $$x=y=1$$ gives $$f(2)=2f(1)^2=2\times4=8=2^3$$, and then for $$x=1,y=2$$ we get $$f(3)=2f(1)f(2)=2\times2\times8=32=2^5$$. Likewise, $$x=1,y=3$$ yields $$f(4)=2f(1)f(3)=2\times2\times32=128=2^7$$, so by induction the pattern is $$f(n)=2^{2n-1}$$, which indeed satisfies $$f(x+y)=2^{2(x+y)-1}=2\cdot2^{2x-1}\cdot2^{2y-1}=2f(x)f(y)\,. $$

Since $$f(n)=2^{2n-1}$$, the sum becomes $$\sum_{k=1}^{10}f(\alpha + k)=\sum_{k=1}^{10}2^{2(\alpha+k)-1}=2^{2\alpha-1}\sum_{k=1}^{10}2^{2k}=2^{2\alpha-1}\sum_{k=1}^{10}4^k$$ which simplifies to $$2^{2\alpha-1}\cdot\frac{4(4^{10}-1)}{4-1}=2^{2\alpha-1}\cdot\frac{4(2^{20}-1)}{3}\,.$$

Equating this to $$\frac{512}{3}(2^{20}-1)$$ gives $$2^{2\alpha-1}\cdot\frac{4(2^{20}-1)}{3}=\frac{512}{3}(2^{20}-1)\,, $$ hence $$2^{2\alpha-1}\cdot4=512\,, $$ or $$2^{2\alpha-1}\cdot2^2=2^9\,, $$ so $$2^{2\alpha+1}=2^9$$ and therefore $$2\alpha+1=9\implies\alpha=4\,. $$

Thus the correct answer is $$4$$.

We need to find $$f^6(6) + f^7(7)$$ where $$f(x) = \frac{x-1}{x+1}$$.

Since $$f^1(x) = \frac{x-1}{x+1}$$, a calculation shows $$f^2(x) = f(f(x)) = \frac{\frac{x-1}{x+1} - 1}{\frac{x-1}{x+1} + 1} = \frac{x - 1 - x - 1}{x - 1 + x + 1} = \frac{-2}{2x} = -\frac{1}{x}$$.

Then applying the function again gives $$f^3(x) = f\left(-\frac{1}{x}\right) = \frac{-\frac{1}{x} - 1}{-\frac{1}{x} + 1} = \frac{-1 - x}{-1 + x} = -\frac{x + 1}{x - 1}$$, and one more iteration yields $$f^4(x) = f\left(-\frac{x+1}{x-1}\right) = \frac{-\frac{x+1}{x-1} - 1}{-\frac{x+1}{x-1} + 1} = \frac{-x-1-x+1}{-x-1+x-1} = \frac{-2x}{-2} = x$$.

Therefore, the function has period 4: $$f^4(x) = x$$.

Now reducing the exponents modulo 4 gives $$f^6 = f^{4+2} = f^2$$ and $$f^7 = f^{4+3} = f^3$$.

Substituting these into the arguments yields $$f^6(6) = f^2(6) = -\frac{1}{6}$$ and $$f^7(7) = f^3(7) = -\frac{7+1}{7-1} = -\frac{8}{6} = -\frac{4}{3}$$.

From this it follows that $$f^6(6) + f^7(7) = -\frac{1}{6} - \frac{4}{3} = -\frac{1}{6} - \frac{8}{6} = -\frac{9}{6} = -\frac{3}{2}$$.

The correct answer is Option B.

Given,

$$f(x)=\sin^{-1}[2x^2-3]+\log_2\left(\log_{\frac12}(x^2-5x+5)\right)$$

For the function to exist, both parts must be defined.

First consider

$$\sin^{-1}[2x^2-3]$$

Since $$\sin^{-1}t$$ is defined for $$-1\le t\le1,$$

we need $$-1\le[2x^2-3]\le1$$

As the GIF gives only integers,

$$[2x^2-3]\in\{-1,0,1\}$$

Hence,

$$-1\le2x^2-3<2$$

$$2\le2x^2<5$$

$$1\le x^2<\frac52$$

Therefore,

$$x\in\left(-\sqrt{\frac52},-1\right]\cup\left[1,\sqrt{\frac52}\right)$$

Now consider

$$\log_2\left(\log_{\frac12}(x^2-5x+5)\right)$$

For the outer logarithm,

$$\log_{\frac12}(x^2-5x+5)>0$$

Since the base

$$\frac12<1,$$

$$\log_{\frac12}t>0\iff0<t<1$$

Thus,

$$0<x^2-5x+5<1$$

Now solve separately.

From $$x^2-5x+5>0,$$

roots are $$x=\frac{5\pm\sqrt5}{2}$$

Hence,

$$x\in\left(-\infty,\frac{5-\sqrt5}{2}\right)\cup\left(\frac{5+\sqrt5}{2},\infty\right)$$

Also,

$$x^2-5x+5<1$$

$$x^2-5x+4<0$$

$$(x-1)(x-4)<0$$

$$1<x<4$$

Combining both,

$$x\in\left(1,\frac{5-\sqrt5}{2}\right)\cup\left(\frac{5+\sqrt5}{2},4\right)$$

Now intersect with

$$x\in\left(-\sqrt{\frac52},-1\right]\cup\left[1,\sqrt{\frac52}\right)$$

Since

$$\sqrt{\frac52}<\frac{5+\sqrt5}{2},$$

only the first interval survives.

Therefore, the domain is

$$\boxed{\left(1,\frac{5-\sqrt5}{2}\right)}$$.

We need to find $$f\left(g\left(\frac{1}{n}\sum_{i=1}^{n} f(a_i)\right)\right)$$ where $$a_1, a_2, \ldots, a_n$$ are in AP with mean zero.

Step 1: Understand the given functions.

$$f(x) = \alpha x^5 + \beta x^3 + \gamma x$$ where $$\alpha, \beta, \gamma > 0$$.

$$g(f(x)) = x$$ for all $$x \in \mathbb{R}$$, so $$g = f^{-1}$$ ($$g$$ is the inverse function of $$f$$).

Step 2: Observe that $$f$$ is an odd function.

$$f(-x) = \alpha(-x)^5 + \beta(-x)^3 + \gamma(-x) = -\alpha x^5 - \beta x^3 - \gamma x = -f(x)$$

So $$f$$ is an odd function: $$f(-x) = -f(x)$$.

Step 3: Use the AP with mean zero property.

Since $$a_1, a_2, \ldots, a_n$$ are in AP with mean zero:

$$\frac{1}{n}\sum_{i=1}^{n} a_i = 0$$

The terms are symmetric about zero, meaning they can be paired as $$a_i$$ and $$a_{n+1-i} = -a_i$$.

Step 4: Evaluate $$\sum f(a_i)$$.

Since the $$a_i$$ are symmetric about 0 and $$f$$ is odd:

$$f(a_i) + f(a_{n+1-i}) = f(a_i) + f(-a_i) = f(a_i) - f(a_i) = 0$$

Each pair sums to zero. If $$n$$ is odd, the middle term is $$a_{(n+1)/2} = 0$$, and $$f(0) = 0$$.

Therefore: $$\sum_{i=1}^{n} f(a_i) = 0$$

Step 5: Compute the final expression.

$$\frac{1}{n}\sum_{i=1}^{n} f(a_i) = \frac{0}{n} = 0$$

$$g(0) = ?$$ Since $$f(0) = 0$$, and $$g = f^{-1}$$, we get $$g(0) = 0$$.

$$f(g(0)) = f(0) = 0$$

The correct answer is Option A: $$0$$

Given,

$$f(x)=\frac{\cos^{-1}\left(\dfrac{x^2-5x+6}{x^2-9}\right)}{\log(x^2-3x+2)}$$

For the function to be defined, we need:

1. $$-1\le \dfrac{x^2-5x+6}{x^2-9}\le1$$

2. $$x^2-3x+2>0$$

3. $$\log(x^2-3x+2)\ne0$$

4. $$x^2-9\ne0$$

Now,

$$\frac{x^2-5x+6}{x^2-9}=\frac{(x-2)(x-3)}{(x-3)(x+3)}=\frac{x-2}{x+3},\qquad x\ne3$$

So,

$$-1\le\frac{x-2}{x+3}\le1$$

First solve

$$\frac{x-2}{x+3}\ge-1$$

$$\frac{x-2+x+3}{x+3}\ge0$$

$$\frac{2x+1}{x+3}\ge0$$

Hence,

$$x\in(-\infty,-3)\cup\left[-\frac12,\infty\right)$$

Now solve

$$\frac{x-2}{x+3}\le1$$

$$\frac{x-2-x-3}{x+3}\le0$$

$$\frac{-5}{x+3}\le0$$

$$x>-3$$

Combining both,

$$x\in\left[-\frac12,\infty\right)$$

Now,

$$x^2-3x+2=(x-1)(x-2)$$

For logarithm to exist,

$$(x-1)(x-2)>0$$

$$x<1\quad\text{or}\quad x>2$$

Also,

$$\log(x^2-3x+2)\ne0$$

$$x^2-3x+2\ne1$$

$$x^2-3x+1\ne0$$

This condition is automatically satisfied in the final interval.

Also,

$$x\ne3$$

Therefore, combining all conditions,

$$x\in\left[-\frac12,1\right)\cup(2,\infty)-\{3\}$$

Hence, the correct answer is

$$\boxed{\left[-\frac12,1\right)\cup(2,\infty)-\{3\}}$$

We need to find the maximum number of the 9 given compound propositions that can be made simultaneously true by assigning truth values to $$p, q, r, s$$.

The 9 propositions are:

1. $$p \vee r \vee s$$   2. $$p \vee r \vee \sim s$$   3. $$p \vee \sim q \vee s$$

4. $$\sim p \vee \sim r \vee s$$   5. $$\sim p \vee \sim r \vee \sim s$$   6. $$\sim p \vee q \vee \sim s$$

7. $$q \vee r \vee \sim s$$   8. $$q \vee \sim r \vee \sim s$$   9. $$\sim p \vee \sim q \vee \sim s$$

Strategy:

There are $$2^4 = 16$$ possible truth value assignments for $$(p, q, r, s)$$. A disjunction (OR) is false only when all its operands are false. We systematically check assignments to maximize the count of true propositions.

Try $$p = T,\; q = T,\; r = F,\; s = F$$:

Evaluate each proposition:

1. $$T \vee F \vee F = T$$ $$\checkmark$$

2. $$T \vee F \vee T = T$$ $$\checkmark$$

3. $$T \vee F \vee F = T$$ $$\checkmark$$

4. $$F \vee T \vee F = T$$ $$\checkmark$$

5. $$F \vee T \vee T = T$$ $$\checkmark$$

6. $$F \vee T \vee T = T$$ $$\checkmark$$

7. $$T \vee F \vee T = T$$ $$\checkmark$$

8. $$T \vee T \vee T = T$$ $$\checkmark$$

9. $$F \vee F \vee T = T$$ $$\checkmark$$

All 9 propositions are true.

Verify this is indeed the maximum:

Since all 9 propositions are satisfied by the assignment $$(p, q, r, s) = (T, T, F, F)$$, the maximum possible number is 9.

The answer is $$\boxed{9}$$.

We need to find the number of functions $$f: A \to B$$ such that $$f(x)\le (x-3)^2 + 1$$ for every $$x\in A$$.

From the inequality $$x^2 - 10x + 9 \le 0$$ we have $$(x-1)(x-9)\le 0$$, which implies $$1\le x\le 9$$. Since $$x\in\mathbb{N}$$, it follows that $$A = \{1,2,3,4,5,6,7,8,9\}$$.

Also, by definition, $$B = \{n^2 : n\in\mathbb{N}\} = \{1,4,9,16,25,36,\ldots\}$$.

For each $$x$$ we require $$f(x)\le (x-3)^2 + 1$$ and $$f(x)\in B$$. We list the bounds and counts:

When $$x=1$$, $$(x-3)^2 + 1 = 4 + 1 = 5$$, so the eligible values in $$B$$ are $$\{1,4\}$$, giving 2 choices.

When $$x=2$$, $$(x-3)^2 + 1 = 1 + 1 = 2$$, so the eligible values in $$B$$ are $$\{1\}$$, giving 1 choice.

When $$x=3$$, $$(x-3)^2 + 1 = 0 + 1 = 1$$, so the eligible values in $$B$$ are $$\{1\}$$, giving 1 choice.

When $$x=4$$, $$(x-3)^2 + 1 = 1 + 1 = 2$$, so the eligible values in $$B$$ are $$\{1\}$$, giving 1 choice.

When $$x=5$$, $$(x-3)^2 + 1 = 4 + 1 = 5$$, so the eligible values in $$B$$ are $$\{1,4\}$$, giving 2 choices.

When $$x=6$$, $$(x-3)^2 + 1 = 9 + 1 = 10$$, so the eligible values in $$B$$ are $$\{1,4,9\}$$, giving 3 choices.

When $$x=7$$, $$(x-3)^2 + 1 = 16 + 1 = 17$$, so the eligible values in $$B$$ are $$\{1,4,9,16\}$$, giving 4 choices.

When $$x=8$$, $$(x-3)^2 + 1 = 25 + 1 = 26$$, so the eligible values in $$B$$ are $$\{1,4,9,16,25\}$$, giving 5 choices.

When $$x=9$$, $$(x-3)^2 + 1 = 36 + 1 = 37$$, so the eligible values in $$B$$ are $$\{1,4,9,16,25,36\}$$, giving 6 choices.

Since the choices at each $$x$$ are independent, the total number of such functions is

$$2 \times 1 \times 1 \times 1 \times 2 \times 3 \times 4 \times 5 \times 6 = 1440$$.

The answer is $$\boxed{1440}$$.

Given $$S = \{1, 2, \ldots, 10\}$$ and $$f: S \to S$$ defined by:

$$f(n) = \begin{cases} 2n & \text{if } n = 1,2,3,4,5 \\ 2n - 11 & \text{if } n = 6,7,8,9,10 \end{cases}$$

So: $$f(1)=2,\; f(2)=4,\; f(3)=6,\; f(4)=8,\; f(5)=10,\; f(6)=1,\; f(7)=3,\; f(8)=5,\; f(9)=7,\; f(10)=9$$

Note that $$f$$ is a bijection. Its inverse is:

$$f^{-1}(k) = \begin{cases} k/2 & \text{if } k \text{ is even} \\ (k+11)/2 & \text{if } k \text{ is odd} \end{cases}$$

We need $$g: S \to S$$ such that $$f \circ g(n) = \begin{cases} n+1 & \text{if } n \text{ is odd} \\ n-1 & \text{if } n \text{ is even} \end{cases}$$

Since $$f$$ is bijective: $$g(n) = f^{-1}(f \circ g(n))$$.

Computing each value:

$$g(1) = f^{-1}(2) = 1, \quad g(2) = f^{-1}(1) = 6, \quad g(3) = f^{-1}(4) = 2$$

$$g(4) = f^{-1}(3) = 7, \quad g(5) = f^{-1}(6) = 3$$

$$g(6) = f^{-1}(5) = 8, \quad g(7) = f^{-1}(8) = 4, \quad g(8) = f^{-1}(7) = 9$$

$$g(9) = f^{-1}(10) = 5, \quad g(10) = f^{-1}(9) = 10$$

Now compute:

$$g(1) + g(2) + g(3) + g(4) + g(5) = 1 + 6 + 2 + 7 + 3 = 19$$

$$g(10) = 10$$

$$g(10) \cdot (g(1) + g(2) + g(3) + g(4) + g(5)) = 10 \times 19 = 190$$

The correct answer is $$\boxed{190}$$.

Given,

$$f(x)=\left(2\left(1-\frac{x^{25}}{2}\right)(2+x^{25})\right)^{\frac{1}{50}}$$

Simplifying,

$$f(x)=\left((2-x^{25})(2+x^{25})\right)^{\frac{1}{50}}$$

$$f(x)=\left(4-x^{50}\right)^{\frac{1}{50}}$$

Now,

$$f(f(x))=\left(4-\left((4-x^{50})^{\frac{1}{50}}\right)^{50}\right)^{\frac{1}{50}}$$

$$=\left(4-(4-x^{50})\right)^{\frac{1}{50}}$$

$$=(x^{50})^{\frac{1}{50}}=x$$

Hence,

$$f(f(x))=x$$

Given,

$$g(x)=f(f(f(x)))+f(f(x))$$

Using $$f(f(x))=x,$$

$$g(x)=f(x)+x$$

Now,

$$f(1)=\left(4-1^{50}\right)^{\frac{1}{50}}=3^{\frac{1}{50}}$$

Therefore,

$$g(1)=3^{\frac{1}{50}}+1$$

Since,

$$1<3^{\frac{1}{50}}<2$$

we get,

$$2<g(1)<3$$

Therefore, the greatest integer less than or equal to $$g(1)$$ is $$\boxed{2}$$.

We need to count functions $$f : S \times S \to S$$ (where $$S = \{1,2,3,4\}$$) that are onto, symmetric ($$f(a,b) = f(b,a)$$), and satisfy $$f(a,b) \geq a$$ for all $$(a,b) \in S \times S$$.

Analyzing the constraint $$f(a,b) \geq a$$, since $$f(a,b) = f(b,a)$$, we also need $$f(a,b) \geq b$$. Therefore:

$$f(a,b) \geq \max(a,b)$$

Next, because $$f(a,b) \in S = \{1,2,3,4\}$$ and $$f(a,b) \geq \max(a,b)$$, we have $$f(4,b) \geq 4$$ for all $$b$$, so $$f(4,b) = 4$$. Similarly, $$f(a,4) = 4$$. This fixes all values involving 4.

Now consider values involving only $$\{1,2,3\}$$. For $$(a,b)$$ with $$a,b \in \{1,2,3\}$$, we have $$f(a,b) \geq \max(a,b)$$. Since $$f$$ is symmetric, we only need to specify $$f(a,b)$$ for $$a \leq b$$.

$$f(3,3) \geq 3$$: choices are $$\{3,4\}$$ (2 choices)

$$f(2,3) = f(3,2) \geq 3$$: choices are $$\{3,4\}$$ (2 choices)

$$f(1,3) = f(3,1) \geq 3$$: choices are $$\{3,4\}$$ (2 choices)

$$f(2,2) \geq 2$$: choices are $$\{2,3,4\}$$ (3 choices)

$$f(1,2) = f(2,1) \geq 2$$: choices are $$\{2,3,4\}$$ (3 choices)

$$f(1,1) \geq 1$$: choices are $$\{1,2,3,4\}$$ (4 choices)

Without the onto constraint, the total number of assignments is $$2 \times 2 \times 2 \times 3 \times 3 \times 4 = 288$$.

The value 4 is already achieved from the forced values. We need the values 1, 2, and 3 to each appear at least once.

Value 1 can only appear at $$f(1,1)$$, so $$f(1,1) = 1$$ is forced. This reduces choices for $$f(1,1)$$ to 1 option.

Value 2 can appear at $$f(1,2)$$ or $$f(2,2)$$, and it must appear in at least one of those entries.

Value 3 can appear at $$f(3,3)$$, $$f(2,3)$$, $$f(1,3)$$, $$f(2,2)$$, or $$f(1,2)$$.

With $$f(1,1) = 1$$, the remaining free choices are:

$$f(3,3)$$: $$\{3,4\}$$, $$f(2,3)$$: $$\{3,4\}$$, $$f(1,3)$$: $$\{3,4\}$$, $$f(2,2)$$: $$\{2,3,4\}$$, $$f(1,2)$$: $$\{2,3,4\}$$.

Without the onto requirement for 2 and 3, there are $$2 \times 2 \times 2 \times 3 \times 3 = 72$$ assignments.

To ensure both 2 and 3 appear, we use inclusion-exclusion. Let $$A$$ be the set of assignments where 2 does not appear, and $$B$$ the set where 3 does not appear.

In $$A$$, we must have $$f(1,2) \in \{3,4\}$$ and $$f(2,2) \in \{3,4\}$$, giving $$|A| = 2 \times 2 \times 2 \times 2 \times 2 = 32$$.

In $$B$$, we must have $$f(3,3) = f(2,3) = f(1,3) = 4$$ and $$f(2,2), f(1,2) \in \{2,4\}$$, giving $$|B| = 1 \times 1 \times 1 \times 2 \times 2 = 4$$.

In $$A \cap B$$, both 2 and 3 are absent, so all five free values must be 4, giving $$|A \cap B| = 1$$.

By inclusion-exclusion, the number of valid assignments is $$72 - 32 - 4 + 1 = 37$$.

The answer is $$\boxed{37}$$.

We need to find the negation of $$\sim p \wedge (p \vee q)$$.

Applying De Morgan's law, the negation is $$\sim(\sim p) \vee \sim(p \vee q)$$, which simplifies to $$p \vee (\sim p \wedge \sim q)$$.

Now we apply the distributive law: $$p \vee (\sim p \wedge \sim q) = (p \vee \sim p) \wedge (p \vee \sim q)$$.

Since $$p \vee \sim p$$ is always true (a tautology), this reduces to $$T \wedge (p \vee \sim q) = p \vee \sim q$$.

Therefore, the negation of $$\sim p \wedge (p \vee q)$$ is $$p \vee \sim q$$.

We need to determine whether statements (a) $$(\sim q \wedge (p \to q)) \to \sim p$$ and (b) $$((p \vee q) \wedge \sim p) \to q$$ are tautologies.

For statement (a), recall that $$p \to q \equiv \sim p \vee q$$. So $$\sim q \wedge (p \to q) = \sim q \wedge (\sim p \vee q) = (\sim q \wedge \sim p) \vee (\sim q \wedge q) = \sim p \wedge \sim q$$, since $$\sim q \wedge q$$ is always false. The statement becomes $$(\sim p \wedge \sim q) \to \sim p$$. Since the hypothesis $$\sim p \wedge \sim q$$ contains $$\sim p$$, the implication is always true. So (a) is a tautology.

For statement (b), we simplify $$(p \vee q) \wedge \sim p = (\sim p \wedge p) \vee (\sim p \wedge q) = F \vee (\sim p \wedge q) = \sim p \wedge q$$. The statement becomes $$(\sim p \wedge q) \to q$$. Since the hypothesis contains $$q$$, the implication is always true. So (b) is also a tautology.

Therefore, both (a) and (b) are tautologies.

We evaluate the truth value of each statement using the rule that a conditional $$p \Rightarrow q$$ is false only when $$p$$ is true and $$q$$ is false.

Statement (A): "If $$3 + 3 = 7$$ then $$4 + 3 = 8$$." The hypothesis $$3 + 3 = 7$$ is false. A conditional with a false hypothesis is vacuously true. So (A) is true.

Statement (B): "If $$5 + 3 = 8$$ then the earth is flat." Here $$5 + 3 = 8$$ is true, but "the earth is flat" is false. A conditional with a true hypothesis and false conclusion is false. So (B) is false.

Statement (C): "If both (A) and (B) are true then $$5 + 6 = 17$$." Since (B) is false, the hypothesis "(A) and (B) are true" is false. Therefore (C) is vacuously true. So (C) is true.

Thus (A) and (C) are true while (B) is false.

We have to simplify the statement $$ (p \wedge q)\;\Rightarrow\;\bigl((r \wedge q)\;\wedge\;p\bigr)\;. $$

First recall the basic implication law:

$$A \Rightarrow B \;\equiv\; \neg A \,\vee\, B.$$

Before using the law, it is convenient to tidy up the right-hand conjunction. Because conjunction is both commutative and associative, we may rearrange the factors freely:

$$ (r \wedge q) \wedge p \;=\; p \wedge q \wedge r. $$

So the whole expression becomes

$$ (p \wedge q)\;\Rightarrow\;(p \wedge q \wedge r). $$

Now set $$A = (p \wedge q),\qquad B = (p \wedge q \wedge r).$$ Applying the implication law we obtain

$$ \neg(p \wedge q) \;\vee\; (p \wedge q \wedge r). $$

Next, use De Morgan’s rule for the negation of a conjunction:

$$ \neg(p \wedge q) \;=\; \neg p \;\vee\; \neg q. $$

Substituting this result we have

$$ (\neg p \;\vee\; \neg q) \;\vee\; (p \wedge q \wedge r). $$

The disjunction symbol “&vee;” is associative, so we can write the whole thing as

$$ \neg p \;\vee\; \neg q \;\vee\; (p \wedge q \wedge r). $$

To see the final simplification clearly, break the analysis into the two possible truth-values of $$q$$.

Case 1: $$q$$ is false. If $$q$$ is false, then $$\neg q$$ is true, and the entire disjunction is automatically true. Therefore the expression is true irrespective of $$p$$ and $$r$$.

Case 2: $$q$$ is true. If $$q$$ is true, the term $$\neg q$$ becomes false, and the disjunction reduces to

$$ \neg p \;\vee\; (p \wedge r). $$

Factor $$p$$ out of the second term:

$$ \neg p \;\vee\; \bigl(p \wedge r\bigr) \;=\; (\neg p \;\vee\; p) \wedge (\neg p \;\vee\; r) \;=\; \text{T} \wedge (\neg p \;\vee\; r) \;=\; \neg p \;\vee\; r. $$

Hence, when $$q$$ is true, the original statement is equivalent to $$\neg p \;\vee\; r$$. Remembering that $$q$$ itself is true in this case, $$\neg p \;\vee\; r \quad\equiv\quad \neg p \;\vee\; (r \wedge q).$$ (The factor $$q$$ can be inserted because $$q$$ is already true and $$r \wedge q$$ therefore has the same truth value as $$r$$.)

Combining both cases, we can write the simplified form valid for all truth-values of $$p,q,r$$ as

$$ \neg p \;\vee\; \neg q \;\vee\; (r \wedge q). $$

Finally recognise this again as an implication, with antecedent $$(p \wedge q)$$ and consequent $$(r \wedge q)$$:

$$ \neg(p \wedge q) \;\vee\; (r \wedge q) \;\equiv\; (p \wedge q) \Rightarrow (r \wedge q). $$

Thus the original Boolean expression is logically equivalent to $$ (p \wedge q) \Rightarrow (r \wedge q). $$

Comparing with the given options, this matches Option C.

Hence, the correct answer is Option C.

We need to simplify $$(p \wedge \sim q) \Rightarrow (q \vee \sim p)$$.

Using the equivalence $$A \Rightarrow B \equiv \sim A \vee B$$, we write: $$\sim(p \wedge \sim q) \vee (q \vee \sim p).$$

By De Morgan's law, $$\sim(p \wedge \sim q) = \sim p \vee q$$. Substituting: $$(\sim p \vee q) \vee (q \vee \sim p) = \sim p \vee q.$$

The expression $$\sim p \vee q$$ is precisely the conditional $$p \Rightarrow q$$.

Therefore $$(p \wedge \sim q) \Rightarrow (q \vee \sim p)$$ is equivalent to $$p \Rightarrow q$$.

We are given $$F_1(A, B, C) = (A \wedge \sim B) \vee [\sim C \wedge (A \vee B)] \vee \sim A$$ and $$F_2(A, B) = (A \vee B) \vee (B \to \sim A)$$.

First, we check whether $$F_1$$ is a tautology. Consider $$A = T$$, $$B = T$$, $$C = T$$: $$(T \wedge F) \vee [F \wedge (T \vee T)] \vee F = F \vee F \vee F = F$$. Since $$F_1$$ evaluates to $$F$$ for this assignment, $$F_1$$ is not a tautology.

Next, we check $$F_2$$. Recall that $$B \to \sim A \equiv \sim B \vee \sim A$$. So $$F_2 = (A \vee B) \vee (\sim B \vee \sim A) = (A \vee \sim A) \vee (B \vee \sim B) = T \vee T = T$$.

More explicitly, rearranging: $$F_2 = A \vee B \vee \sim B \vee \sim A = (A \vee \sim A) \vee (B \vee \sim B) = T$$. So $$F_2$$ is always true, making it a tautology.

Therefore $$F_1$$ is not a tautology but $$F_2$$ is a tautology.

We have to simplify the Boolean expression $$ (p \Rightarrow q)\;\wedge\; (q \Rightarrow \sim p)\,. $$

First, recall the standard logical equivalence for an implication. The implication formula states:

$$ a \Rightarrow b \;\equiv\; \sim a \,\vee\, b. $$

Applying this to each implication in our expression, we replace the arrows by disjunctions:

For the first part, $$p \Rightarrow q \equiv \sim p \vee q.$$

For the second part, $$q \Rightarrow \sim p \equiv \sim q \vee \sim p.$$

Substituting these two results back into the original conjunction, we obtain

$$ (\sim p \vee q)\;\wedge\;(\sim q \vee \sim p). $$

Now we notice that both disjunctions contain the common literal $$\sim p.$$ To combine the two clauses, we use the distributive law of Boolean algebra, which says

$$ (A \vee B)\;\wedge\;(A \vee C)\;=\;A \;\vee\;(B \wedge C). $$

Here, we match the symbols as follows:

$$A = \sim p,\quad B = q,\quad C = \sim q.$$

Substituting into the distributive formula, we get

$$ (\sim p \vee q)\;\wedge\;(\sim p \vee \sim q)\;=\;\sim p\;\vee\;(q \wedge \sim q). $$

The expression $$q \wedge \sim q$$ is always false, because a statement and its negation can never be true at the same time. Hence

$$ q \wedge \sim q = \text{False}. $$

Therefore, our entire expression simplifies to

$$ \sim p \;\vee\; \text{False} \;=\; \sim p. $$

So the Boolean expression $$ (p \Rightarrow q) \wedge (q \Rightarrow \sim p) $$ is logically equivalent to $$\sim p.$$

Looking at the given options, $$\sim p$$ appears as Option D.

Hence, the correct answer is Option D.

The given statement is of the form "If $$P$$, then $$Q$$", where $$P$$ = "you will work" and $$Q$$ = "you will earn money".

The contrapositive of "If $$P$$, then $$Q$$" is "If not $$Q$$, then not $$P$$".

Substituting, the contrapositive is: "If you will not earn money, you will not work".

We need to identify which statement is a tautology (always true regardless of truth values of $$A$$ and $$B$$).

Option D is $$(A \wedge (A \to B)) \to B$$. This is the well-known rule of Modus Ponens.

Let us verify using a truth table. When $$A$$ is True and $$B$$ is True: $$A \to B$$ is True, so $$A \wedge (A \to B)$$ is True, and $$\text{True} \to \text{True}$$ is True.

When $$A$$ is True and $$B$$ is False: $$A \to B$$ is False, so $$A \wedge (A \to B)$$ is False, and $$\text{False} \to \text{False}$$ is True.

When $$A$$ is False and $$B$$ is True: $$A \to B$$ is True, so $$A \wedge (A \to B)$$ is False, and $$\text{False} \to \text{True}$$ is True.

When $$A$$ is False and $$B$$ is False: $$A \to B$$ is True, so $$A \wedge (A \to B)$$ is False, and $$\text{False} \to \text{False}$$ is True.

In all cases the result is True, confirming this is a tautology.

Hence, the correct answer is Option D.

We need to identify which Boolean expression is NOT a tautology. Recall that $$p \Rightarrow q \equiv \sim p \vee q$$.

Option A: $$(p \Rightarrow q) \vee (\sim q \Rightarrow p) \equiv (\sim p \vee q) \vee (q \vee p) \equiv (\sim p \vee p) \vee q \equiv T \vee q \equiv T$$. This is a tautology.

Option B: $$(q \Rightarrow p) \vee (\sim q \Rightarrow p) \equiv (\sim q \vee p) \vee (q \vee p) \equiv (\sim q \vee q) \vee p \equiv T \vee p \equiv T$$. This is a tautology.

Option C: $$(p \Rightarrow \sim q) \vee (\sim q \Rightarrow p) \equiv (\sim p \vee \sim q) \vee (q \vee p) \equiv (\sim p \vee p) \vee (\sim q \vee q) \equiv T \vee T \equiv T$$. This is a tautology.

Option D: $$(\sim p \Rightarrow q) \vee (\sim q \Rightarrow p) \equiv (p \vee q) \vee (q \vee p) \equiv p \vee q$$. When $$p = F$$ and $$q = F$$, this equals $$F$$. So this is NOT a tautology.

The answer is Option D.

We need to check which compound statement is a tautology. Option 2 is $$((P \Rightarrow Q) \wedge \sim Q) \Rightarrow \sim P$$. This is precisely the law of Modus Tollens, a well-known logical tautology: if $$P$$ implies $$Q$$ and $$Q$$ is false, then $$P$$ must be false.

To verify, suppose the antecedent $$(P \Rightarrow Q) \wedge \sim Q$$ is true. Then $$\sim Q$$ is true, so $$Q$$ is false. Since $$P \Rightarrow Q$$ is also true and $$Q$$ is false, $$P$$ must be false (if $$P$$ were true, $$P \Rightarrow Q$$ would be false). Hence $$\sim P$$ is true, and the implication holds. If the antecedent is false, the implication is vacuously true. Therefore the statement is always true, confirming it is a tautology.

We have to select the two connectives $$*$$ and $$\square$$ from the set $$\{\wedge ,\vee\}$$ in such a way that the statement

$$ (\,p * \sim q\,)\;\Rightarrow\;(p \;\square\; q) $$

is a tautology, that is, it must be true for every possible truth-value assignment of the propositional variables $$p$$ and $$q$$.

First recall the standard equivalence that rewrites an implication:

$$ a \Rightarrow b \;\equiv\; \lnot a \;\vee\; b. $$

Applying this formula to our expression we obtain

$$ (\,p * \sim q\,)\;\Rightarrow\;(p \square q) \;\equiv\; \lnot\,(p * \sim q) \;\vee\; (p \square q). $$

Because $$*$$ and $$\square$$ may each be either $$\wedge$$ (AND) or $$\vee$$ (OR), four distinct combinations are possible. We test them one by one, showing every algebraic step.

Case 1: $$*=\wedge,\;\square=\vee$$

Substituting these symbols gives

$$ \lnot\,(p \wedge \sim q) \;\vee\; (p \vee q). $$

We now push the negation inside the parenthesis with De Morgan’s law:

$$ \lnot (p \wedge \sim q) \;\equiv\; (\lnot p) \;\vee\; (\lnot\!\sim q) \;=\; (\lnot p) \;\vee\; q. $$

Therefore the whole expression becomes

$$ \big((\lnot p) \;\vee\; q\big) \;\vee\; (p \;\vee\; q). $$

Using associativity and commutativity of $$\vee$$ we group the disjuncts that involve $$p$$:

$$ \big((\lnot p) \;\vee\; p\big) \;\vee\; q. $$

Since a proposition OR its negation is always true, we have

$$ (\lnot p) \;\vee\; p \;=\; \text{T} \quad\text{(a tautology).} $$

Thus the entire expression simplifies to

$$ \text{T} \;\vee\; q \;=\; \text{T}. $$

So for $$*=\wedge,\;\square=\vee$$ the implication is always true; hence this choice makes the given statement a tautology.

Case 2: $$*=\wedge,\;\square=\wedge$$

The expression now is

$$ \lnot\,(p \wedge \sim q) \;\vee\; (p \wedge q) \;=\; (\lnot p \;\vee\; q) \;\vee\; (p \wedge q). $$

To check whether this is a tautology, take the assignment $$p=\text{T},\;q=\text{F}$$ (i.e. $$p=1,q=0$$).

Then we get

$$ \lnot p = 0,\quad q = 0,\quad p \wedge q = 1\wedge 0 = 0. $$

So the whole disjunction becomes $$0\;\vee\;0=0$$, which is false. Hence this case fails to be a tautology.

Case 3: $$*=\vee,\;\square=\vee$$

We have

$$ \lnot\,(p \vee \sim q) \;\vee\; (p \vee q). $$

Again by De Morgan,

$$ \lnot\,(p \vee \sim q) \;=\; (\lnot p) \wedge q. $$

Hence the whole statement is

$$ \big((\lnot p) \wedge q\big) \;\vee\; (p \vee q). $$

Choose $$p=\text{F},\;q=\text{F}$$. Then

$$ (\lnot p)\wedge q = 1\wedge 0 = 0,\quad p\vee q = 0\vee 0 = 0, $$

so the overall value is $$0\vee 0 = 0$$—not a tautology.

Case 4: $$*=\vee,\;\square=\wedge$$

The expression becomes

$$ \lnot\,(p \vee \sim q) \;\vee\; (p \wedge q) \;=\; ((\lnot p) \wedge q) \;\vee\; (p \wedge q). $$

With the assignment $$p=\text{F},\;q=\text{F}$$ we get

$$ (\lnot p)\wedge q = 1\wedge 0 = 0,\quad p\wedge q = 0\wedge 0 = 0, $$

so the disjunction is again $$0\vee 0 = 0$$. Hence this case also fails to be a tautology.

Among all four possibilities, only Case 1—namely $$*=\wedge$$ and $$\square=\vee$$—produces a statement that is true under every possible truth assignment. All other choices admit at least one counter-example, so they are not tautologies.

Hence, the correct answer is Option C.

We begin with the given statement of implication

$$ (p \vee r) \Rightarrow (q \vee r). $$

First, we recall the logical equivalence that an implication $$A \Rightarrow B$$ can be rewritten as $$\sim A \,\vee\, B.$$ Stating this clearly:

$$ A \Rightarrow B \;\; \text{is equivalent to} \;\; \sim A \vee B. $$

Here, the role of $$A$$ is played by $$(p \vee r)$$ and the role of $$B$$ is played by $$(q \vee r).$$ Applying the formula gives

$$ (p \vee r) \Rightarrow (q \vee r) \;\; \equiv \;\; \sim(p \vee r) \,\vee\, (q \vee r). $$

Now we want the negation of the entire implication. So we place a negation sign in front of the whole expression we just obtained:

$$ \sim\!\bigl[(p \vee r) \Rightarrow (q \vee r)\bigr] \;=\; \sim\!\bigl[\;\sim(p \vee r) \,\vee\, (q \vee r)\bigr]. $$

Next, we use De Morgan’s law, which states that the negation of a disjunction is the conjunction of the negations:

$$ \sim(A \vee B) \;\equiv\; \sim A \,\wedge\, \sim B. $$

Applying this law with $$A = \sim(p \vee r)$$ and $$B = (q \vee r),$$ we get

$$ \sim\!\bigl[\;\sim(p \vee r) \,\vee\, (q \vee r)\bigr] \;=\; \bigl[\;\sim\!\bigl(\sim(p \vee r)\bigr)\bigr] \,\wedge\, \bigl[\;\sim(q \vee r)\bigr]. $$

Simplify each part separately. First, note that the double negation law tells us $$\sim(\sim X) = X.$$ Therefore,

$$ \sim\!\bigl(\,\sim(p \vee r)\bigr) \;=\; (p \vee r). $$

For the second part, we again apply De Morgan’s law:

$$ \sim(q \vee r) \;=\; (\sim q) \,\wedge\, (\sim r). $$

Substituting these simplified pieces back, the negation becomes

$$ (p \vee r) \,\wedge\, \bigl[(\sim q) \wedge (\sim r)\bigr]. $$

Associativity of $$\wedge$$ allows us to drop parentheses inside the conjunction, giving

$$ (p \vee r) \,\wedge\, \sim q \,\wedge\, \sim r. $$

Now we must distribute $$(p \vee r)$$ over the other conjunctive factors. Using the distributive property $$ (X \vee Y)\,\wedge\,Z \;\equiv\; (X \wedge Z) \vee (Y \wedge Z), $$ with $$X = p,\; Y = r,\; Z = (\sim q \wedge \sim r),$$ we have

$$ (p \vee r) \,\wedge\, \sim q \,\wedge\, \sim r \;=\; \bigl[p \wedge (\sim q) \wedge (\sim r)\bigr] \;\vee\; \bigl[r \wedge (\sim q) \wedge (\sim r)\bigr]. $$

Observe that the second term contains the factor $$(r \wedge \sim r),$$ which is always false. Hence that entire term vanishes, leaving only

$$ p \,\wedge\, \sim q \,\wedge\, \sim r. $$

Thus, the negation of $$(p \vee r) \Rightarrow (q \vee r)$$ simplifies completely to

$$ p \wedge \sim q \wedge \sim r. $$

Comparing with the options provided, this matches Option C.

Hence, the correct answer is Option C.

First recall some basic logical equivalences that we will use repeatedly.

$$p \rightarrow q$$ is equivalent to $$\neg p \vee q$$   $$-(1)$$
De Morgan’s laws:   $$\neg(p \wedge q)=\neg p \vee \neg q$$   and   $$\neg(p \vee q)=\neg p \wedge \neg q$$.

We will analyse every option one by one and see whether the statement is always true (tautology) or not.

Case A:

Expression: $$ (p \wedge q) \vee (p \vee q) $$

The term $$p \vee q$$ already contains every situation covered by $$p \wedge q$$, so using the absorption law
$$ (p \wedge q) \vee (p \vee q)=p \vee q $$.
$$p \vee q$$ is not always true (for $$p=\text{False},\,q=\text{False}$$ it is False).
Hence Option A is NOT a tautology.

Case B:

Expression: $$ (p \wedge q) \vee (p \rightarrow q) $$

Using $$-(1)$$, rewrite:
$$ (p \wedge q) \vee (\neg p \vee q) $$.
Group the disjunctions:
$$ =(\neg p \vee q) \vee (p \wedge q) $$.
Construct a truth-table for the three literals $$\neg p,\,q,\,p \wedge q$$ (or test the four combinations of $$p,q$$):

 • $$p=\text{T},\,q=\text{T}$$: expression T
 • $$p=\text{T},\,q=\text{F}$$: $$\neg p \vee q=\text{F},\,p\wedge q=\text{F}$$ ⇒ overall F
As it becomes False, the statement is not a tautology. Hence Option B is NOT a tautology.

Case C:

Expression: $$ (p \wedge q) \wedge (p \rightarrow q) $$

Again substitute $$p \rightarrow q=\neg p \vee q$$:
$$ (p \wedge q) \wedge (\neg p \vee q) $$.
If $$p=\text{F},\;q=\text{T}$$ then the first part $$p \wedge q=\text{F}$$, so whole expression is False. Therefore it is NOT a tautology.

Case D:

Expression: $$ (p \wedge q) \rightarrow (p \rightarrow q) $$

First convert the implication using $$-(1)$$:
$$ (p \wedge q) \rightarrow (p \rightarrow q)=\neg(p \wedge q) \vee (p \rightarrow q) $$

Apply De Morgan’s law on the first term and $$-(1)$$ on the second term:
$$ =(\neg p \vee \neg q) \vee (\neg p \vee q) $$

Associativity and commutativity of $$\vee$$ let us group the literals:
$$ =\neg p \vee \neg q \vee q $$

But $$\neg q \vee q$$ is always True, so

$$\neg p \vee (\neg q \vee q)=\neg p \vee \text{True}=\text{True}$$

Since the expression reduces to True for every choice of $$p$$ and $$q$$, it is a tautology.

Therefore, only Option D $$\big((p \wedge q) \rightarrow (p \rightarrow q)\big)$$ is a tautology.

Final Answer: Option D.

We have to find an option whose logical value is always identical to the expression $$p \wedge \sim q$$. The expression $$p \wedge \sim q$$ is true exactly when $$p$$ is true and $$q$$ is false, and false in every other situation. We shall translate each option into a form that can be compared directly with this conjunction.

First, we recall and state the fundamental implication law: for any statements $$x$$ and $$y$$,

$$x \rightarrow y \;\equiv\; \sim x \;\vee\; y.$$

Using this single law and ordinary De Morgan simplifications $$\sim(x \vee y)=\sim x \wedge \sim y$$ and $$\sim(x \wedge y)=\sim x \vee \sim y,$$ we convert every option step by step.

Option A is $$\sim p \rightarrow \sim q$$. Applying the implication law with $$x=\sim p$$ and $$y=\sim q$$ gives

$$\sim p \rightarrow \sim q \;\equiv\; \sim(\sim p) \;\vee\; \sim q.$$

Now $$\sim(\sim p)=p$$, so

$$\sim p \rightarrow \sim q \;\equiv\; p \;\vee\; \sim q.$$

Thus Option A simplifies to $$p \vee \sim q$$, a disjunction, not the required conjunction $$p \wedge \sim q$$. Hence Option A is not equivalent.

Option B is $$\sim\bigl(q \rightarrow p\bigr)$$. First write the inner implication:

$$q \rightarrow p \;\equiv\; \sim q \;\vee\; p.$$

Substituting, we get

$$\sim\bigl(q \rightarrow p\bigr) \;=\; \sim\bigl(\sim q \;\vee\; p\bigr).$$

Using De Morgan’s law for negation of a disjunction,

$$\sim(\sim q \;\vee\; p) \;=\; \sim(\sim q) \;\wedge\; \sim p.$$

Now $$\sim(\sim q)=q,$$ so

$$\sim\bigl(q \rightarrow p\bigr) \;\equiv\; q \;\wedge\; \sim p.$$

This is the conjunction $$q \wedge \sim p,$$ which is different from $$p \wedge \sim q$$. Therefore Option B is also not equivalent.

Option C is $$\sim\bigl(p \rightarrow q\bigr).$$ Again we expand the implication:

$$p \rightarrow q \;\equiv\; \sim p \;\vee\; q.$$

Substituting gives

$$\sim\bigl(p \rightarrow q\bigr) \;=\; \sim\bigl(\sim p \;\vee\; q\bigr).$$

Using De Morgan’s law,

$$\sim(\sim p \;\vee\; q) \;=\; \sim(\sim p) \;\wedge\; \sim q.$$

Simplifying $$\sim(\sim p)=p,$$ we obtain

$$\sim\bigl(p \rightarrow q\bigr) \;\equiv\; p \;\wedge\; \sim q.$$

This is exactly the original expression we wanted. So Option C matches $$p \wedge \sim q$$ perfectly.

Option D is $$\sim\bigl(p \rightarrow \sim q\bigr).$$ Start with the implication:

$$p \rightarrow \sim q \;\equiv\; \sim p \;\vee\; \sim q.$$

Negating it,

$$\sim\bigl(p \rightarrow \sim q\bigr) \;=\; \sim\bigl(\sim p \;\vee\; \sim q\bigr).$$

Applying De Morgan’s law,

$$\sim(\sim p \;\vee\; \sim q) \;=\; \sim(\sim p) \;\wedge\; \sim(\sim q).$$

This simplifies to

$$p \;\wedge\; q.$$

Since $$p \wedge q$$ differs from $$p \wedge \sim q,$$ Option D is not equivalent.

Among all four options, only Option C reduces to $$p \wedge \sim q$$. Hence, the correct answer is Option 3.

We are given that $$(p \Rightarrow q) \Leftrightarrow (q * (\sim p))$$ is a tautology, and we need to find what $$p * (\sim q)$$ is equivalent to.

Recall the standard logical equivalence: $$p \Rightarrow q \equiv \sim p \lor q$$. This means "if p then q" is the same as "not-p or q".

For the biconditional $$(p \Rightarrow q) \Leftrightarrow (q * (\sim p))$$ to be a tautology, both sides must always have the same truth value. So we need $$\sim p \lor q \equiv q * (\sim p)$$ for every combination of truth values of $$p$$ and $$q$$.

Let us substitute $$a = \sim p$$ to simplify. Then we need $$a \lor q \equiv q * a$$ for all truth values of $$a$$ and $$q$$. Since $$a$$ and $$q$$ range over all truth values independently (as $$p$$ and $$q$$ do), this means the operation $$*$$ must be exactly the same as $$\lor$$ (OR).

To verify with a truth table: when $$a = T, q = T$$: $$a \lor q = T$$ so $$q * a = T$$. When $$a = T, q = F$$: $$a \lor q = T$$ so $$q * a = T$$. When $$a = F, q = T$$: $$a \lor q = T$$ so $$q * a = T$$. When $$a = F, q = F$$: $$a \lor q = F$$ so $$q * a = F$$. This matches the OR truth table exactly.

Now we evaluate $$p * (\sim q)$$. Since $$*$$ is $$\lor$$, we get $$p * (\sim q) = p \lor (\sim q)$$.

Using the equivalence $$q \Rightarrow p \equiv \sim q \lor p = p \lor (\sim q)$$, we see that $$p * (\sim q) \equiv q \Rightarrow p$$.

This matches Option A: $$q \Rightarrow p$$.

The function $$f : A \to A$$ is defined as $$f(k) = k + 1$$ if $$k$$ is odd, and $$f(k) = k$$ if $$k$$ is even. So: $$f(1) = 2$$, $$f(2) = 2$$, $$f(3) = 4$$, $$f(4) = 4$$, $$f(5) = 6$$, $$f(6) = 6$$, $$f(7) = 8$$, $$f(8) = 8$$, $$f(9) = 10$$, $$f(10) = 10$$.

The range of $$f$$ is $$\{2, 4, 6, 8, 10\}$$. We need $$g \circ f = f$$, i.e., $$g(f(k)) = f(k)$$ for all $$k \in A$$.

This means $$g(2) = 2$$, $$g(4) = 4$$, $$g(6) = 6$$, $$g(8) = 8$$, and $$g(10) = 10$$. These five values of $$g$$ are completely determined.

However, the values $$g(1)$$, $$g(3)$$, $$g(5)$$, $$g(7)$$, and $$g(9)$$ are not constrained by the condition $$g \circ f = f$$ (since 1, 3, 5, 7, 9 are not in the range of $$f$$). Each of these can be any element of $$A = \{1, 2, \ldots, 10\}$$.

Therefore the number of possible functions $$g$$ is $$10 \times 10 \times 10 \times 10 \times 10 = 10^5$$.

We have to inspect the compound statement $$\bigl(p \wedge (p \rightarrow q) \wedge (q \rightarrow r)\bigr) \rightarrow r$$ and decide whether it is always true (a tautology), always false (a fallacy), or equivalent to some simpler implication.

First of all, recall the standard equivalence for an implication:

$$x \rightarrow y \;\;\text{is logically equal to}\;\; \sim x \vee y.$$

Using this rule, we rewrite each implication inside the larger expression.

$$p \rightarrow q \equiv \sim p \vee q,$$

$$q \rightarrow r \equiv \sim q \vee r.$$

Substituting these into the original statement gives

$$\Bigl(p \wedge (\sim p \vee q) \wedge (\sim q \vee r)\Bigr) \rightarrow r.$$

Let us now simplify the conjunction in the antecedent, step by step.

First combine the first two factors:

$$p \wedge (\sim p \vee q) = (p \wedge \sim p) \vee (p \wedge q).$$

But $$p \wedge \sim p$$ is a contradiction, hence equal to false. So

$$(p \wedge \sim p) \vee (p \wedge q) = \text{false} \vee (p \wedge q) = p \wedge q.$$

Thus the antecedent has now become

$$(p \wedge q) \wedge (\sim q \vee r).$$

We keep going. Group the last two factors:

$$q \wedge (\sim q \vee r) = (q \wedge \sim q) \vee (q \wedge r).$$

Again $$q \wedge \sim q$$ is a contradiction, so this reduces to

$$(q \wedge \sim q) \vee (q \wedge r) = \text{false} \vee (q \wedge r) = q \wedge r.$$

Hence the whole antecedent simplifies neatly to

$$p \wedge (q \wedge r) = p \wedge q \wedge r.$$

The complete statement is therefore

$$(p \wedge q \wedge r) \rightarrow r.$$

Apply the implication equivalence once more:

$$(p \wedge q \wedge r) \rightarrow r \equiv \sim(p \wedge q \wedge r) \vee r.$$

By De Morgan’s law,

$$\sim(p \wedge q \wedge r) = \sim p \vee \sim q \vee \sim r.$$

So the entire disjunction becomes

$$\bigl(\sim p \vee \sim q \vee \sim r\bigr) \vee r.$$

Now observe that $$\sim r \vee r$$ is a tautology (it is always true). Because that tautology is one of the disjuncts, the whole expression is invariably true, no matter what truth-values $$p, q,$$ and $$r$$ may take.

Therefore the original statement is always true; that is, it is a tautology.

Hence, the correct answer is Option A.

We need to find the connectives $$\circledast$$ and $$\otimes$$ such that $$(p \wedge q) \circledast (p \otimes q)$$ is a tautology.

Let us check Option A: $$\circledast$$ is $$\to$$ and $$\otimes$$ is $$\to$$. The expression becomes $$(p \wedge q) \to (p \to q)$$.

Recall that an implication $$A \to B$$ is false only when $$A$$ is true and $$B$$ is false. So $$(p \wedge q) \to (p \to q)$$ is false only when $$(p \wedge q)$$ is true and $$(p \to q)$$ is false.

If $$(p \wedge q)$$ is true, then both $$p$$ and $$q$$ are true. But when $$p$$ is true and $$q$$ is true, $$(p \to q)$$ is also true. So the antecedent being true forces the consequent to be true as well.

Therefore $$(p \wedge q) \to (p \to q)$$ can never be false, making it a tautology.

Let us verify the other options fail. For Option B ($$\circledast = \wedge$$, $$\otimes = \vee$$): the expression is $$(p \wedge q) \wedge (p \vee q)$$. When $$p = T, q = F$$: $$(T \wedge F) \wedge (T \vee F) = F \wedge T = F$$. Not a tautology.

For Option C ($$\circledast = \vee$$, $$\otimes = \to$$): the expression is $$(p \wedge q) \vee (p \to q)$$. When $$p = T, q = F$$: $$(T \wedge F) \vee (T \to F) = F \vee F = F$$. Not a tautology.

For Option D ($$\circledast = \wedge$$, $$\otimes = \to$$): the expression is $$(p \wedge q) \wedge (p \to q)$$. When $$p = F, q = F$$: $$(F \wedge F) \wedge (F \to F) = F \wedge T = F$$. Not a tautology.

The answer is $$\to, \to$$, which is Option A.

We have $$f(x) = 2x - 1$$ and $$g(x) = \frac{x - \frac{1}{2}}{x - 1}$$, with domain of $$g$$ being $$\mathbb{R} - \{1\}$$.

The composition is $$f(g(x)) = 2 \cdot \frac{x - \frac{1}{2}}{x - 1} - 1 = \frac{2x - 1}{x - 1} - 1 = \frac{2x - 1 - (x - 1)}{x - 1} = \frac{x}{x - 1}$$.

The domain of $$f(g(x))$$ is $$\mathbb{R} - \{1\}$$.

To check one-one: suppose $$\frac{x_1}{x_1 - 1} = \frac{x_2}{x_2 - 1}$$. Then $$x_1(x_2 - 1) = x_2(x_1 - 1)$$, giving $$x_1 x_2 - x_1 = x_1 x_2 - x_2$$, so $$x_1 = x_2$$. The function is one-one.

To check onto: let $$\frac{x}{x - 1} = y$$. Then $$x = y(x - 1) = yx - y$$, so $$x(1 - y) = -y$$, giving $$x = \frac{y}{y - 1}$$. This requires $$y \neq 1$$. So the range is $$\mathbb{R} - \{1\}$$, which is not all of $$\mathbb{R}$$.

Therefore $$f(g(x))$$ is one-one but not onto.

Hence, the correct answer is Option B.

We have $$g(x) = \dfrac{x^2 - x - 2}{2x^2 - x - 6} = \dfrac{(x-2)(x+1)}{(2x+3)(x-2)}$$. For $$x \neq 2$$, this simplifies to $$g(x) = \dfrac{x + 1}{2x + 3}$$.

Since $$g(2) = \displaystyle\lim_{x \to 2} g(x) = \dfrac{3}{7}$$, the function $$g$$ is defined as $$g(x) = \dfrac{x+1}{2x+3}$$ for all $$x \neq -\dfrac{3}{2}$$.

For $$f \circ g$$ to be defined, we need $$-1 \leq g(x) \leq 1$$ (the domain of $$\sin^{-1}$$).

Condition 1: $$g(x) \geq -1$$, i.e., $$\dfrac{x+1}{2x+3} \geq -1$$. This gives $$\dfrac{x + 1 + 2x + 3}{2x+3} = \dfrac{3x + 4}{2x + 3} \geq 0$$. By sign analysis, this holds when $$x \leq -\dfrac{3}{2}$$ or $$x \geq -\dfrac{4}{3}$$.

Condition 2: $$g(x) \leq 1$$, i.e., $$\dfrac{x+1}{2x+3} \leq 1$$. This gives $$\dfrac{x + 1 - 2x - 3}{2x + 3} = \dfrac{-(x+2)}{2x+3} \leq 0$$, equivalently $$\dfrac{x + 2}{2x + 3} \geq 0$$. By sign analysis, this holds when $$x \leq -2$$ or $$x > -\dfrac{3}{2}$$.

Taking the intersection (and excluding $$x = -\dfrac{3}{2}$$ where $$g$$ is undefined): $$\left(x \leq -\dfrac{3}{2} \text{ or } x \geq -\dfrac{4}{3}\right) \cap \left(x \leq -2 \text{ or } x > -\dfrac{3}{2}\right) = (-\infty, -2] \cup \left[-\dfrac{4}{3}, \infty\right)$$.

Therefore the domain of $$f \circ g$$ is $$\left(-\infty, -2\right] \cup \left[-\dfrac{4}{3}, \infty\right)$$.

We begin by rewriting the definition of the given function in the language of congruence-classes because this will make every later step completely transparent. For any natural number $$x\in \mathbb N$$ we can write $$x=3n+1,\;3n+2$$ or $$3n+3$$ for a unique $$n\ge 0$$. Using this unique representation we have

So, whenever $$x\equiv 1\pmod 3$$ or $$x\equiv 2\pmod 3$$, the value of $$g$$ is simply $$x+1$$, while for $$x\equiv 0\pmod 3$$ the value becomes $$x-2$$. Placing the three cases side by side we see that every set

forms a closed cycle under $$g$$:

Because the cycle length is exactly three, applying $$g$$ three times brings us back to the starting point:

In compact notation this is written as $$g^3 = \operatorname{id}_{\mathbb N}$$. Therefore $$g^3$$ is the identity, while $$g$$ itself is not the identity (no natural number is fixed). We will now test each option one after another.

Testing Option A. We must find a surjective (onto) function $$f:\mathbb N\to\mathbb N$$ satisfying

Let us first analyse the algebraic condition. Substituting the three possible forms of $$x$$ we get

Combining the three equalities we obtain the single rule

In words, each entire 3-cycle must be sent to one and the same value. But we are still free to choose which value that is, for every individual cycle. A very convenient (and classical) choice is

Now we check surjectivity. Given any natural number $$k\in\mathbb N$$, set $$n=k-1$$ (allowed because $$k\ge 1$$). Then $$f(3n+1)=k$$, so $$k$$ lies in the image. Thus the function hits every natural number, i.e. it is onto. Because the function also satisfies the required equality, Option A is true.

Testing Option B. The same algebraic argument used above tells us that any $$f$$ with $$f\circ g=f$$ must be constant on each 3-cycle. Hence there will always be at least three distinct inputs having the same output, immediately contradicting the definition of a one-one (injective) function. Therefore Option B is false.

Testing Option C. We have already observed that $$g^3=\operatorname{id}_{\mathbb N}$$, so

because no element is fixed by $$g$$. Hence $$gogog=g$$ is incorrect, making Option C false.

Testing Option D. Here we seek an $$f$$ such that $$g\circ f=f$$. Writing this out we have

This means every value taken by $$f$$ must be a fixed point of $$g$$. Yet we saw earlier that $$g$$ possesses no fixed point at all. Consequently no such $$f$$ can exist, and Option D is also false.

Only Option A survives every check. Hence, the correct answer is Option A.

We need to solve $$[e^x]^2 + [e^x + 1] - 3 = 0$$, where $$[y]$$ is the greatest integer (floor) function.

Using the property $$[y + n] = [y] + n$$ for any integer $$n$$: $$[e^x + 1] = [e^x] + 1$$.

The equation becomes $$[e^x]^2 + [e^x] + 1 - 3 = 0$$, i.e., $$[e^x]^2 + [e^x] - 2 = 0$$.

Let $$t = [e^x]$$. Then $$t^2 + t - 2 = 0$$, so $$(t+2)(t-1) = 0$$, giving $$t = 1$$ or $$t = -2$$.

Since $$e^x > 0$$ for all real $$x$$, we have $$[e^x] \geq 0$$, so $$t = -2$$ is rejected.

Thus $$[e^x] = 1$$, meaning $$1 \leq e^x < 2$$.

Taking natural logarithm: $$0 \leq x < \log_e 2$$.

The values of $$x$$ lie in $$[0, \log_e 2)$$, which is Option D.

We need to determine which statement is logically equivalent to $$A \to (B \to A)$$.

First, we simplify $$A \to (B \to A)$$. Since $$B \to A \equiv \neg B \vee A$$, we have $$A \to (B \to A) \equiv \neg A \vee (\neg B \vee A) \equiv (\neg A \vee A) \vee \neg B \equiv \text{True}$$.

So $$A \to (B \to A)$$ is a tautology (always true regardless of the truth values of $$A$$ and $$B$$).

Now we check the options. For $$A \to (A \vee B)$$: this equals $$\neg A \vee (A \vee B) \equiv (\neg A \vee A) \vee B \equiv \text{True}$$. This is also a tautology.

Since both $$A \to (B \to A)$$ and $$A \to (A \vee B)$$ are tautologies, they are logically equivalent (both are always true).

Therefore, $$A \to (B \to A)$$ is equivalent to $$A \to (A \vee B)$$.

We need to find the domain of $$f(x) = \frac{\cos^{-1}\sqrt{x^2 - x + 1}}{\sqrt{\sin^{-1}\left(\frac{2x-1}{2}\right)}}$$.

Condition 1 (square root inside cos inverse): We need $$x^2 - x + 1 \geq 0$$. The discriminant is $$1 - 4 = -3 < 0$$ with positive leading coefficient, so this is always positive. Always satisfied.

Condition 2 (argument of cos inverse in $$[-1,1]$$): Since $$\sqrt{x^2 - x + 1} \geq 0$$, we need $$\sqrt{x^2 - x + 1} \leq 1$$, i.e., $$x^2 - x + 1 \leq 1$$, so $$x^2 - x \leq 0$$, giving $$x(x-1) \leq 0$$, i.e., $$0 \leq x \leq 1$$.

Condition 3 (argument of sin inverse in $$[-1,1]$$): $$-1 \leq \frac{2x-1}{2} \leq 1$$ gives $$-\frac{1}{2} \leq x \leq \frac{3}{2}$$.

Condition 4 (denominator positive): $$\sin^{-1}\left(\frac{2x-1}{2}\right) > 0$$. Since $$\sin^{-1}(t) > 0$$ iff $$t > 0$$, we need $$\frac{2x-1}{2} > 0$$, i.e., $$x > \frac{1}{2}$$.

Intersecting all conditions: $$0 \leq x \leq 1$$ and $$x > \frac{1}{2}$$ gives $$\frac{1}{2} < x \leq 1$$, i.e., the interval $$\left(\frac{1}{2}, 1\right]$$.

So $$\alpha = \frac{1}{2}$$ and $$\beta = 1$$, giving $$\alpha + \beta = \frac{3}{2}$$.

The answer is $$\frac{3}{2}$$, which is Option A.

We are given a function $$f : \mathbb N \rightarrow \mathbb N$$ satisfying the additive property

$$f(m+n)=f(m)+f(n) \quad \text{for all } m,n\in\mathbb N.$$

First, we recall the standard fact that if a function on the natural numbers is additive in this way, then its value at any positive integer can be built up by repeatedly adding the value at $$1$$. More precisely, we use the principle of mathematical induction:

For $$n=1$$ the statement $$f(1)=f(1)$$ is trivially true. Now assume $$f(k)=k\,f(1)$$ for some $$k\in\mathbb N$$. Then, using the given property,

$$f(k+1)=f(k)+f(1).$$

Substituting the induction hypothesis $$f(k)=k\,f(1)$$, we obtain

$$f(k+1)=k\,f(1)+f(1)=(k+1)\,f(1).$$

Thus by induction the general formula

$$f(n)=n\,f(1) \quad\text{holds for every } n\in\mathbb N.$$

Now we use the given numerical information. We are told that

$$f(6)=18.$$

Applying the formula $$f(6)=6\,f(1)$$ we write

$$6\,f(1)=18.$$

Dividing both sides by $$6$$, we get

$$f(1)=\frac{18}{6}=3.$$

With $$f(1)$$ known, we can find the required values:

For $$n=2$$,

$$f(2)=2\,f(1)=2\times 3=6.$$

For $$n=3$$,

$$f(3)=3\,f(1)=3\times 3=9.$$

The question asks for the product $$f(2)\cdot f(3)$$. Substituting the values just found,

$$f(2)\cdot f(3)=6\times 9=54.$$

Hence, the correct answer is Option A.

We have $$f(x) = \frac{5x + 3}{6x - \alpha}$$. We need $$(f \circ f)(x) = x$$ for all $$x \in R \setminus \{\frac{\alpha}{6}\}$$.

Computing $$f(f(x))$$ directly: let $$u = f(x) = \frac{5x+3}{6x-\alpha}$$, then $$f(u) = \frac{5u + 3}{6u - \alpha} = \frac{5\cdot\frac{5x+3}{6x-\alpha} + 3}{6\cdot\frac{5x+3}{6x-\alpha} - \alpha}.$$

Multiplying numerator and denominator by $$(6x - \alpha)$$: $$f(f(x)) = \frac{5(5x+3) + 3(6x-\alpha)}{6(5x+3) - \alpha(6x-\alpha)} = \frac{25x + 15 + 18x - 3\alpha}{30x + 18 - 6\alpha x + \alpha^2}.$$

For this to equal $$x$$ for all valid $$x$$, we need $$\frac{(25+18)x + (15-3\alpha)}{(30-6\alpha)x + (18+\alpha^2)} = x$$, i.e., $$(43)x + (15-3\alpha) = x\left[(30-6\alpha)x + (18+\alpha^2)\right].$$

For this to hold as an identity, the coefficient of $$x^2$$ on the right must be zero: $$30 - 6\alpha = 0$$, but this gives $$\alpha = 5$$. Let us verify: with $$\alpha = 5$$, the numerator becomes $$(25+18)x + (15-15) = 43x$$ and the denominator becomes $$(30-30)x + (18+25) = 43$$, so $$f(f(x)) = \frac{43x}{43} = x$$. ✓

Therefore $$\alpha = 5$$.

The problem defines $$[x]$$ as the greatest integer function. The function under consideration is $$f(x) = \sqrt{\dfrac{|[x]|-2}{|[x]|-3}}$$, which requires $$\dfrac{|[x]|-2}{|[x]|-3} \ge 0$$ with $$|[x]| \ne 3$$.

Setting $$n = [x]$$ (an integer), the condition $$\dfrac{|n|-2}{|n|-3} \ge 0$$ holds when $$|n| \le 2$$ or $$|n| \ge 4$$ (excluding $$|n| = 3$$).

We translate each integer condition back to $$x$$, using $$[x] = n \iff x \in [n, n+1)$$.

For $$|n| \le 2$$: $$n \in \{-2,-1,0,1,2\}$$, so $$x \in [-2,-1) \cup [-1,0) \cup [0,1) \cup [1,2) \cup [2,3) = [-2, 3)$$.

For $$|n| = 3$$ (excluded): $$n = -3$$ gives $$x \in [-3,-2)$$, and $$n = 3$$ gives $$x \in [3,4)$$. These are excluded from the domain.

For $$|n| \ge 4$$: $$n \le -4$$ gives $$x \in (-\infty,-3)$$, and $$n \ge 4$$ gives $$x \in [4,\infty)$$.

Therefore the domain is $$(-\infty,-3) \cup [-2,3) \cup [4,\infty)$$.

Matching with the given form $$(-\infty, a) \cup [b,c) \cup [4,\infty)$$: we identify $$a = -3$$, $$b = -2$$, $$c = 3$$.

Hence $$a + b + c = -3 + (-2) + 3 = -2$$.

Given $$f(n + 1) = f(n) + f(1)$$ for all $$n \in \mathbb{N}$$, we can deduce that $$f(n) = n \cdot f(1)$$ for all $$n \in \mathbb{N}$$ (by induction). Let $$f(1) = k$$ where $$k \in \mathbb{N}$$, so $$f(n) = kn$$.

Checking Option A: If $$f$$ is onto, then every natural number must be in the range. Since $$f(n) = kn$$, the range is $$\{k, 2k, 3k, \ldots\}$$. For this to equal $$\mathbb{N}$$, we need $$k = 1$$, so $$f(n) = n$$. This statement is TRUE.

Checking Option C: $$f$$ is one-one. Since $$f(n) = kn$$ and $$k \geq 1$$, if $$f(m) = f(n)$$ then $$km = kn$$ so $$m = n$$. This statement is TRUE.

Checking Option D: If $$f \circ g$$ is one-one, then $$g$$ is one-one. If $$g(m) = g(n)$$, then $$f(g(m)) = f(g(n))$$, and since $$f \circ g$$ is one-one, we must have $$m = n$$. This statement is TRUE.

Checking Option B: If $$g$$ is onto, then $$f \circ g$$ is one-one. Consider the counterexample where $$k = 1$$ (so $$f(n) = n$$) and define $$g$$ as: $$g(1) = g(2) = 1$$ and $$g(n) = n - 1$$ for $$n \geq 3$$. Then $$g$$ is onto (every natural number is in the range), but $$f(g(1)) = f(1) = 1 = f(1) = f(g(2))$$ while $$1 \neq 2$$, so $$f \circ g$$ is NOT one-one. This statement is NOT TRUE.

Therefore, the statement that is NOT true is: if $$g$$ is onto, then $$f \circ g$$ is one-one.

We have to evaluate the finite sum

$$\sum_{n=8}^{100}\left[\frac{(-1)^n\,n}{2}\right]$$

where $$[x]$$ denotes the greatest integer less than or equal to $$x$$ (the “floor” of $$x$$).

First we separate the index $$n$$ into the two possible parities because the factor $$(-1)^n$$ behaves differently for even and odd indices.

For an even integer $$n$$ we can write $$n=2k$$. Then $$(-1)^n = 1$$, so

$$\frac{(-1)^n\,n}{2}= \frac{1\cdot 2k}{2}=k,$$ and since $$k$$ is already an integer, its greatest-integer value is

$$\left[\frac{(-1)^n\,n}{2}\right]=k=\frac{n}{2}.$$

For an odd integer $$n$$ we can write $$n=2k+1$$. Then $$(-1)^n = -1$$, so

$$\frac{(-1)^n\,n}{2}= \frac{-1\cdot(2k+1)}{2}= -\left(k+\tfrac12\right).$$

The number $$-\!\left(k+\tfrac12\right)$$ lies strictly between the two consecutive integers $$-(k+1)$$ and $$-k$$ and is less than both of them, hence its greatest-integer value is

$$\left[\frac{(-1)^n\,n}{2}\right]=-(k+1)= -\frac{n+1}{2}.$$

So, term by term,

$$ \left[\frac{(-1)^n\,n}{2}\right]= \begin{cases} \dfrac{n}{2}, & n \text{ even},\\[6pt] -\dfrac{n+1}{2}, & n \text{ odd}. \end{cases} $$

Now we split the required sum into its even and odd parts:

$$ \sum_{n=8}^{100}\Bigl[\tfrac{(-1)^n n}{2}\Bigr] =\sum_{\substack{n=8\\ n\text{ even}}}^{100}\frac{n}{2} +\sum_{\substack{n=8\\ n\text{ odd}}}^{100}\!\!\!\!\left(-\frac{n+1}{2}\right). $$

The even values of $$n$$ run from 8 to 100: 8, 10, 12, …, 100.

There are

$$\frac{100-8}{2}+1=\frac{92}{2}+1=46+1=47$$

such even integers. For each of them we take half, giving the consecutive integers

$$4,\,5,\,6,\ldots,\,50$$

(because $$8/2=4$$ and $$100/2=50$$). Their sum is obtained by the formula for an arithmetic progression:

$$ \sum_{\text{even }n}\frac{n}{2} =\frac{\text{number of terms}}{2}\,( \text{first term}+\text{last term}) =\frac{47}{2}\,(4+50) =\frac{47}{2}\times54 =47\times27 =1269. $$

The odd values of $$n$$ run from 9 to 99:

9, 11, 13, …, 99.

There are 46 of them, because the total count 93 minus the 47 evens gives 46 odds. For each odd $$n$$ we substitute $$-\dfrac{n+1}{2}$$. Now $$(n+1)/2$$ for these odds is the consecutive list

$$5,\,6,\,7,\ldots,\,50,$$

and taking the negative gives

$$-5,\,-6,\,-7,\ldots,\,-50.$$

Again using the arithmetic-progression formula, the sum of these 46 integers is

$$ \sum_{\text{odd }n}\left(-\frac{n+1}{2}\right) =-\,\frac{46}{2}\,(5+50) =-\,23\times55 =-1265. $$

Combining the even and odd contributions:

$$ \sum_{n=8}^{100}\Bigl[\tfrac{(-1)^n n}{2}\Bigr] =1269 + (-1265)=4. $$

Hence, the correct answer is Option B.

Given $$f(x) = \frac{x-2}{x-3}$$, we find $$f^{-1}$$. Setting $$y = \frac{x-2}{x-3}$$, we solve for $$x$$: $$y(x-3) = x - 2$$, so $$yx - 3y = x - 2$$, giving $$x(y-1) = 3y - 2$$, hence $$x = \frac{3y - 2}{y - 1}$$. Therefore $$f^{-1}(x) = \frac{3x - 2}{x - 1}$$.

Given $$g(x) = 2x - 3$$, we find $$g^{-1}(x) = \frac{x + 3}{2}$$.

The equation $$f^{-1}(x) + g^{-1}(x) = \frac{13}{2}$$ becomes $$\frac{3x-2}{x-1} + \frac{x+3}{2} = \frac{13}{2}$$. Multiplying through by $$2(x-1)$$: $$2(3x-2) + (x+3)(x-1) = 13(x-1)$$. Expanding: $$6x - 4 + x^2 + 2x - 3 = 13x - 13$$. This simplifies to $$x^2 + 8x - 7 = 13x - 13$$, so $$x^2 - 5x + 6 = 0$$, which factors as $$(x-2)(x-3) = 0$$.

Thus $$x = 2$$ or $$x = 3$$. Both values are in the domain of $$f^{-1}$$ (since $$x \neq 1$$). The sum of all values is $$2 + 3 = 5$$.

We begin with the information that $$f:A\rightarrow B$$ and $$g:B\rightarrow C$$ are two real-valued functions such that the composite function $$g\circ f:A\rightarrow C$$ possesses an inverse, written $$(g\circ f)^{-1}$$.

A function admits an inverse if and only if it is bijective; that is, it must be simultaneously one-one (injective) and onto (surjective). Hence we may immediately state

$$g\circ f \text{ is one-one and onto.}$$

We now translate the bijectivity of $$g\circ f$$ into separate conditions on $$f$$ and $$g$$.

Injectivity part: To prove that $$f$$ is one-one, let us take any two elements $$x_{1},x_{2}\in A$$ and assume that their images under $$f$$ coincide, i.e.

$$f(x_{1})=f(x_{2}).$$

Applying $$g$$ on both sides we obtain

$$g\!\left(f(x_{1})\right)=g\!\left(f(x_{2})\right) \;\Longrightarrow\; (g\circ f)(x_{1})=(g\circ f)(x_{2}).$$

Since $$g\circ f$$ is injective, equality of the outputs forces equality of the inputs, so

$$x_{1}=x_{2}.$$

This derivation shows that no two distinct elements of $$A$$ can share the same image under $$f$$; therefore $$f$$ is one-one (injective).

Surjectivity part: To show that $$g$$ is onto, start with an arbitrary element $$c\in C$$. Because $$g\circ f$$ is surjective, there exists some $$a\in A$$ satisfying

$$(g\circ f)(a)=c.$$

Writing the composition explicitly, this reads

$$g\!\bigl(f(a)\bigr)=c.$$

Denote $$b=f(a)\in B$$; then the equation becomes $$g(b)=c$$. We have thus produced an element $$b\in B$$ whose image under $$g$$ is the pre-selected $$c$$, verifying that every element of $$C$$ is attained by $$g$$. Consequently, $$g$$ is onto (surjective).

Putting the two deductions together, we have established that

$$f \text{ is one-one and } g \text{ is onto.}$$

None of the other combinations is compelled by the mere bijectivity of $$g\circ f$$. Hence, the correct answer is Option C.

We need the inverse function of $$y = 5^{\log x}$$.

Taking logarithm (base 10) on both sides: $$\log y = \log x \cdot \log 5$$.

Solving for $$\log x$$: $$\log x = \frac{\log y}{\log 5}$$.

So $$x = 10^{\frac{\log y}{\log 5}}$$.

We can write this as $$x = \left(10^{\log y}\right)^{1/\log 5} = y^{1/\log 5}$$, since $$10^{\log y} = y$$.

For the inverse function, we swap $$x$$ and $$y$$: $$y = x^{1/\log 5}$$.

This matches Option C: $$y = x^{\frac{1}{\log 5}}$$.

We are given the set $$S=\{1,2,3,4,5,6,7\}$$ and we must count all functions $$f:S\to S$$ that satisfy the multiplicative rule

$$f(m\cdot n)=f(m)\cdot f(n)$$

for every pair $$m,n\in S$$ for which the product $$m\cdot n$$ is still inside $$S$$. We begin by looking at the special element $$1$$.

Taking $$m=1$$ in the defining property gives the formula

$$f(1\cdot n)=f(1)\cdot f(n).$$

Because $$1\cdot n=n$$, this simplifies to

$$f(n)=f(1)\,f(n)\quad\text{for every }n\in S.$$

The only way this can hold for all $$n$$ (with the values always lying in $$S$$ and hence never being $$0$$) is when

$$f(1)=1.$$

So the value at $$1$$ is completely fixed.

Next we list all products inside $$S$$ that are still in $$S$$. Besides the trivial $$1\cdot n=n$$, only three distinct products appear:

$$2\cdot2=4,\qquad 2\cdot3=6,\qquad 3\cdot2=6.$$

No other product of two elements of $$S$$ stays within $$S$$, because, for instance, $$3\cdot3=9\notin S$$ and $$2\cdot4=8\notin S$$. Therefore the multiplicative rule imposes conditions only on the numbers $$4$$ and $$6$$. Let us introduce the convenient names

$$a=f(2),\qquad b=f(3).$$

Using the rule on each admissible product, we obtain:

1. For $$2\cdot2=4$$ we must have

$$f(4)=f(2\cdot2)=f(2)^2=a^2.$$ 2. For $$2\cdot3=6$$ (and equally for $$3\cdot2=6$$) we must have

$$f(6)=f(2\cdot3)=f(2)\,f(3)=ab.$$

Thus the values of $$f$$ at $$4$$ and $$6$$ are forced once we pick $$a=f(2)$$ and $$b=f(3)$$. In contrast, the values at $$5$$ and $$7$$ are not tied to any product lying in $$S$$, so they may be chosen freely from $$\{1,2,3,4,5,6,7\}$$ later on.

The crucial point is that every value of the function must lie in $$S$$. Hence we need

$$a^2\le 7\quad\text{and}\quad ab\le 7.$$

We now enumerate all possibilities for $$a$$ and $$b$$ satisfying these two inequalities.

• If $$a=1$$, then $$a^2=1\le7$$ automatically, and $$ab=b\le7$$ for every $$b\in\{1,2,3,4,5,6,7\}$$. So with $$a=1$$, we have $$7$$ admissible choices for $$b$$.

• If $$a=2$$, then $$a^2=4\le7$$ is still allowed, but the inequality $$ab\le7$$ becomes $$2b\le7$$, i.e. $$b\le3$$. Hence $$b$$ can be $$1,2,$$ or $$3$$, giving $$3$$ choices.

• If $$a\ge3$$, then $$a^2\ge9>7$$, which violates $$a^2\le7$$, so no further values of $$a$$ are possible.

Combining the two admissible cases, we have in total

$$7+3=10$$

legal pairs $$(a,b)=(f(2),f(3)).$$

After fixing such a pair, the values at $$4$$ and $$6$$ are already determined by $$f(4)=a^2$$ and $$f(6)=ab$$, and they automatically lie in $$S$$ by the way we chose $$a$$ and $$b$$. The final freedom rests with $$f(5)$$ and $$f(7)$$, each of which can be any of the $$7$$ members of $$S$$. Therefore the total number of distinct functions is

$$10\;\times\;7\;\times\;7 \;=\;10\;\times\;49 \;=\;490.$$

So, the answer is $$490$$.

We are given $$a + \alpha = 1$$, $$b + \beta = 2$$, and the functional equation $$af(x) + \alpha f\!\left(\frac{1}{x}\right) = bx + \frac{\beta}{x}$$ for $$x \neq 0$$.

Replacing $$x$$ with $$\frac{1}{x}$$ in the functional equation gives $$af\!\left(\frac{1}{x}\right) + \alpha f(x) = \frac{b}{x} + \beta x$$.

Adding the original equation and this new equation: $$(a + \alpha)f(x) + (a + \alpha)f\!\left(\frac{1}{x}\right) = (b + \beta)x + (b + \beta)\frac{1}{x}$$, which simplifies to $$(a + \alpha)\left[f(x) + f\!\left(\frac{1}{x}\right)\right] = (b + \beta)\left(x + \frac{1}{x}\right)$$.

Since $$a + \alpha = 1$$ and $$b + \beta = 2$$, we get $$f(x) + f\!\left(\frac{1}{x}\right) = 2\left(x + \frac{1}{x}\right)$$.

Therefore, $$\frac{f(x) + f\!\left(\frac{1}{x}\right)}{x + \frac{1}{x}} = 2$$.

The correct answer is $$2$$.

First, we identify the logical form of the given English sentence. The sentence “If I reach the station in time, then I will catch the train” can be written in symbolic logic as $$P \rightarrow Q,$$ where

$$P : \text{“I reach the station in time,”}$$ $$Q : \text{“I will catch the train.”}$$

Now, we recall the definition of a contrapositive. For any implication of the form

$$P \rightarrow Q,$$

the contrapositive is obtained by simultaneously negating both parts and reversing their order. Symbolically, the contrapositive is

$$\neg Q \;\rightarrow\; \neg P.$$

Here $$\neg$$ (read “not”) is the logical negation operator. So we proceed step by step.

First we negate the consequent $$Q$$: $$\neg Q : \text{“I will not catch the train.”}$$

Next we negate the antecedent $$P$$: $$\neg P : \text{“I do not reach the station in time.”}$$

Finally, we reverse their order in the implication. Thus the contrapositive of $$P \rightarrow Q$$ is written in words as

“If I will not catch the train, then I do not reach the station in time.”

This matches exactly with Option D.

Hence, the correct answer is Option D.

To decide which given statement is a tautology, we consider every option one by one and simplify it algebraically.

First, recall the basic logical equivalence for an implication:

$$a \rightarrow b \;\equiv\; \sim a \;\vee\; b.$$

We will convert each implication using this rule and then simplify.

Option A is $$p \,\vee\, (\sim q) \;\rightarrow\; p \,\wedge\, q.$$

Using the implication rule we have

$$\sim\bigl(p \,\vee\, (\sim q)\bigr) \;\vee\; (p \,\wedge\, q).$$

Apply De Morgan’s law to the negation:

$$\sim p \;\wedge\; q \;\vee\; p \;\wedge\; q.$$

Now take $$q$$ common:

$$q \;\wedge\;(\sim p \;\vee\; p).$$

Inside the parentheses, $$\sim p \;\vee\; p$$ is a tautology (it is always true). Therefore the whole expression reduces to

$$q \;\wedge\; \text{(True)} \;=\; q.$$

This final result $$q$$ is not always true (it is false when $$q$$ is false), so Option A is not a tautology.

Option B is $$\sim(p \,\wedge\, \sim q) \;\rightarrow\; p \,\vee\, q.$$

Convert the implication:

$$\sim\!\bigl(\sim(p \,\wedge\, \sim q)\bigr) \;\vee\; (p \,\vee\, q).$$

The double negation simplifies immediately:

$$(p \,\wedge\, \sim q) \;\vee\; p \;\vee\; q.$$

Group the $$p$$ terms:

$$p \;\vee\; (p \,\wedge\, \sim q) \;\vee\; q.$$

Using the absorption law $$p \,\vee\, (p \,\wedge\, r)=p,$$ we obtain

$$p \;\vee\; q.$$

The expression $$p \;\vee\; q$$ can be false (specifically when $$p$$ and $$q$$ are both false). Thus Option B is not a tautology.

Option C is $$\sim(p \,\vee\, \sim q) \;\rightarrow\; p \,\wedge\, q.$$

Again convert the implication:

$$\sim\!\bigl(\sim(p \,\vee\, \sim q)\bigr) \;\vee\; (p \,\wedge\, q).$$

The double negation gives

$$(p \,\vee\, \sim q) \;\vee\; (p \,\wedge\, q).$$

Take $$p$$ common from the last two pieces:

$$p \;\vee\; \sim q \;\vee\; (p \,\wedge\, q).$$

Using absorption again, $$p \;\vee\; (p \,\wedge\, q)=p$$, so we get

$$p \;\vee\; \sim q.$$

This can be false whenever $$p$$ is false and $$q$$ is true. Hence Option C is not a tautology.

Option D is $$\sim(p \,\vee\, \sim q) \;\rightarrow\; p \,\vee\, q.$$

Convert the implication exactly as before:

$$\sim\!\bigl(\sim(p \,\vee\, \sim q)\bigr) \;\vee\; (p \,\vee\, q).$$

Simplify the double negation:

$$(p \,\vee\, \sim q) \;\vee\; (p \,\vee\, q).$$

Rearrange the terms:

$$p \;\vee\; \sim q \;\vee\; p \;\vee\; q.$$

Using idempotent law $$p \,\vee\, p = p,$$ merge the repeated $$p$$ terms:

$$p \;\vee\; \sim q \;\vee\; q.$$

Now notice that $$\sim q \;\vee\; q$$ is always true (law of excluded middle), so write

$$p \;\vee\; \text{(True)}.$$

Finally, $$\alpha \;\vee\; \text{(True)} = \text{True}$$ for any statement $$\alpha$$. Therefore the whole expression is always true, no matter what truth-values $$p$$ and $$q$$ take.

Thus Option D represents a statement that is true in every possible case; that is exactly the definition of a tautology.

Hence, the correct answer is Option D.

We begin with the original conditional statement:

“For an integer $$n$$, if $$n^{3}-1$$ is even, then $$n$$ is odd.”

To form the contrapositive, we first identify its two parts clearly.

Let

$$P : n^{3}-1 \text{ is even}$$

$$Q : n \text{ is odd}$$

The original statement in symbolic form is $$P \Rightarrow Q$$, which reads “If $$P$$, then $$Q$$.”

Now, we recall the definition of a contrapositive. For any implication $$P \Rightarrow Q$$, the contrapositive is obtained by negating both parts and reversing the direction of implication. In symbols, the contrapositive is:

$$\neg Q \Rightarrow \neg P$$

We now write down each negation explicitly.

Negation of $$Q$$:

$$\neg Q : n \text{ is not odd}$$

But for integers, “not odd” is the same as “even.” So we can re-express it as

$$\neg Q : n \text{ is even}$$

Negation of $$P$$:

$$\neg P : n^{3}-1 \text{ is not even}$$

Again, for integers, “not even” means “odd.” Hence we rewrite

$$\neg P : n^{3}-1 \text{ is odd}$$

Substituting these negations into the symbolic form $$\neg Q \Rightarrow \neg P$$, we obtain the contrapositive sentence:

“For an integer $$n$$, if $$n$$ is even, then $$n^{3}-1$$ is odd.”

Looking at the options provided, this matches Option A.

Hence, the correct answer is Option A.

We begin with the original statement, which is an implication:
    $$P : \text{The function } f \text{ is differentiable at } a$$
    $$Q : \text{The function } f \text{ is continuous at } a$$
Stated in words, the implication is “If $$P$$, then $$Q$$” or symbolically $$P \Rightarrow Q$$.

To find its contrapositive, we recall the logical rule:
    Formula for contrapositive: For any implication $$P \Rightarrow Q$$, the contrapositive is $$\lnot Q \Rightarrow \lnot P$$.
That is, we first negate the conclusion $$Q$$, then make it the new hypothesis; simultaneously we negate the original hypothesis $$P$$ and make it the new conclusion.

Applying this rule step by step:

1. Negate the conclusion $$Q$$.
    The negation of “$$f$$ is continuous at $$a$$” is “$$f$$ is not continuous at $$a$$”.
    So $$\lnot Q : \text{The function } f \text{ is not continuous at } a$$.

2. Negate the hypothesis $$P$$.
    The negation of “$$f$$ is differentiable at $$a$$” is “$$f$$ is not differentiable at $$a$$”.
    So $$\lnot P : \text{The function } f \text{ is not differentiable at } a$$.

3. Form the new implication $$\lnot Q \Rightarrow \lnot P$$.
    In sentence form this reads: “If the function $$f$$ is not continuous at $$a$$, then $$f$$ is not differentiable at $$a$$.”

We now compare this derived contrapositive with the given options:

Option A: “If $$f$$ is continuous at $$a$$, then it is not differentiable at $$a$$.” (This is $$Q \Rightarrow \lnot P$$, not the contrapositive.)

Option B: “If $$f$$ is not continuous at $$a$$, then it is not differentiable at $$a$$.” (This matches $$\lnot Q \Rightarrow \lnot P$$ exactly.)

Option C: “If $$f$$ is not continuous at $$a$$, then it is differentiable at $$a$$.” (This is $$\lnot Q \Rightarrow P$$, the inverse of Option A, still incorrect.)

Option D: “If $$f$$ is continuous at $$a$$, then it is differentiable at $$a$$.” (This is simply the converse of the original statement.)

Only Option B reproduces the contrapositive. Hence, the correct answer is Option B.

We begin with the compound statement

$$ (p \rightarrow q)\;\wedge\;(q \rightarrow \sim p). $$

First, recall the fundamental implication formula

$$ p \rightarrow q \;\equiv\; \sim p \,\vee\, q, $$

which says “if p then q” is logically the same as “either not p or q.” Applying this rule to each implication separately, we obtain

$$ (p \rightarrow q)\;=\;\sim p \,\vee\, q, $$

and

$$ (q \rightarrow \sim p)\;=\;\sim q \,\vee\, \sim p. $$

Therefore our original statement becomes

$$ (\sim p \,\vee\, q)\;\wedge\;(\sim q \,\vee\, \sim p). $$

Observe that both disjunctions contain the term $$\sim p$$, so it is natural to use the distributive law of Boolean algebra, which states

$$ (A \,\vee\, B)\;\wedge\;(A \,\vee\, C) \;\equiv\; A \,\vee\, (B \,\wedge\, C). $$

Here we identify

$$A = \sim p,\qquad B = q,\qquad C = \sim q.$$

Substituting these into the distributive formula, we have

$$ (\sim p \,\vee\, q)\;\wedge\;(\sim p \,\vee\, \sim q) \;\equiv\; \sim p \,\vee\, (q \,\wedge\, \sim q). $$

Now, $$q \,\wedge\, \sim q$$ is a contradiction: one cannot have $$q$$ and “not $$q$$” true at the same time. Thus

$$ q \,\wedge\, \sim q \;\equiv\; \text{False}. $$

Using the identity $$A \,\vee\, \text{False} \equiv A$$, we finally arrive at

$$ \sim p \,\vee\, \text{False} \;\equiv\; \sim p. $$

Hence the entire logical statement simplifies to $$\sim p$$.

Looking at the given options, $$\sim p$$ corresponds to Option C.

Hence, the correct answer is Option C.

We begin by translating the given English sentence into symbols. Let $$P$$ denote the statement “$$\sqrt{5}$$ is an integer.” Let $$Q$$ denote the statement “$$5$$ is irrational.” The original sentence is “$$\sqrt{5}$$ is an integer or 5 is irrational,” which is symbolically $$P \lor Q$$.

To find the negation, we recall the logical rule known as De Morgan’s law. The law states: for any two statements $$P$$ and $$Q$$, $$\neg(P \lor Q) \;=\; (\neg P) \land (\neg Q).$$ In words, “not (P or Q)” is equivalent to “(not P) and (not Q).”

Applying this formula to our case, we have $$\neg(P \lor Q) \;=\; (\neg P) \land (\neg Q).$$

Now we translate each piece back into ordinary language. The negation $$\neg P$$ is “$$\sqrt{5}$$ is not an integer.” The negation $$\neg Q$$ is “5 is not irrational,” i.e. “5 is rational” (indeed 5 is an integer, hence rational).

Combining these two with the connective “and,” we obtain “$$\sqrt{5}$$ is not an integer and 5 is not irrational.”

Examining the choices, this wording matches exactly with Option B.

Hence, the correct answer is Option B.

Let us first write clearly what every symbol means. The symbol $$\vee$$ stands for “or”, $$\wedge$$ for “and”, $$\sim$$ for “not”, $$\to$$ for “if … then …”, and $$\leftrightarrow$$ for “if and only if” (biconditional). A statement is a tautology when it is true for every possible truth-value assignment of its component variables, while it is a fallacy when it is false for every assignment.

We have two compound statements:

$$S_1 :\;(q \vee p) \to (p \leftrightarrow \sim q)$$

$$S_2 :\;\sim q \wedge (\sim p \leftrightarrow q)$$

To test whether $$S_1$$ is a tautology, we enumerate all possible truth-values of $$p$$ and $$q$$. There are four possibilities: $$(p,q) = (T,T),\,(T,F),\,(F,T),\,(F,F).$$ For every pair we shall compute each sub-expression step by step.

Case 1: $$p=T,\;q=T$$

First, $$q \vee p = T \vee T = T.$$

Next, $$\sim q = \sim T = F,$$ so $$p \leftrightarrow \sim q = T \leftrightarrow F = F$$ because a biconditional is true only when both sides have the same truth value.

Finally, $$(q \vee p) \to (p \leftrightarrow \sim q) = T \to F = F,$$ since “true implies false” is false.

Case 2: $$p=T,\;q=F$$

Here, $$q \vee p = F \vee T = T.$$

Also, $$\sim q = \sim F = T,$$ hence $$p \leftrightarrow \sim q = T \leftrightarrow T = T.$$

Thus $$(q \vee p) \to (p \leftrightarrow \sim q) = T \to T = T.$$

Case 3: $$p=F,\;q=T$$

We obtain $$q \vee p = T \vee F = T.$$

Further, $$\sim q = \sim T = F,$$ so $$p \leftrightarrow \sim q = F \leftrightarrow F = T.$$

Consequently, $$(q \vee p) \to (p \leftrightarrow \sim q) = T \to T = T.$$

Case 4: $$p=F,\;q=F$$

Now $$q \vee p = F \vee F = F.$$

Whenever the antecedent of an implication is false, the whole implication is automatically true, because $$F \to X = T$$ for any $$X$$. Therefore we do not even need to calculate the biconditional; still, for completeness we note $$\sim q = T$$ and $$p \leftrightarrow \sim q = F \leftrightarrow T = F.$$ The implication is $$F \to F = T.$$

Collecting the truth values of $$S_1$$ we have: $$F,\;T,\;T,\;T.$$ Because one row (the very first) gives false, $$S_1$$ is not a tautology. So statement $$S_1$$ is incorrect.

Next we examine $$S_2 = \sim q \wedge (\sim p \leftrightarrow q)$$ to see whether it is a fallacy. Again we test all four assignments.

Case 1: $$p=T,\;q=T$$

Compute $$\sim q = F.$$

Compute $$\sim p = F,$$ so $$\sim p \leftrightarrow q = F \leftrightarrow T = F.$$

Hence $$\sim q \wedge (\sim p \leftrightarrow q) = F \wedge F = F.$$

Case 2: $$p=T,\;q=F$$

We get $$\sim q = T.$$

Also $$\sim p = F,$$ and $$\sim p \leftrightarrow q = F \leftrightarrow F = T.$$

Therefore $$\sim q \wedge (\sim p \leftrightarrow q) = T \wedge T = T.$$ This row gives true.

Case 3: $$p=F,\;q=T$$

Here $$\sim q = F.$$

Then $$\sim p = T,$$ and $$\sim p \leftrightarrow q = T \leftrightarrow T = T.$$

Thus $$\sim q \wedge (\sim p \leftrightarrow q) = F \wedge T = F.$$

Case 4: $$p=F,\;q=F$$

Now $$\sim q = T.$$

Also $$\sim p = T,$$ giving $$\sim p \leftrightarrow q = T \leftrightarrow F = F.$$

Hence $$\sim q \wedge (\sim p \leftrightarrow q) = T \wedge F = F.$$

The truth values obtained for $$S_2$$ are $$F,\;T,\;F,\;F.$$ Because at least one row (the second) is true, the statement is not always false; therefore it is not a fallacy. Hence statement $$S_2$$ is also incorrect.

We have shown that $$S_1$$ is not a tautology and $$S_2$$ is not a fallacy. Thus both statements asserted in the question are wrong.

Hence, the correct answer is Option A.

We are told that the proposition $$p \to (p \wedge \sim q)$$ is false.

First, we recall the fundamental rule for a material implication. The compound statement $$A \to B$$ is false only in the single case when the antecedent $$A$$ is true and the consequent $$B$$ is false. Symbolically,

$$A \to B = F \;\; \text{iff} \;\; A = T \text{ and } B = F.$$

In our problem, the antecedent is $$p$$ and the consequent is $$p \wedge \sim q.$$ So, for the whole implication to be false, we must have

$$p = T \quad\text{and}\quad (p \wedge \sim q) = F.$$

We already know from the first part that

$$p = T.$$

Now we must analyze the conjunctive statement $$p \wedge \sim q.$$ A conjunction $$X \wedge Y$$ is true only when both $$X$$ and $$Y$$ are true; otherwise it is false. Here,

$$X = p, \qquad Y = \sim q.$$

We have just found $$p = T.$$ Let us substitute this known value:

$$p \wedge \sim q = T \wedge \sim q.$$

We are told that this entire conjunction is actually false:

$$T \wedge \sim q = F.$$

The only way an expression of the form $$T \wedge Y$$ can be false is when the second factor $$Y$$ is false. Hence,

$$\sim q = F.$$

A negation $$\sim q$$ is false precisely when the original statement $$q$$ is true. Therefore,

$$q = T.$$

Collecting our results, we have obtained

$$p = T, \qquad q = T.$$

This matches Option C, which lists the truth values as $$T, T.$$

Hence, the correct answer is Option 3.

First, recall that the inverse relation $$R^{-1}$$ is obtained by interchanging the coordinates of every ordered pair of $$R$$. Consequently, every first component (the “domain”) of $$R^{-1}$$ comes from a second component of some ordered pair in $$R$$. In other words,

$$\text{Domain}(R^{-1})=\text{Range}(R).$$

Therefore we must determine all integer values of $$y$$ for which there exists at least one integer $$x$$ satisfying the inequality that defines $$R$$, namely

$$x^2+3y^2\le 8.$$

We systematically test every integer value of $$y$$ whose square cannot make the left-hand side exceed 8.

Let us examine successive values of $$y$$.

1. Take $$y=0$$. We have $$y^2=0$$, so the inequality becomes

$$x^2+3(0)^2\le 8 \;\Longrightarrow\; x^2\le 8.$$

The integer squares not exceeding 8 are $$0,1,4$$. Thus the corresponding integer values of $$x$$ are

$$x=\,-2,\,-1,\,0,\,1,\,2.$$

Since at least one such $$x$$ exists, $$y=0$$ appears in the range of $$R$$.

2. Next take $$y=1$$. We have $$y^2=1$$, giving

$$x^2+3(1)^2\le 8 \;\Longrightarrow\; x^2+3\le 8 \;\Longrightarrow\; x^2\le 5.$$

The integer squares not exceeding 5 are again $$0,1,4$$, so

$$x=\,-2,\,-1,\,0,\,1,\,2.$$

Hence $$y=1$$ also lies in the range of $$R$$.

3. Because the expression depends on $$y^2$$, the same result holds for $$y=-1$$. Indeed, $$(-1)^2=1$$, so the identical calculation shows at least one suitable $$x$$ exists. Thus $$y=-1$$ is present in the range of $$R$$.

4. Now consider $$y=2$$ or $$y=-2$$. Here $$y^2=4$$, leading to

$$x^2+3(4)\le 8 \;\Longrightarrow\; x^2+12\le 8 \;\Longrightarrow\; x^2\le -4.$$

The inequality $$x^2\le -4$$ has no integer solution because a square is always non-negative. Therefore $$y=\pm2$$ do not belong to the range of $$R$$.

5. For any $$|y|\ge 2$$, the value of $$3y^2$$ is at least 12, which already exceeds 8, so no integer $$x$$ can satisfy the defining inequality. Thus no further $$y$$ values are possible.

Collecting all successful $$y$$ values, we obtain

$$\text{Range}(R)=\{-1,\,0,\,1\}.$$

Therefore

$$\text{Domain}(R^{-1})=\{-1,\,0,\,1\}.$$

Hence, the correct answer is Option D.

We have the compound statement $$ (p \wedge q) \to (\sim q \vee r) $$ whose truth value is given to be $$F$$ (false).

First, recall the truth rule for an implication. The statement $$A \to B$$ is false only in one situation: when the antecedent $$A$$ is true and simultaneously the consequent $$B$$ is false. In every other case the implication is true.

Here the antecedent is $$p \wedge q$$ and the consequent is $$\sim q \vee r$$. Because the whole implication is false, we must have

$$ p \wedge q = T \quad \text{and} \quad \sim q \vee r = F. $$

Now we analyse each part in turn.

From $$ p \wedge q = T $$, the truth table of conjunction tells us that a conjunction is true only when both components are true. Hence

$$ p = T \quad \text{and} \quad q = T. $$

Next, look at the consequent $$\sim q \vee r$$. We already know $$q = T$$, so its negation is

$$ \sim q = \sim T = F. $$

The consequent simplifies to

$$ \sim q \vee r = F \vee r. $$

The truth table for a disjunction states that $$A \vee B$$ is false only when both $$A$$ and $$B$$ are false. We already have $$F$$ in place of $$A$$, so to make the entire disjunction false we must also have

$$ r = F. $$

Collecting the results, we find

$$ p = T, \; q = T, \; r = F. $$

Checking the option list, this matches Option A.

Hence, the correct answer is Option A.

We are asked to find the negation of the Boolean expression $$p \vee (\sim p \wedge q)$$ and then match the simplified result with one of the given options.

First we write the required negation explicitly:

$$\neg\bigl[p \vee (\sim p \wedge q)\bigr].$$

We recall De Morgan’s first law, which states that for any two propositions $$A$$ and $$B$$, the negation of their disjunction is

$$\neg(A \vee B)=\neg A \wedge \neg B.$$

In our expression, we can treat $$A$$ as $$p$$ and $$B$$ as $$(\sim p \wedge q)$$. Applying the law we get

$$\neg\bigl[p \vee (\sim p \wedge q)\bigr] \;=\; \bigl(\neg p\bigr) \;\wedge\; \neg\bigl(\,\sim p \wedge q\,\bigr).$$

Now we turn to the second negation inside. Again, by De Morgan’s second law, which says

$$\neg(X \wedge Y)=\neg X \vee \neg Y,$$

we let $$X=\sim p$$ and $$Y=q$$. Thus

$$\neg\bigl(\,\sim p \wedge q\,\bigr)=\neg(\sim p) \;\vee\; \neg q.$$

The double negation rule $$\neg(\sim p)=p$$ simplifies the first term, so we have

$$\neg\bigl(\,\sim p \wedge q\,\bigr)=p \;\vee\; \neg q.$$

Substituting this back into the earlier expression yields

$$\bigl(\neg p\bigr) \wedge \bigl(p \;\vee\; \neg q\bigr).$$

We now use the distributive law of conjunction over disjunction, namely

$$X \wedge (Y \vee Z) = (X \wedge Y) \vee (X \wedge Z).$$

Here $$X=\neg p,$$ $$Y=p,$$ and $$Z=\neg q.$$ Distributing gives

$$\bigl(\neg p \wedge p\bigr) \;\vee\; \bigl(\neg p \wedge \neg q\bigr).$$

Observing that $$\neg p \wedge p$$ is always false (a contradiction), we replace it by the constant false value, traditionally denoted by $$0$$. Since $$0 \vee$$ (anything) is just that “anything,” the expression simplifies to

$$\neg p \wedge \neg q.$$

Thus the negation of $$p \vee (\sim p \wedge q)$$ is exactly $$\sim p \wedge \sim q$$, which corresponds to Option B.

Hence, the correct answer is Option B.

We have to find the negation (logical NOT) of the Boolean statement $$x \leftrightarrow \sim y$$ and then express the result in a simplified disjunctive form so that it can be matched with one of the given options.

First of all, let us recall the standard equivalence for a biconditional. The formula for any two propositions $$p$$ and $$q$$ is:

$$p \leftrightarrow q \;=\; (p \wedge q)\;\vee\;(\sim p \wedge \sim q).$$

Now we apply this rule to the concrete biconditional $$x \leftrightarrow \sim y$$ by replacing $$p$$ with $$x$$ and $$q$$ with $$\sim y$$. Doing so gives:

$$x \leftrightarrow \sim y \;=\; (x \wedge \sim y)\;\vee\;(\sim x \wedge \sim(\sim y)).$$

The double negation inside the second conjunct can be removed with the rule $$\sim(\sim y)=y$$. Hence we get:

$$x \leftrightarrow \sim y \;=\; (x \wedge \sim y)\;\vee\;(\sim x \wedge y).$$

We must now take the negation of the whole expression. So we write:

$$\sim\bigl(x \leftrightarrow \sim y\bigr) \;=\; \sim\Bigl[(x \wedge \sim y)\;\vee\;(\sim x \wedge y)\Bigr].$$

To push the negation inside the parentheses, we invoke De Morgan’s law, which states:

$$\sim(A \vee B)\;=\;(\sim A) \wedge (\sim B).$$

Here $$A$$ is $$x \wedge \sim y$$ and $$B$$ is $$\sim x \wedge y$$. Therefore:

$$\sim\Bigl[(x \wedge \sim y)\;\vee\;(\sim x \wedge y)\Bigr] \;=\; \bigl[\sim(x \wedge \sim y)\bigr] \;\wedge\; \bigl[\sim(\sim x \wedge y)\bigr].$$

The next step is to remove each inner negation. Again, we apply De Morgan’s law, this time in the form $$\sim(A \wedge B) = (\sim A) \vee (\sim B).$$ For the first bracket we have $$A = x$$ and $$B = \sim y$$:

$$\sim(x \wedge \sim y) \;=\; (\sim x) \;\vee\; (\sim(\sim y)) \;=\; (\sim x) \vee y.$$

For the second bracket we have $$A = \sim x$$ and $$B = y$$:

$$\sim(\sim x \wedge y) \;=\; (\sim(\sim x)) \;\vee\; (\sim y) \;=\; x \vee (\sim y).$$

So the entire expression is now

$$\bigl[(\sim x) \vee y\bigr] \;\wedge\; \bigl[x \vee (\sim y)\bigr].$$

To simplify, we distribute the conjunction across the disjunctions. The distributive law tells us:

$$(A \vee B)\wedge(C \vee D) \;=\; (A \wedge C)\;\vee\;(A \wedge D)\;\vee\;(B \wedge C)\;\vee\;(B \wedge D).$$

Letting $$A = \sim x,\; B = y,\; C = x,\; D = \sim y,$$ we compute each product term one by one:

1. $$(\sim x) \wedge x = \text{False},$$ because a proposition and its negation cannot both be true.

2. $$(\sim x) \wedge (\sim y) = \sim x \wedge \sim y.$$

3. $$y \wedge x = x \wedge y.$$

4. $$y \wedge (\sim y) = \text{False},$$ again because a variable and its negation cannot both be true.

Combining the surviving non-false terms under disjunction gives:

$$(\sim x \wedge \sim y)\;\vee\;(x \wedge y).$$

Conventionally we write the positive term first, so we obtain the final simplified negation as:

$$ (x \wedge y)\;\vee\;(\sim x \wedge \sim y). $$

Now we match this with the options provided. Option B is exactly

$$(x \wedge y) \vee (\sim x \wedge \sim y).$$

Hence, the correct answer is Option 2.

We have to simplify the proposition $$p \to \sim(p \wedge \sim q)$$ and then compare the final result with the four given options.

First recall the logical equivalence for an implication. The standard formula is:

$$a \to b \equiv (\sim a) \vee b.$$

In our problem the role of $$a$$ is played by $$p$$ and the role of $$b$$ is played by $$\sim(p \wedge \sim q).$$ Substituting into the formula we obtain

$$p \to \sim(p \wedge \sim q) \equiv (\sim p) \vee \bigl[\;\sim(p \wedge \sim q)\bigr].$$

Now we tackle the term $$\sim(p \wedge \sim q)$$. To simplify a negation of a conjunction, we use De Morgan’s law, which states:

$$\sim(A \wedge B) \equiv (\sim A) \vee (\sim B).$$

Here $$A$$ is $$p$$ and $$B$$ is $$\sim q$$. Applying the law carefully gives

$$\sim(p \wedge \sim q) \equiv (\sim p) \vee \bigl[\sim(\sim q)\bigr].$$

A double negation disappears, because $$\sim(\sim q) \equiv q$$. Hence the entire right-hand side becomes

$$\sim(p \wedge \sim q) \equiv (\sim p) \vee q.$$

We now substitute this back into the expression we obtained after the implication step:

$$(\sim p) \vee \bigl[\;\sim(p \wedge \sim q)\bigr] \equiv (\sim p) \vee \bigl[(\sim p) \vee q\bigr].$$

The disjunction is associative and idempotent, meaning that repeating the same statement inside a string of ORs does not change the result. Explicitly,

$$ (\sim p) \vee (\sim p) \equiv \sim p, \quad\text{and}\quad (\sim p) \vee \bigl[(\sim p) \vee q\bigr] \equiv (\sim p) \vee q.$$

Therefore the original proposition simplifies all the way down to

$$p \to \sim(p \wedge \sim q) \equiv (\sim p) \vee q.$$

This final expression matches exactly what is written in Option B. Hence, the correct answer is Option B.

We want to know which of the given compound statements is a tautology, that is, which statement is always true no matter which truth-values (T or F) we assign to the simple propositions $$p$$ and $$q$$. The straightforward way is to construct truth-tables for every option, but often we can save time by using well-known logical equivalences. Nevertheless, for absolute clarity we shall give every algebraic step and also exhibit at least one counter-example for all the non-tautological options.

First recall some standard logical facts that we shall need:

1. Implication equivalence $$p \rightarrow q \;\;\text{is equivalent to}\;\; \neg p \vee q$$ (Because an implication is false only when $$p$$ is true and $$q$$ is false.)

2. Distributive, associative, commutative and absorption laws for $$\vee$$ (OR) and $$\wedge$$ (AND) behave exactly like their algebraic counterparts, but with the understanding that TRUE behaves like 1 and FALSE like 0.

3. The absorption laws that we shall use twice are $$p \vee (p \wedge q) \equiv p$$ $$p \wedge (p \vee q) \equiv p.$$ These tell us that mixing a statement with itself inside an OR or an AND does not add any new logical content: $$p$$ already captures the whole truth of the compound.

Now we treat every option one by one.

Option A : $$ (p \wedge (p \rightarrow q)) \rightarrow q $$

We start by rewriting the implication inside the antecedent ($$p \rightarrow q$$) using fact 1.

So, $$p \rightarrow q \equiv \neg p \vee q.$$

Substituting this inside the main expression gives

$$ (p \wedge (\,\neg p \vee q\,)) \rightarrow q. $$

Next we simplify the bracket $$p \wedge (\,\neg p \vee q\,)$$ by distributing $$p$$ over the parentheses:

$$ p \wedge (\,\neg p \vee q\,) \equiv (p \wedge \neg p) \;\vee\; (p \wedge q). $$

But $$p \wedge \neg p$$ is always FALSE (a contradiction), hence

$$ (p \wedge \neg p) \;\vee\; (p \wedge q) \equiv \text{FALSE} \;\vee\; (p \wedge q) \equiv p \wedge q, $$

because OR with FALSE leaves the other term unchanged.

So up to now we have reduced the whole statement to

$$ (p \wedge q) \rightarrow q. $$

Once again we use fact 1 on the outer implication, i.e.

$$ (p \wedge q) \rightarrow q \equiv \neg (p \wedge q) \;\vee\; q. $$

We apply De Morgan to the negation:

$$ \neg (p \wedge q) \equiv \neg p \vee \neg q, $$

so the expression becomes

$$ (\neg p \vee \neg q) \;\vee\; q. $$

OR is associative and commutative, therefore we can rearrange freely. We now gather the two clauses that contain $$q$$:

$$ (\neg p \vee \neg q \vee q) \equiv (\neg p \vee \text{TRUE}), $$

because $$\neg q \vee q$$ is always TRUE. And of course $$\neg p \vee \text{TRUE}$$ simplifies to TRUE (OR with TRUE is TRUE).

Thus the entire compound statement has simplified to the constant truth-value TRUE for every truth assignment to $$p$$ and $$q$$. Therefore option A is a tautology.

Option B : $$ q \rightarrow (p \wedge (p \rightarrow q)) $$

Again write $$p \rightarrow q$$ as $$\neg p \vee q$$ and substitute:

$$ q \rightarrow \bigl(p \wedge (\neg p \vee q)\bigr). $$

As in option A we know that $$p \wedge (\neg p \vee q) \equiv p \wedge q.$$ So we obtain

$$ q \rightarrow (p \wedge q). $$

This implication is equivalent to $$\neg q \vee (p \wedge q).$$ Now choose a valuation with $$q = \text{T}$$ and $$p = \text{F}.$$ Then $$\neg q = \text{F}$$ and $$p \wedge q = \text{F}$$, so the whole expression evaluates to FALSE. Hence option B is not a tautology.

Option C : $$ p \wedge (p \vee q) $$

Using the absorption law stated earlier, we get directly

$$ p \wedge (p \vee q) \equiv p. $$

The value of $$p$$ obviously depends on our choice of truth-values; it is TRUE for $$p = \text{T}$$ and FALSE for $$p = \text{F}$$. Therefore this compound statement sometimes becomes FALSE, so it is not a tautology.

Option D : $$ p \vee (p \wedge q) $$

This is the other absorption law:

$$ p \vee (p \wedge q) \equiv p. $$

Again the truth of the whole expression varies with the truth of $$p$$, so option D is not a tautology.

After examining every choice we see that only option A is always TRUE, i.e. is a tautology.

Hence, the correct answer is Option A.

We have to decide the logical nature of the formula $$S=(p\to(q\to p))\to(p\to(p\vee q)).$$

First, we recall the standard implication-disjunction equivalence:

$$a\to b\;\equiv\;\sim a\;\vee\;b.$$

We apply this rule to the innermost implication $$q\to p.$$

So, $$q\to p\;\equiv\;\sim q\;\vee\;p.$$

Substituting this result back into the left part of $$S$$ we get

$$p\to(q\to p)\;\equiv\;p\to(\sim q\vee p).$$

Once again using the same rule $$a\to b\equiv\sim a\vee b$$ on the implication whose antecedent is $$p$$, we have

$$p\to(\sim q\vee p)\;\equiv\;\sim p\;\vee\;(\sim q\;\vee\;p).$$

Now we re-arrange the disjunction because disjunction is associative and commutative:

$$\sim p\;\vee\;(\sim q\;\vee\;p)\;=\;(\sim p\;\vee\;p)\;\vee\;\sim q.$$

We know that the law of excluded middle tells us $$\sim p\;\vee\;p\equiv \text{True}.$$ Therefore,

$$(\sim p\;\vee\;p)\;\vee\;\sim q\;\equiv\;\text{True}\;\vee\;\sim q\;\equiv\;\text{True}.$$

Hence, the whole sub-formula $$p\to(q\to p)$$ is always true; it is a tautology. We can now write

$$p\to(q\to p)\;\equiv\;\text{True}.$$

Next, we turn to the right part of $$S$$, namely $$p\to(p\vee q).$$ Again we use $$a\to b\equiv\sim a\vee b.$$ So,

$$p\to(p\vee q)\;\equiv\;\sim p\;\vee\;(p\vee q).$$

Associating the disjunctions yields

$$\sim p\;\vee\;p\;\vee\;q.$$

By the law of excluded middle, $$\sim p\;\vee\;p\equiv\text{True},$$ hence

$$\sim p\;\vee\;p\;\vee\;q\;\equiv\;\text{True}\;\vee\;q\;\equiv\;\text{True}.$$

Therefore, the consequent $$(p\to(p\vee q))$$ is also a tautology. We can thus rewrite $$S$$ purely in terms of truth values:

$$S\;\equiv\;(\text{True})\;\to\;(\text{True}).$$

Finally, we know that $$\text{True}\to\text{True}\equiv\text{True},$$ because an implication with a true antecedent and a true consequent is true.

So the entire statement $$S$$ is always true for every possible truth-value assignment of $$p$$ and $$q$$. That is, $$S$$ is a tautology.

Hence, the correct answer is Option 4.

First we recall that the composite function $$(f \circ f)(x)$$ means “apply $$f$$ twice”, i.e. $$(f \circ f)(x)=f\bigl(f(x)\bigr).$$ The statement of the question says that this composite equals $$x$$ itself. Symbolically, for every real $$x \neq -a$$ with $$f(x)\neq -a,$$ we have the requirement

$$f\bigl(f(x)\bigr)=x.$$

Such a function is called an involution. We shall use this condition to determine the constant $$a$$ and then evaluate $$f\!\left(-\dfrac12\right).$$

The given function is

$$f(x)=\dfrac{a-x}{\,a+x\,}.$$

We first compute $$y=f(x):$$

$$y=\dfrac{a-x}{a+x}.$$

Now we apply $$f$$ again, replacing its argument by $$y$$. Thus

$$f\bigl(f(x)\bigr)=\dfrac{a-y}{\,a+y\,}.$$

Substituting $$y=\dfrac{a-x}{a+x}$$ we obtain

$$f\bigl(f(x)\bigr)=\dfrac{a-\dfrac{a-x}{a+x}}{\,a+\dfrac{a-x}{a+x}\,}.$$

To simplify, we bring every term over the common denominator $$(a+x).$$ For the numerator:

$$a-\dfrac{a-x}{a+x}=\dfrac{a(a+x)-(a-x)}{a+x}=\dfrac{a^2+ax-a+x}{a+x}.$$

Collecting like terms gives

$$a^2+ax-a+x=a^2-a+x(a+1).$$

So the numerator equals

$$\dfrac{a^2-a+x(a+1)}{a+x}.$$

For the denominator:

$$a+\dfrac{a-x}{a+x}=\dfrac{a(a+x)+(a-x)}{a+x}=\dfrac{a^2+ax+a-x}{a+x}.$$

Collecting like terms gives

$$a^2+ax+a-x=a^2+a+x(a-1).$$

Hence the denominator equals

$$\dfrac{a^2+a+x(a-1)}{a+x}.$$

Therefore

$$f\bigl(f(x)\bigr)=\dfrac{\dfrac{a^2-a+x(a+1)}{a+x}}{\dfrac{a^2+a+x(a-1)}{a+x}}=\dfrac{a^2-a+x(a+1)}{a^2+a+x(a-1)}.$$

The requirement $$f\bigl(f(x)\bigr)=x$$ now becomes the identity

$$\dfrac{a^2-a+x(a+1)}{a^2+a+x(a-1)}=x \quad \text{for all admissible } x.$$

Cross-multiplying (which is valid because denominators are non-zero away from the excluded points) we get

$$a^2-a+x(a+1)=x\bigl[a^2+a+x(a-1)\bigr].$$

Expanding the right-hand side yields

$$a^2-a+x(a+1)=a^2x+ax+(a-1)x^2.$$

Now we bring every term to the right so that the left side becomes zero, collecting like powers of $$x$$:

$$0=a^2x+ax+(a-1)x^2-\bigl(a^2-a+ax+x\bigr).$$

Distributing the negative sign and combining terms one by one, we obtain

$$0=\underbrace{(a-1)}_{\text{coefficient of }x^2}x^2+\underbrace{\bigl(a^2-1\bigr)}_{\text{coefficient of }x}x+\underbrace{\bigl(-a(a-1)\bigr)}_{\text{constant term}}.$$

For the polynomial to be identically zero for all real $$x,$$ each individual coefficient must vanish:

$$a-1=0,\qquad a^2-1=0,\qquad -a(a-1)=0.$$

The first of these directly gives $$a=1.$$ Substituting $$a=1$$ into the second and third equations confirms that they are also satisfied. Thus the only admissible constant is

$$a=1.$$

With this value, the original function becomes

$$f(x)=\dfrac{1-x}{\,1+x\,}.$$

We are required to evaluate $$f\!\left(-\dfrac12\right).$$ Substituting $$x=-\dfrac12$$ gives

$$f\!\left(-\dfrac12\right)=\dfrac{1-\left(-\dfrac12\right)}{1+\left(-\dfrac12\right)}=\dfrac{1+\dfrac12}{1-\dfrac12}=\dfrac{\dfrac32}{\dfrac12}.$$

Dividing the two fractions, we obtain

$$f\!\left(-\dfrac12\right)=\dfrac32 \times \dfrac21=3.$$

Hence, the correct answer is Option D.

We have $$g(x)=x^2+x-1$$ and the composition $$\bigl(g\circ f\bigr)(x)=g\!\bigl(f(x)\bigr)=4x^2-10x+5.$$

Writing $$y=f(x),$$ the definition of composition gives us

$$g\!\bigl(f(x)\bigr)=g(y)=y^2+y-1.$$

But by the statement of the problem the same quantity also equals $$4x^2-10x+5.$$ Hence we must have

$$y^2+y-1=4x^2-10x+5.$$

Substituting back $$y=f(x)$$ we get a quadratic equation in $$f(x):$$

$$\bigl(f(x)\bigr)^2+f(x)-1=4x^2-10x+5.$$

Now we bring every term to the left‐hand side:

$$\bigl(f(x)\bigr)^2+f(x)-1-4x^2+10x-5=0,$$

so

$$\bigl(f(x)\bigr)^2+f(x)-4x^2+10x-6=0.$$

This is a quadratic in the unknown $$f(x).$$ For a quadratic $$at^2+bt+c=0$$ the roots are given by the quadratic formula $$t=\dfrac{-b\pm\sqrt{\,b^2-4ac\,}}{2a}.$$

Here $$a=1,\;b=1,\;c=-4x^2+10x-6,$$ so

$$f(x)=\dfrac{-1\pm\sqrt{\,1-4\bigl(-4x^2+10x-6\bigr)\,}}{2}.$$

Inside the square root we simplify step by step:

$$1-4\bigl(-4x^2+10x-6\bigr)=1+16x^2-40x+24.$$

Combining like terms gives

$$16x^2-40x+25.$$

Recognising a perfect square, we note $$16x^2-40x+25=(4x-5)^2.$$ Hence

$$f(x)=\dfrac{-1\pm|4x-5|}{2}.$$

The two possible expressions that come out are obtained very easily:

If $$|4x-5|=4x-5$$ we get $$f(x)=\dfrac{-1+(4x-5)}{2}=2x-3,$$ and if $$|4x-5|=5-4x$$ we get $$f(x)=\dfrac{-1+(5-4x)}{2}=2-2x.$$

Thus for every $$x$$ the value of $$f(x)$$ is either $$2x-3$$ or $$2-2x.$$ Both of these, when put back into $$g(x),$$ reproduce $$4x^2-10x+5,$$ so the composition condition is satisfied in either case.

Now we specifically need $$f\!\left(\dfrac54\right).$$ Substituting $$x=\dfrac54$$ in either expression:

Using $$f(x)=2x-3,$$ we get $$f\!\left(\dfrac54\right)=2\left(\dfrac54\right)-3=\dfrac{10}{4}-3=\dfrac{5}{2}-3=-\dfrac12.$$

Using $$f(x)=2-2x,$$ we get $$f\!\left(\dfrac54\right)=2-2\left(\dfrac54\right)=2-\dfrac{10}{4}=2-\dfrac{5}{2}=-\dfrac12.$$

Both routes give the same numerical answer, so unambiguously

$$f\!\left(\dfrac54\right)=-\dfrac12.$$

Hence, the correct answer is Option B.

We recall that a statement is called a tautology when it remains true for every possible truth-value assignment of its component propositions. We shall examine each option one by one, reducing the compound statement with standard logical identities and looking for a possible counter-example. If even a single assignment makes the statement false, then it is not a tautology.

We begin with Option A:

$$ (\sim p)\;\wedge\;(p \vee q)\;\to\; q $$

First we simplify the antecedent "$$(\sim p)\wedge (p\vee q)$$". Using the distributive law

$$ a\wedge(b\vee c)\;=\;(a\wedge b)\;\vee\;(a\wedge c), $$

with $$a=\sim p,\; b=p,\; c=q,$$ we obtain

$$ (\sim p)\wedge(p\vee q)\;=\;[(\sim p)\wedge p] \;\vee\;[(\sim p)\wedge q]. $$

Now $$ (\sim p)\wedge p $$ is a contradiction, i.e. it is always false, so it may be dropped from the disjunction:

$$ (\sim p)\wedge(p\vee q)\;=\;(\sim p)\wedge q. $$

Therefore Option A becomes

$$ [(\sim p)\wedge q]\;\to\;q. $$

Next we recall the definition of implication:

$$ a\to b\;\equiv\;\sim a\;\vee\;b. $$

Replacing $$a$$ with $$ (\sim p)\wedge q $$ and $$b$$ with $$q$$ gives

$$ [(\sim p)\wedge q]\;\to\;q\;\equiv\;\sim[(\sim p)\wedge q]\;\vee\;q. $$

Apply De Morgan’s law to the negation inside:

$$ \sim[(\sim p)\wedge q]\;=\;\sim(\sim p)\;\vee\;\sim q\;=\;p\;\vee\;\sim q. $$

So the whole expression is

$$ (p\;\vee\;\sim q)\;\vee\;q. $$

The associative and commutative laws for disjunction permit us to regroup and obtain

$$ p\;\vee\;(q\;\vee\;\sim q). $$

Within the parentheses, $$ q\;\vee\;\sim q $$ is the law of excluded middle, which is always true, i.e. a tautology $$T$$. Hence we have

$$ p\;\vee\;T\;=\;T. $$

Thus Option A simplifies to a statement that is invariably true, making it a tautology.

Now we test Option B:

$$ (q\to p)\;\vee\;\sim(p\to q). $$

First expand each implication using $$a\to b \equiv \sim a\;\vee\;b$$:

$$ (q\to p) = (\sim q)\;\vee\;p, \qquad (p\to q) = (\sim p)\;\vee\;q.$$ Hence

$$ \sim(p\to q) = \sim[(\sim p)\vee q] = p\wedge\sim q $$ by De Morgan’s law. Therefore Option B becomes

$$ (\sim q\;\vee\;p)\;\vee\;(p\wedge\sim q). $$

We look for a truth-value combination that makes the whole disjunction false. A disjunction is false only when each of its components is false. So we need

$$ \sim q\;\vee\;p = F, \quad\text{and}\quad p\wedge\sim q = F. $$

Take $$p = F$$ and $$q = T$$. Then

$$\sim q = F,\quad p = F,\quad \therefore \sim q\;\vee\;p = F.$$ Also $$p\wedge\sim q = F\wedge F = F.$$ Both parts are false, so the entire statement is false for this assignment. Hence Option B is not a tautology.

Next we inspect Option C:

$$ (\sim q)\;\vee\;(p\wedge q)\;\to\;q. $$

Choose $$q = F$$ (i.e. $$q$$ is false). Then $$\sim q = T$$, so the antecedent $$(\sim q)\;\vee\;(p\wedge q)$$ is true irrespective of $$p$$. The consequent $$q$$, however, is false. Therefore the implication becomes $$T\to F,$$ which is false. So Option C is also not a tautology.

Finally Option D:

$$ (p\to q)\;\wedge\;(q\to p). $$

This is logically equivalent to the biconditional $$p\leftrightarrow q,$$ which is true only when $$p$$ and $$q$$ share the same truth value. Take $$p = T,\, q = F$$. Then $$p\to q = F$$, so the whole conjunction is false. Consequently Option D is not a tautology.

Only Option A survives every test and is always true.

Hence, the correct answer is Option A.

We have $$A = \{a,\; b,\; c\}$$ so $$|A| = 3$$, and $$B = \{1,\; 2,\; 3,\; 4\}$$ so $$|B| = 4$$. A function $$f : A \to B$$ assigns to every element of $$A$$ exactly one element of $$B$$.

First, recall that the total number of all possible functions from a set with $$m$$ elements to a set with $$n$$ elements is $$n^{\,m}$$ because each element of the domain has $$n$$ independent choices. Here, $$m = 3$$ and $$n = 4$$, so the total number of functions is

$$4^{\,3} = 64.$$

We are interested only in those functions which satisfy two simultaneous conditions:

(i) $$2 \in f(A)\;,$$ meaning that at least one element of $$A$$ is mapped to $$2$$.

(ii) $$f$$ is not one-one (not injective), i.e. at least two elements of $$A$$ share the same image.

We shall count functions satisfying (i) first, then remove those that violate (ii). Let us denote by $$N_1$$ the number of functions satisfying (i) and by $$N_{1,\,\text{inj}}$$ the number of those functions which are also injective. The desired count will therefore be

$$N = N_1 \;-\; N_{1,\,\text{inj}}.$$

Counting $$N_1$$ - functions with $$2$$ in the image.

Consider the complementary set of functions where $$2$$ never appears. For every element of $$A$$ we then have only $$\{1,3,4\}$$ available, that is $$3$$ choices per element. Hence the number of such functions is

$$3^{\,3} = 27.$$

By subtraction, the number of functions that do include $$2$$ is

$$N_1 = 64 \;-\; 27 = 37.$$

Counting $$N_{1,\,\text{inj}}$$ - injective functions with $$2$$ in the image.

A function from a set of $$3$$ elements to a set of $$4$$ elements is injective exactly when it assigns three distinct images. The total number of injective functions is therefore the number of permutations of any $$3$$ distinct elements chosen from $$4$$, which is the permutation number

$$P(4,3) = 4 \times 3 \times 2 = 24.$$

Among these, let us identify how many avoid the value $$2$$ entirely; such functions must use the three values $$1,3,4$$. There is only one way to choose these three values, and they can be arranged in $$P(3,3) = 3 \times 2 \times 1 = 6$$ ways. Hence

$$\text{Injective functions without }2 = 6.$$

Consequently, the injective functions that do contain $$2$$ are

$$N_{1,\,\text{inj}} = 24 \;-\; 6 = 18.$$

Final subtraction.

We now exclude these $$18$$ injective functions from the $$37$$ functions that contain $$2$$:

$$N = 37 \;-\; 18 = 19.$$

So, the answer is $$19$$.