a+b+c=30 and ab+bc+ca=192
=> b(a+c)+ca=192
b=30-(a+c)
=> [30-(a+c)](a+c)+ac=192
=> $$30(a+c)-(a+c)^2+ca=192$$
=> $$30a+30c-a^2-c^2-ca=192$$
=> $$c^2+(a-30)c+(a^2-30a+192)=0$$
Now, c is real. This means that discriminant of the above equation in c should be $$\geq0$$.
=> $$(a-30)^2-4(a^2-30a+192)\geq0$$
=>$$a^2+900-60a-4a^2+120a-4X192\geq0$$
=> $$a^2+900-60a-4a^2+120a-4X192\geq0$$
=> $$3a^2-60a-132\leq0$$
=> $$a^2-20a-44\leq0$$
=> $$(a-22)(a+2)\leq0$$
=> $$a-22\geq 0=a\geq22$$ and $$a+2\leq 0=a\leq-2$$, which is not possible
Or $$a-22\leq 0=a\leq22$$ and $$a+2\geq 0=a\geq-2$$
Thus, max possible value is 22.

f(a) = $$\frac{6}{a} - 1$$
f(b) = $$\frac{6}{b} - 1$$
f(c) = $$\frac{6}{c} - 1$$
f(a)*f(b)*f(c) = $$\frac{6-a}{a}*\frac{6-b}{b}*\frac{6-c}{c}$$

f(a)*f(b)*f(c) = $$\frac{216 - 36(a+b+c) + 6(ab + bc + ca) - abc}{abc}$$

=> $$\frac{ 6(ab + bc + ca)}{abc} - 1$$

=> 6 (1/a + 1/b + 1/c) - 1

We know that

$$AM \geq \textrm{HM for positive numbers}$$

So, $$\frac{(6/a + 6/b + 6/c)}{3} \geq \frac{3}{a/6 + b/6 + c/6}$$

=> $$\frac{(6/a + 6/b + 6/c)}{3} \geq 3$$

$$6/a + 6/b + 6/c \geq 9$$

Thus

f(a)*f(b)*f(c) ≥ 9 -1 = 8

Thus the minimum value is 8

$$f(1) = 1$$
$$f(2) = 2*1+3 = 5 = 1+4 = f(1) + 4$$
$$f(3) = 2*5+3 = 13 = 5+8 = f(2) + 8$$
$$f(4) = 2*13+3 = 29 = 13+16 = f(3) + 16$$
and so on.
$$f(25) = f(24)+2^{25} = f(23)+2^{24}+2^{25} = f(22)+2^{23}+2^{24}+2^{25}$$
$$f(25)-f(22) = 2^{23}+2^{24}+2^{25} = 2^{23}(1+2+4) = 7*2^{23}$$

$$ 4A^{2}(x)-2B(x)B(-x)=B^{2}(x)+B^{2}(-x)$$
$$=> 4A^{2}(x)=[B(x)+B(-x)]^{2}$$
$$=>4A^{2}(x)=4A^{2}(-x)$$
$$=>A(x)= \pm A(-x)$$
$$=>A(-4) = \pm A(4) = \pm 24$$

A reflection in the like y = k followed by a reflection in the like x = p is effectively a reflection of the original graph in the point (p,k). So, to get back the original graph, the new graph has to be reflected in the point (p,k).

So, the reflection of h(x) in the point (1,1) gives f(x).

f(n) is neither always increasing nor always decreasing. For example, f(109)>f(108) and f(110)<f(109). Hence, answer is d)

Every F(X) is divisible by 3 and 8, so every F(X) is divisible by 24.
The HCF of F(7), F(9) and F(11) is 24, so the HCF of the set K is less than or equal to 24.
Hence, HCF is 24

F(2) = 40, F(3) = 20 and so on. F(n) = 40 if n is even and F(n) = 20 if n is odd. So average is 30

Consider option d:

$$g(1-x)$$ = $$(1-x)^2 * (1-(1-x)^2)$$ = $$(1-x)^2*x^2$$ = g(x).

No other option satisfies the condition.

So, option d) is the correct answer.

f(5) = -4*f(2) 25a + 5b + c = -4(4a + 2b + c) => 41a + 13b +5c =0

f(7) =0 => 49a + 7b + c =0

Two equations, three unknowns without any special condition. Hence the answer cannot be determined.

2x + 1, 4x - 3 and 5 - 2x are three straight lines.The minimum value of f(x) will occur at the intersection point of any of the two lines.

When 2x + 1 intersects 4x - 3, 2x + 1 = 4x - 3 => x = 2, then f(x) = 5

When 2x + 1 intersects 5 - 2x, x = 1 and f(x) = 3

When 4x - 3 intersects 5 - 2x, x = 4/3 and f(x) = 11/3.

Thus f(x) = 3 is the minimum value of f(x).

f(x) = $$x^4 - 6x^3 + 4x^2 - 2x + p^2$$
f(1) = $$1 - 6 + 4 - 2 + p^2$$ = $$p^2-3$$
f(2) = $$16 - 48 + 16 - 4 + p^2$$ = $$p^2-20$$
f(2) < f(1) => f(2) is negative and f(1) is positive.
=> $$p^2-20$$ < 0 and $$p^2-3$$ > 0
$$(p-2\sqrt{5})(p+2\sqrt{5}) < 0 $$
=> $$-2\sqrt{5} < p < 2\sqrt{5}$$
$$(p-\sqrt{3})(p+\sqrt{3}) > 0$$
=> $$p > \sqrt{3}$$ and $$p < -\sqrt{3}$$
=> $$-2\sqrt{5} < p < -\sqrt{3}$$ and $$\sqrt{3} < p < 2\sqrt{5}$$
=> p = -4, -3, -2, 2, 3, 4
=> p can take 6 values.

f(0.25)+2f(4) = 0.5, f(4)+2f(0.25)=8. Solving simultaneous equations, f(0.25)= 5.167

As log function is defined for only positive values, x<2. As the square root of a number is real only if it is positive $$16-x^2>0$$ Hence -4<x<4. Using both conditions -4<x<2

For the function to be a real function, the log value should be >0 i.e.$$ X^2-8>0$$ => $$x>2\sqrt{2}$$ and $$x<-2\sqrt{2}$$. As the denominator cannot be zero, x!= 5. Hence, using these two conditions we get option B

-f(x) implies f(x) reflected in the x-axis. g(-x) implies g(x) reflected in the y-axis.
The function of f(x) is symmetrical around the y-axis.
Hence g(x) is also symmetrical and hence g(x)=h(x).
i(x)=-h(x) implies reflecting in the x axis again hence f(x)=i(x)

Let $$( x^3 + 2x^2 + 2x + 1)^6 = f(x)$$.
Therefore, $$f(1) = a_0+a_1+a_2+...+a_{18}$$

$$f(-1) = a_0-a_1+a_2-...+a_{18}$$

Therefore, $$a_0+a_2+a_4+...+a_{18} = {f(1) + f(-1)}/2.$$
$$= ({(1+2+2+1)^6 +(-1+2-2+1)^6})/2 = ({6^6 + 0})/2 = 23328.$$

For a function to be onto, the cardinality of co-domain must be less than that of domain.
Here, the cardinality of domain is 5 and co-domain is 3.

By inclusion exclusion method we can find out the value of onto function.

Number of onto functions from a set of cardinality m to a set of cardinality n = $$n^m - ^nC_1(n-1)^m + ^nC_2(n-2)^m - ......$$=

$$n^m$$ is the number of ways every element of set P can link to any element of set R

$$ ^nC_1(n-1)^m $$ is number of ways every element of set P can link to any element of set R except one element.

$$  ^nC_2(n-2)^m $$ is number of ways every element of set P can link to any element of set R except two element.

..

.

 $$3^5 - ^3C_1(3-1)^5 + ^3C_2(3-2)^5 - ^3C_3(3-3)^5$$

= 243 - 96 + 3
= 150

h(1) = 14400
h(1) + h(2) = 4h(2) => h(2) = 14400/3 = 4800
h(1) + h(2) + h(3) = 9h(3) => h(3) = 19200/8 = 2400
h(1) + h(2) + h(3) + h(4) = 16h(4) => h(4) = 21600/15 = 1440
h(1) + h(2) + h(3) + h(4) + h(5) = 25h(5) => h(5) = 23040/24 = 960

f’(x)=3cosx-4sinx
f’’(x)=-3sinx-4cosx
f’(x)=0 => tan x = 3/4
f’’(x) will be negative when sin x=3/5 and cos x=4/5 and f’’(x) will be positive when sin x=-3/5 and cos x=-4/5
Thus, sin x=3/5 and cos x=4/5 corresponds to maximum value and sin x=-3/5 and cos x=-4/5 corresponds to minimum value.
f(x) max= 5 and f(x) min=-5