CAT Functions, Graphs and Statistics Questions

Set A={2,3,5,7,11,13} so |A|=6
Set B={1, 8, 27} so |B|=3

Without any restrictions, each element in A can map to any of the 3 elements in B. Thus, the total number of functions is: $$3^6=729$$

Excluding Functions That Miss One Element in B: If a function does not map to an element in B, there are 2 elements in B left for mapping. The total number of such functions (for each specific element not mapped) is: $$2^6=64$$

Since there are 3 elements in B, the total number of such functions is:3x64=192

Adding Back Functions That Miss Two Elements in B: If a function misses two elements in B, there is only 1 element left for mapping. The total number of such functions is: 1^6=1. 

Since there are $$^3C_2$$ ways to choose which two elements are missed, the total number of such functions is: 3

Using the inclusion-exclusion principle, the number of functions where all elements of B are mapped by at least one element of A is:
729-192+3=540. 

Looking at the additional information about the prime numbers should make one realise that they are the key to solving the question. 
f(16000) can be written as $$f\left(2^8\times\ 5^4\right)$$

Now, we can try to find these individual values:

For any prime p: f(p)=1
$$f\left(p^2\right)=f\left(p\right)f\left(p\right)+f\left(p\right)+f\left(p\right)=1+1+1=3$$
$$f\left(p^3\right)=f\left(p^2\right)f\left(p\right)+f\left(p^2\right)+f\left(p\right)=3+3+1=7$$

This way, we can find the function output for any prime number raised to a power. 

We can see that each new exponent is twice the previous output +1, solving this way till prime raised to power 8
$$f\left(p^4\right)=7+7+1=15$$
$$f\left(p^5\right)=15+15+1=31$$
$$f\left(p^6\right)=31+31+1=63$$
$$f\left(p^7\right)=63+63+1=127$$
$$f\left(p^8\right)=127+127+1=255$$

Using these values in the original expression of $$f\left(2^8\times\ 5^4\right)=f\left(2^8\right)f\left(5^4\right)+f\left(2^8\right)+f\left(5^4\right)$$ we get
$$f\left(2^8\times\ 5^4\right)=\left(255\times\ 15\right)+255+15=4095$$

Therefore, Option A is the correct answer. 

The expression on the right-hand side will have two critical points: 2 and -2

For any value of x greater than equal to 2, the equation changes to x+2-(x-2) = 4

the value of min{x,2} would be 2, so we would want max{x,2} to be 4+2 = 6

Therefore, x=6 works. 

For any value of x less than equal to -2, the equation changes to -(x+2)+(x-2) = -4
the value of max{x,2} would be 2, so we would want min{x,2} to be 6 again; this cannot be the case.
In general, subtracting the smaller (min) of the two values from the bigger (max) can not lead to a negative number. The max it can lead to is a 0

When x lies between -2 and 2,  the equation becomes 2x
The maximum function will give back 2, and the minimum function will give back x, with the right-hand side giving 2x
Solving this we would get 2-x = 2x which is x=2/3

Therefore, there are two real values of x for which the given equation holds. 

Hence, 2 is the correct answer.

The graph of the function on the right hand side can be visualised as:

Hence, 2 is the correct answer. 

We are given, $$f(x) + 2f \left(\cfrac{1}{x}\right) = 3x$$
Substituting $$\frac{1}{x}\ for\ x$$

$$f\left(\dfrac{1}{x}\right)+2f\left(x\right)=\dfrac{3}{x}$$
Multiplying the second equation by 2 we will have 

$$2f\left(\dfrac{1}{x}\right)+4f\left(x\right)=\dfrac{6}{x}$$

Subtracting the first equation from the second equation we have, 

$$3f\left(x\right)=\frac{6}{x}-3x$$

$$f\left(x\right)=\frac{2}{x}-x$$
We want the sum of values when this function equals 3, 

$$\frac{2}{x}-x=3$$
$$x^2+3x-2=0$$
Since the discriminant is greater than zero, both values of x will be real, and we can directly take the sum of values of $$x$$ to be, 
$$-\frac{3}{1}$$

Answer is -3. 

From the inequality and nature of x, and y, we get the given diagram:

We need to find the area of the quadrilateral ABDE = area of rectangle ABCD + area of triangle CDE

=> Area of ABCD = (7-3)*5 = 20 units, and the area of triangle CDE = (1/2)*10*5 = 25 units.

Hence, the area of the quadrilateral ABDE = (20+25) = 45 units.

Given that f(3x + 2y, 2x - 5y) = 19x.

Let us assume the function f(a,b) is a linear combination of a and b.

=> f(3x+2y, 2x-5y) = m(3x+2y) + n(2x-5y) = 19x

=> 3m + 2n = 19 and 2m - 5n = 0

Solving we get m = 5 and n = 2

=> f(a,b) = 5a+2b

=> f(x,2x) = 5x + 2(2x) = 9x = 27 => x = 3.

x>=a, so |x-a| = x-a

x<100, so |x-100| = 100-x

f(x) = (x-a) + (100-x) + |x-a-50| =100

or, |x-a-50| = a

From the graph we can can see that when x=a then

|x-a-50|=a

or, a= 50

Similarly when x=a+100

|x-a-50|=a

or, a= 50

So value of a is 50 when f(x) is 100.

$$f(x) \geq 0$$for all real numbers $$x$$, so D<=0

Since f(2)=0 therefore x=2 is a root of f(x)

Since the discriminant of f(x) is less than equal to 0 and 2 is a root so we can conclude that D=0

Therefore f(x) = $$a\left(x-2\right)^2$$

f(4)=6

or, 6 = $$a\left(x-2\right)^2$$

a= 3/2

$$f\left(-2\right)=\ -\dfrac{3}{2}\left(-4\right)^2=24$$

Given, 

$$f\left(x\right)+f\left(x-1\right)=1$$ ...... (1)

$$f\left(x^2-x\right)=5$$ ......  (2)

$$g\left(x\right)=x^2$$

Substituting x = 1 in (1) and (2), we get

f(0) = 5

f(1) + f(0) = 1

f(1) = 1 - 5 = -4

f(2) + f(1) = 1

f(2) = 1 + 4 = 5

f(n) = 5 if n is even and f(n) = -4 if n is odd

f(g(5)) + g(f(5)) = f(25) + g(-4) = -4 + 16 = 12

When x< r

f(x) = r

f(x) = f(f(x))

r = f(r)

r= 2r-r

r=r

When x>=r

f(x) = 2x-r

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

2x-r = f(2x-r)

2x-r = 2(2x-r) - r

2x-r = 4x-3r

or, x=r

Therefore x<= r

It is given,

$$\Sigma_{n=1}^N\ \left[\dfrac{1}{5}+\dfrac{n}{25}\right]=25$$

$$\Sigma_{n=1}^N\ \left[\dfrac{5+n}{25}\right]=25$$

For n = 1 to n = 19, value of function is zero.

For n = 20 to n = 44, value of function will be 1.

44 = 20 + n - 1

n = 25 which is equal to given value.

This implies N = 44

Now we have :
$$f(g(x))-3x$$
so we get f(x+3)-3x
= $$\left(x+3\right)^2-7\left(x+3\right)-3x$$
=$$x^2-4x-12$$
Now minimum value of expression = $$-\frac{D}{4a}$$ $$ \frac{\left(4ac-b^2\right)}{4a}$$
We get - (16+48)/4
= -16

$$f\left(x\right)=\ x^2+ax+b$$ 

$$f\left(x+1\right)=x^2+2x+1+ax+a+b$$

$$f\left(x-1\right)=x^2-2x+1+ax-a+b$$

$$ g(x)=f(x+1)-f(x-1)= 4x+2a$$

Now $$g(20) = 72$$ from this we get $$a = -4$$ ; $$f\left(x\right)=x^2-4x\ +b$$

For this expression to be greater than zero it has to be a perfect square which is possible for $$b\ge\ 4$$

Hence the smallest value of 'b' is 4.

The graphs of $$2^{x}+2^{-x} and 2-(x-2)^{2}$$ never intersect. So, number of solutions=0.

Alternate method:

We notice that the minimum value of the term in the LHS will be greater than or equal to 2 {at x=0; LHS = 2}. However, the term in the RHS is less than or equal to 2 {at x=2; RHS = 2}. The values of x at which both the sides become 2 are distinct; hence, there are zero real-valued solutions to the above equation.

The area of the region contained by the lines $$\mid x\mid-y\leq1,y\geq0$$ and $$y\leq1$$ is the white region.

Total area = Area of rectangle + 2 * Area of triangle = $$2+\left(\frac{1}{2}\times\ 2\times\ 1\right)\ =3$$

Hence, 3 is the correct answer.

Let x(1) be the least number and x(10) be the largest number. Now from the condition given in the question , we can say that

x(2)+x(3)+x(4)+........x(10)= 47*9=423...................(1)

Similarly x(1)+x(2)+x(3)+x(4)................+x(9)= 42*9=378...............(2)

Subtracting both the equations we get x(10)-x(1)=45

Now, the sum of the 10 observations from equation (1) is 423+x(1)

Now the minimum value of x(10) will be 47 and the minimum value of x(1) will be 2 . Hence minimum average 425/10=42.5

Maximum value of x(1) is 42. Hence maximum average will be 465/10=46.5

Hence difference in average will be 46.5-42.5=4 which is the correct answer

The required figure is a trapezium with vertices A(0,0), B(2,0), C(2,4) and D(0,6)

AB = 2 BC = 4 and AD = 6

Area of trapezium = $$\frac{1}{2}\left(sum\ of\ the\ opposite\ sides\right)\cdot height$$ = $$\frac{1}{2}\left(4+6\right)\cdot2\ =\ 10$$

The given function is equivalent to f(x) = $$a^x$$

Given, f(5) = 4

=> $$a^5=4=>\ a=4^{\frac{1}{5}}$$

=> f(x) = $$4^{\frac{x}{5}}$$

f(10) - f(-10) = 16 - 1/16 = 15.9375

Let 'r' be the root of the function. It follows that f(r) = 0. We can represent this as $$f\left(r\right)=f\left\{5-\left(5-r\right)\right\}$$

Based on the relation: $$f\left(5-x\right)=f\left(5+x\right)$$; $$f\left(r\right)=f\left\{5-\left(5-r\right)\right\}=f\left\{5+\left(5-r\right)\right\}$$

$$\therefore\ f\left(r\right)=f\left(10-r\right)$$

Thus, every root 'r' is associated with another root '(10-r)' [these form a pair]. For even distinct roots, in this case four, let us assume the roots to be as follows: $$r_1,\ \left(10-r_1\right),\ r_2,\ \left(10-r_2\right)$$

The sum of these roots = $$r_1\ +\left(10-r_1\right)+\ r_2+\ \left(10-r_2\right)\ =\ 20$$

Hence, Option D is the correct answer.

$$x^{2}-2\mid x\mid+\mid a-2\mid=0$$

where x>= 0 and x>=2

$$x^2-2x+a-2\ =0$$ Using quadratic equation we have $$x=\ 1+\sqrt{\ 3-a}\ and\ x=1-\sqrt{\ 3-a}$$ Only two integer values are possible

a=2 and a=3. So corresponding "x" values are x=1 and a=3, x=2 and a=2, x=0 and a=2

where x>=0 and x<2

Applying the above process we get x=1 and a=1

where x<0 and x>=2 we get a=3 and x=-1 , a=2 and x=-2

where x<0 and x<2 we get a=1 and x=-1

Hence there are total 7 values possible

f (x + y) = f (x) f (y)

Hence, f(2)=f(1+1)=f(1)*f(1)=2*2=4

f(3)=f(2+1)=f(2)*f(1)=4*2=8

f(4)=f(3+1)=f(3)*f(1)=8*2=16

.......=> f(x)=$$2^x$$

Now, f(a + 1) +f (a + 2) + ... + f(a + n) = 16 (2$$^n$$ - 1)

On putting n=1 in the equation we get, f(a+1)=16   => f(a)*f(1)=16  (It is given that f (x + y) = f (x) f (y))

=> $$2^a$$*2=16

=> a=3

Assuming m is even, then 8f(m+1)-f(m)=2

m+1 will be odd

So, 8(m+1+3)-m(m+1)=2

=> 8m+32-$$m^2-m$$=2

=> $$m^2-7m-30=0$$

=> m=10,-3 

Rejecting the negative value, we get m=10

Assuming m is odd, m+1 will be even.

then, 8(m+1)(m+2)-m-3=2

=> 8($$m^2+3m+2$$)-m-3=2

=> $$8m^2+23m+11=0$$

Solving this, m = -2.26 and -0.60

Hence, the value of m is not integral. Hence this case will be rejected.

$$2 \cos (x(x + 1)) = 2^x + 2^{-x}$$

The maximum value of LHS is 2 when $$\cos (x(x + 1))$$ is 1 and the minimum value of RHS is 2 using AM $$\geq$$ GM 

Hence LHS and RHS can only be equal when both sides are 2. For LHS, cosx(x+1)=1   => x(x+1)=0   => x=0,-1

For RHS minimum value, x=0

Hence only one solution x=0

Sum of the area of region I and II is the required area.

Now, required area =  $$\ 4\times\frac{\ 1}{2}\times\ 1\times\ 1$$ = 2

Given, f(mn) = f(m)f(n)
when m= n= 1, f(1) = f(1)*f(1) ==> f(1) = 1

when m=1,  n= 2, f(2) = f(1)*f(2) ==> f(1) = 1

when m=n= 2, f(4) = f(2)*f(2) ==> f(4) = $$[f(2)]^2$$

Similarly f(8) = f(4)*f(2) =$$[f(2)]^3$$

f(24) = 54

$$[f(2)]^3$$ * $$[f(3)]$$ = $$3^3*2$$

On comparing LHS and RHS, we get 

f(2) = 3 and f(3) = 2

Now we have to find the value of f(18)

f(18) = $$[f(2)]$$ * $$[f(3)]^2$$

= 3*4=12

The minimum value of the function will occur when the expressions inside the function are equal.
So, 5$$x$$ = $$52 - 2x^2$$
or, $$2x^2 + 5x - 52$$ = 0
On solving, we get $$x$$ = 4 or $$-\dfrac{13}{2}$$
But, it is given that $$x$$ is a positive number.
So, $$x$$ = 4
And the minimum value = 5*4 = 20
Hence, 20 is the correct answer.

f(x) = min ($${2x^{2},52-5x}$$)
The maximum possible value of this function will be attained at the point in which $$2x^2$$ is equal to $$52-5x$$.

$$2x^2 = 52-5x$$
$$2x^2+5x-52=0$$
$$(2x+13)(x-4)=0$$
=> $$x=\frac{-13}{2}$$ or $$x = 4$$
It has been given that $$x$$ is a positive real number. Therefore, we can eliminate the case $$x=\frac{-13}{2}$$.
$$x=4$$ is the point at which the function attains the maximum value. $$4$$ is not the maximum value of the function. 
Substituting $$x=4$$ in the original function, we get, $$2x^2 = 2*4^2= 32$$.
f(x) = $$32$$.
Therefore, 32 is the right answer.

$$f(x + 2) = f(x) + f(x + 1)$$
As we can see, the value of a term is the sum of the 2 terms preceding it. 

It has been given that $$f(11) = 91$$ and $$f(15) = 617$$.
We have to find the value of $$f(10)$$.

Let $$f(10)$$ = b
$$f(12)$$ = b + 91
$$f(13)$$ = 91 + b + 91 = 182 + b
$$f(14)$$ = 182+b+91+b = 273+2b
$$f(15)$$ = 273+2b+182+b = 455+3b
It has been given that 455+3b = 617
3b = 162
=> b = 54

Therefore, 54 is the correct answer. 

$$f[f(g(1)) + g(f(1))]$$ 
= $$f[f(2^1) + g(1^2)]$$
= $$f[f(2) + g(1)]$$
= $$f[2^2 + 2^1]$$
= $$f(6)$$
= $$6^2 = 36$$

$$f_{1}(x)=x^{2}+11x+n$$ and $$f_{2}(x) = x$$
$$f_{1}(x)=f_{2}(x)$$
=> $$x^{2}+11x+n = x$$
=> $$ x^2 + 10x + n = 0 $$
=> For this equation to have distinct real roots, b$$^2$$-4ac>0
$$ 10^2 > 4n$$
=> n < 100/4 
=> n < 25
Thus, largest integral value that n can take is 24.

f(1 * 1) = f(1)f(1)
=> f(1) = f(1)f(1)
=> f(1) = 0 or f(1) = 1
Hence maximum value of f(1) is 1

$$|f(x)+ g(x)| = |f(x)| + |g(x)|$$ if and only if

case 1: $$f(x) \geq 0$$ and $$g(x) \geq 0$$
<=> $$ 2x-5 \geq 0 $$ and $$7-2x \geq 0$$
<=> $$ x \geq \frac{5}{2}$$ and $$ \frac{7}{2} \geq x$$ 

<=> $$\frac{5}{2}\leq x\leq\frac{7}{2}$$

case 2: $$f(x) \leq 0$$ and $$g(x) \leq 0$$

<=> $$ 2x-5 \leq 0 $$ and $$7-2x \leq 0$$
<=> $$ x \leq \frac{5}{2}$$ and $$ \frac{7}{2} \leq x$$
 So x<=5/2 and x>=7/2 which is not possible.

Hence, answer is 

<=> $$\frac{5}{2}\leq x\leq\frac{7}{2}$$

$$f(3) = \frac{15 + 2}{9 - 5} = \frac{17}{4}$$
$$f(f(3)) = \frac{5*17/4 + 2}{3*17/4 - 5} = \frac{93/4}{31/4} = \frac{93}{31} = 3$$
$$g(f(f(3))) = 3^2 - 3*2 - 1 = 2$$

$$f(1)^2$$ = f(1) => f(1) = 1

f(2)*(f(1/2) = f(1) => 4x = 1

So, f(1/2) = 1/4

For seed (n) = 9, all the numbers below 500 must have a digit sum of 9.

These numbers are all divisible by 9.

So total number of numbers below 500 and divisible by 9 is 55.

Consider the first term:
$$\sqrt{1+1/1^2+1/2^2}$$ = $$\sqrt{9/4}$$ = 3/2
Second term: $$\sqrt{1+1/2^2+1/3^2}$$ = $$\sqrt{49/36}$$ = 7/6
First term + Second term = 3/2 + 7/6 = (9+7)/6 = 16/6 = 8/3 = 3 - 1/3
.
.
​Required sum = 2008 - 1/2008

f(3) = 9a + 3b + c = 0 f(5) = 25a + 5b + c
f(2) = 4a + 2b + c
f(5) = -3f(2) => 25a + 5b + c = -12a -6b -3c
=> 37a + 11b + 4c = 0 --> (1)
4(9a + 3b + c) = 36a + 12b + 4c = 0 --> (2)
From (1) and (2), a - b = 0 => a = b
=> c = -12a
The equation is, therefore, $$ax^2 + ax - 12a = 0 => x^2 + x - 12 = 0$$
=> -4 is a root of the equation.

f(3) = 9a + 3b + c = 0 f(5) = 25a + 5b + c
f(2) = 4a + 2b + c
f(5) = -3f(2) => 25a + 5b + c = -12a -6b -3c
=> 37a + 11b + 4c = 0 --> (1)
4(9a + 3b + c) = 36a + 12b + 4c = 0 --> (2)
From (1) and (2), a - b = 0 => a = b
=> c = -12a
The equation is, therefore, $$ax^2 + ax - 12a = 0 => x^2 + x - 12 = 0$$
a + b + c = a + a - 12a = -10a.
But the value of a is not given. Therefore, the value cannot be determined.

According to given conditions we get f(2)=f(1)/3 , then f(3)=f(1)/6, then  f(4)=f(1)/10 , then f(5)=f(1)/15 .

We can see the pattern here that the denominator goes on increasing from 3,3+3,6+4,10+5,15+6,.. so for the f(9) the denominator will be same as 15+6+7+8+9=45 .

So f(9)=3600/45 = 80

$$a_n + b_n$$ for even n = $$p*b_{n-1} + q*b_{n-1}$$

= $$(p+q)*b_{n-1}$$

$$b_{n-1} = q*a_{n-2} = qp*b_{n-3}$$

= $$q^2*p*a_{n-4} = q^2p^2*b_{n-5}$$

.

.

= $$(qp)^{n/2-1}*b_1 = (qp)^{n/2-1}*q$$

So, $$a_n + b_n$$ = $$q(pq)^{(n/2) -1}(p+q)$$

$$a_{n} + b_{n}$$ (n is odd) = $$p^{\frac{n+1}{2}}*q^{\frac{n-1}{2}} + p^{\frac{n -1}{2}}*q^{\frac{n+1}{2}}$$ = $$(p + q)pq^{\frac{n-1}{2}}$$

Substituting the values of p and q we get

$$a_{n} + b_{n}$$ = $$(\frac{2}{9})^{\frac{n-1}{2}}$$

Now substitute the values of n and check. 

We can see that the lowest value of n for which 

$$a_{n} + b_{n}$$ < .01 is 9

For n, the number of elements in set S is $$^nC_2$$.

Lets say the 2 friends are (x,a) and (y,a)

These two friends have 3 numbers in total and 1 common element(say a) (as both elements cannot be exactly same)

They have 2 non common elements(x, y)  

The number of common friends is formed by the non-common elements of the friends (x,y) + the number of elements in the set which have the common element other than the two friends (a,c), (a,d) and so on = 1 + (n-1 - 2) = n-2.

For the example in question, if the friends are (1,2) and (1,3), then common friends are (2,3) and all other elements with 1

All elements with 1 = n-1= 3 which are (1,2) (1,3) (1,4) excluding the friends (1,2) and (1,3) only 1 other friend is common. Hence it is 1+(n-1)-2=n-2

Any ordered pair has 2 elements => There are n-2 elements that are not present in the ordered pair.

The number of enemies of any ordered pair is all the ordered pairs in the set formed using the numbers other than these two elements = $$^{n-2}C_2$$ = $$1/2 * (n^2 - 5n + 6)$$.

For a normal graph with y and x-axis, the equation of the line passing through the origin is y =mx where m is the slope of the line.

m is +ve if the angle made by the line with the x-axis is < $$90^{\circ\ }$$

$$\therefore\ $$ The equation of the line in the given graph would be y-x = k( y+ x) since the axes are y-x and y+x and the line is passing through the origin.

k > 1 because the angle is greater than 45$$^{\circ\ }$$

$$y=\dfrac{x\left(k+1\right)}{1-k}$$

Since k>1

Therefore y<0 for x>-1 and y>0 for x<-1

Option d correctly satisfy this condition

The minimum value is obtained when 2x+1 = 3-4x => 6x = 2 => x = 1/3
So, f((x) = 2*1/3 + 1 = 5/3

Using given condition we find $$a_2$$ = 5 and $$a_3$$ = 21 and so on.

We see that the numbers are of form $$3^n-(2*n)$$

So for 100 we have $$3^{100}-200$$

If the moduli are removed, the equations formed are

x+y+x-y = 4 => x=2

x+y-x+y = 4 => y =2

-x-y+x-y = 4 => y=-2

-x-y-x+y = 4 => x=-2

The area enclosed by these equations is a square with vertices at (2,2), (-2,2), (-2,-2), (2,-2) as shown in figure.

The required area = 4*4 = 16

According to given condition we have , g(x+1) = -g(x-1) + g(x)

Putting x=x+1 we get g(x+2) = g(x+1) - g(x) = -g(x-1)

Putting x=x+2 we get g(x+3)=-g(x)

Similarly g(x+4)=-g(x+1), g(x+5)=-g(x+2)=-g(x+1) + g(x)  and g(x+6) = g(x+1)-g(x+2)=g(x).

So p=6.

f(1) = 1-4+p = p-3
f(0) = p
Since they are of opposite signs, p(p-3) < 0
=> 0 < p < 3

$$ y = \frac{1}{2+\frac{1}{3+\frac{1}{2+\frac{1}{3+…}}}}$$

which is equal to $$ y = \frac{1}{2+\frac{1}{3+y}} $$ solving we get 

$$ y = \frac{3+y}{7+2y} $$ we get

$$2y^2+6y-3=0$$ .

Solution of this equation is $$\frac{\sqrt{15}-3}{2}$$. 

$$f(x) = ax^2 - b|x|$$. When $$x=0, f(x) = 0$$
When $$a > 0$$ and $$b < 0$$,
For x > 0, $$f(x) = ax^2 - bx$$, will be greater than 0 as $$ax^2 > 0$$ and $$bx<0$$ as $$b$$ is negative and $$x$$ is positive.
For x < 0, $$f(x) = ax^2 + bx$$ will again be greater than 0 as $$ax^2 >0$$ and $$bx>0$$ as both $$b$$ and $$x$$ are negative. 

Therefore, the function $$f(x)$$ is positive when $$x<0$$ and when $$x>0$$ but becomes 0 when $$x=0$$.

Therefore, for $$a > 0$$ and $$b < 0$$, f(x) will attain its minimum value at $$x = 0$$.

a = r(b+c)

b = r(a+c)

c = r(a+b)

On adding all the equations,

a+b+c = 2r(a+b+c)

If r = 1/2, a+b+c = a+b+c (valid)

If r = -1, a+b+c = -2(a+b+c) => a+b+c = 0 => b+c = -a and a/(b+c) = a/(-a) = -1 (valid)

So, r can take the values 1/2 or -1

Checking for different values of x . Suppose x= -0.5 we get

$$f_1(x)f_2(x) = 0*0.5 = 0$$

$$f_2(x)f_4(x) = 0.5*0 = 0$$ .

But $$f_2(x)f_3(x)$$ is not equal to zero.

Hence two functions are necessarily equal to zero and two products given above are equal to zero.

Relation between f3 and f1 would be

$$f_3(x) = -f_1(-x)$$.

Put x= -x we get

$$f_3(-x) = -f_1(x)$$ so multiply by -1 we get

$$-f_3(-x) = f_1(x)$$.

$$2^x - x - 1 = 0$$ for this equation only 0 and 1 i.e 2 non-negative solutions are possible. Or we can plot the graph of $$2^x$$ and x+1 and determine the number of points of intersection and hence the solutin.

Graph of logx goes on increasing in 1st quadrant and graph of 1/x goes no decreasing with both intersecting only once

Smallest possible value would be at 5-x = x+2 i.e. x= 1.5 as shown

 Substituting we get smallest value as 3.5. 

f(x) = |x - 2| + |2.5 - x| + |3.6 - x|

For x belonging to (-infinity to 2), f(x) = 2-x + 2.5-x + 3.6-x = 8.1-3x

This attains the minimum value at x=2. Value = 2.1

For x belonging to (2 to 2.5), f(x) = x-2 + 2.5-x + 3.6-x = 4.1-x

Attains the minimum value at x = 2.5. Value = 1.6

For x belonging to (2.5 to 3.6), f(x) = x-2 + x-2.5 + 3.6-x = x-0.9

Attains the minimum at x=2.5, value = 1.6

For x > 3.6, f(x) = x-2+x-2.5+x-3.6 = 3x - 8.1

Attains the minimum at x= 3.6, value = 2.7

So, min value of the function is 1.6 at x=2.5

Equate the 2 equations we get value of x = 1 and -1 . Also we notice that there is intersection at x=0 . hence D

Consider different values of x and y:

x = -1.5 and y = -1.5; x = 1.5 and y = -1.5; x = -1.5 and y = 1.5; x = 1.5 and y = 1.5.

For these possibilities, options A,B and C gets satisfied , but it is impossible to find any two positive real numbers x and y for which L(x, y) > R(x, y).

y=a+bx is a linear function and clearly the function shown in the table is not linear.
Lets take option B:

Suppose $$y=f(x)=a+bx+cx^2$$
f(1)=4 => a+b+c=4 -- (1)
f(2)=8 => a+2b+4c=8 -- (2)
f(3)=14 => a+3b+9c=14 -- (3)

Let us solve the equations and see if $$f(4), f(5),f(6)$$ also satisfy the given equation.

(2)-(1) => b+3c=4
(3)-(2) => b+5c=6
This gives c=1 and b=1 ==> a=2

So, $$f(x)=2+x+x^2$$
$$f(4)=2+4+4^2 = 22$$
$$f(5)=2+5+5^2 = 32$$
$$f(6)=2+6+6^2 = 44$$

As all the three also satisfy the numbers given in the table, it can be inferred that the relationship between x and y is quadratic and the correct option is option (b)

|x| = 1 and |y| = 1 form a square of area = 2*2 = 4 sq units

|x+y| = 1 forms a set of parallel lines cutting the axes at (1,0), (0,1), (-1,0) and (0,-1). The graph is as shown:

The area bounded by the three curves is 2*2 - 1/2*1*1*2 = 4 - 1 = 3 sq units

The union of the two sets is the set of positive integers. Also, given the increasing nature of elements, either f(1) or g(1) must be equal to 1. If g(1) = 1, then f(f(1)) =0 which cannot be under the given conditions.

Hence, f(1) = 1

g(1) = f(f(1))+1 = 2

For f(1,2). First consider x=0 and y=1 and use 3rd given equation, we get f(0,f(1,1)) now for f(1,1) take x=0 and y=0 we get f(0,f(1,0)), for f(1,0) which we use 2nd equation we get f(0,1) whose value is 2. So we have f(0,f(1,0))= f(0,2) whose value is 3 then put this in f(0,f(1,1)) we get f(0,3) we get as 4

f(2) = 1/3

$$f^2(2)$$ = f(1/3) = 3/4

$$f^3(2)$$ = f(3/4) = 4/7

$$f^4(2)$$ = f(4/7) = 7/11

$$f^5(2)$$ = f(7/11) = 11/18

So, product = 1/18

f(-2) = -1

$$f^2(-2)$$ = f(f(-2)) = f(-1) = 0

$$f^3(-2)$$ = f(f(-2)) = f(0) = 1

So, sum = 0

f(x,y) is always non-negative because $$(x+y)^2$$ is always positive

g(x, y) = $$(x + y)^2$$, if $$\sqrt{(x + y)}$$ is real = Always positive
g(x, y) = $$- (x + y)$$ otherwise = Always positive because this happen when (x+y)<0 and -(x+y) is always greater than zero.

f(x,y)+g(x,y) = Always positive

When both x and y are less than -1, g(x,y) > 2 and f(x,y) > 4 and $$f(x,y) = g(x,y)^2$$
So, f(x,y) > g(x,y)

In the given graph, the value of y is constant irrespective of the value of x. Hence value of y would be same for a particular x and -x.

Hence f(x) = f(-x)

In this graph, f(-1) = 1 and f(1) = 2 and since the two lines pass origin, their values increase linearly.

3f(x) = 6f(-x)

In this graph, we can see that the graph on LHS of y axis is symmetrical to the graph on RHS of y axis about origin.

Hence f(x) = -f(-x)

From the given functions we can make out that function h and m give max value , function n and j give min value , function f and g give middle  value. From this equation (f(x, y, z) + h(x, y, z)-g(x, y, z))/j(x, y, z) , numerator is always max value and denominator is min value . So this will always be greater than 1 .

Suppose x>y>z

f(x,y,z) = y

g(x,y,z) = y

h(x,y,z) = x

j(x,y,z) = z

Option d = x/z >1

From the given functions we can make out that function h and m give max value , function n and j give min value , function f and g give middle  value. So according to equation (f(x, y, z)- m(x, y, z))/(g(x, y, z)-h(x,y, z)) , value of numerator and denominator is equal and hence ratio is equal to 1. 

Suppose x>y>z

f(x,y,z) = y

g(x,y,z) = y

h(x,y,z) = x

j(x,y,z) = z

Option a = (y-x)/(y-x) = 1

From the given functions we can make out that function h and m give max value , function n and j give min value , function f and g give middle  value.So in option B , j cancels out n and h cancels out m . So the denominator becomes 0 and value is indeterminable. 

Suppose x>y>z

f(x,y,z) = y

g(x,y,z) = y

h(x,y,z) = x

j(x,y,z) = z

m(x,y,z) = x

n(x,y,z) = z

The denominator of the second option becomes 0, hence making it indeterminate.

Let p and q be any two elements of the set A.
For the computation of the GCF of elements of the set A, we can replace both p and q by just the GCF(p,q) and the result is unchanged.

So, for every application of the function h, we are reducing the number of elements of the set A by 1. (In this case two numbers p and q are replaced by one number GCF(p,q)).
Expanding this concept further, the minimum number of times the function h should be called is n-1

The correct relation between the two is: F(x) = | F1(x) |

So, all the three options a), b) and c) can be ruled out. Option d) is the correct answer.

The value of F(x) for x < 0 is the same as the value of F1(x) for x > 0.

So, F1(x) = F(-x)

Option b) is the correct answer.

The value of F(x) for x > 0 is the same as the value of F1(x) for x < 0.

So, F1(x) = F(-x)

Option b) is the correct answer.

F(0) = 1 ; F1(0) = -1

F(1) = 0 ; F1(-1) = 0

F(2) = -1 ; F1(-2) = 1

=> F1(x) = -F(-x).

Before, the third instruction, the point on which the robot is present is (6,2).

Before, the second instruction, the point on which the robot is present is (4,2).

Hence, the values of x and y are 4 and 2 respectively.

WALKX(-x) and WALKY(y) are the commands to be used for the robot to reach origin in any order.

Hence, the answer is 2.

f(x,y) = |x+y|

F(f(x,y)) = -f(x,y) = -|x+y|

G(f(x,y)) = -F(f(x,y)) = |x+y|

Option A: F(f(x, y)) . G(f(x, y)) = -F(f(x, y)) . G(f(x, y))  =>LHS = -|x+y|$$^2$$, RHS = |x+y|$$^2$$  Hence false.

Option B: F(f(x, y)) . G(f(x, y)) > -F(f(x, y)) . G(f(x, y)), Since, LHS is smaller than RHS. False

Option C: F(f(x, y)) . G(f(x, y)) $$\neq $$ G(f(x, y)) . F(f(.x, y)), Here LHS=RHS. Hence false.

Option D:

=> G(f(x,y)) + F(f(x,y)) = 0

f(x,y) = f(-x,-y)

=> G(f(x,y)) + F(f(x,y)) + f(x,y) = f(-x.-y)

F(f(x,y)) = -f(x,y)

G(f(x,y)) = -F(f(x,y)) = f(x,y)

G(f(1,0)) = 1

F(f(1,2)) = -3

G(f(1,2)) = 3

f(F(f(1,2)),G(f(1,2))) = 0

f(G(f(1, 0)), f(F(f(1, 2)), G(f(1, 2)))) = 1 + 0 = 1

F(f(x,-x)) = 0 and G(f(x,-x)) = 0

F(f(x,x)) = -2x and G(f(x,x)) = 2x

F(f(x,x)).G(f(x,x)) = -$$4x^2$$

$$\log_216$$ = 4

=> $$\frac{-F(f(x,x)).G(f(x,x))}{\log_216}$$ = $$x^2$$

Given expression can be reduced to 

M ( x(xy+y-x) , (y+x) ) 

Or x(x+y)(xy+y-x)

After putting value of x and y , expression will reduce to a value =  70.

Given expression can be reduced to:

$$S[M(\frac{a+b}{2}),(\frac{a+b}{2})), M((\frac{a- b}{2}, (\frac{a-b}{2})]$$

= $$S[(\frac{(a+b)}{2})^2, (\frac{(a-b)}{2})^2]$$

= $$(\frac{a+b}{2})^2 - (\frac{(a-b)}{2})^2 $$

= ab

Putting y as x and x as y in given options, only options B satisfies the function property as follows.

x = $$\frac{2y+3}{3y-2}$$

or 3xy - 2x = 2y+3

or y(3x-2) = 2x+3

or $$y = \frac{2x+3}{3x-2}$$

On solving both equations, we will get $$x = \frac{23}{2+2k}$$
now for having no solutions to system 2+2k should be 0.
Hence k=-1

Among all options only D satisfies the given equations as follows:
$$f(y) = \frac{1-y}{1+y}$$
and for x:
$$y + xy = 1-x$$
$$x(1+y) = 1-y$$
$$x= \frac{1-y}{1+y}$$
Hence $$x=f(y)$$

For x<0 ; y=x+2

for 0<x<$$\frac{1}{2}$$ ; $$y=x+2$$

for $$x>\frac{1}{2}$$ ; $$y=3-x$$

Hence, $$y$$ attains its maxima at $$x=\frac{1}{2}$$  i.e. $$y$$ = 2.5