Question

Solve the question in attachment.
Please elaborate all the steps involved.

Answer 1

$\large\underline{\sf{Solution-}}$

Given that f(x) is a polynomial of second degree.

So, let assume that

$\rm \: f(x) = {ax}^{2} + bx + c$

Further given that,

$\red{\rm \: f(1) = f( - 1) \: }$

$\rm \: a {(1)}^{2} + b(1) + c = a {( - 1)}^{2} + b( - 1) + c$

$\rm \: a + b + c = a - b + c$

$\rm \: b = - b$

$\rm \: 2b = 0$

$\rm\implies \:b = 0$

Further, given that, a, b, c are in AP

$\rm\implies \:b - a = c - b$

$\rm\implies \:2b = a + c$

Now, as b = 0, So, given quadratic polynomial can be rewritten as

$\rm \: f(x) = {ax}^{2} + c$

On differentiating both sides w. r. t. x, we get

$\rm \: \dfrac{d}{dx}f(x) = \dfrac{d}{dx}({ax}^{2} + c)$

$\rm\implies \:f'(x) = 2ax$

So,

$\rm\implies \:f'(a) = 2 {a}^{2}$

$\rm\implies \:f'(b) = 2ab$

$\rm\implies \:f'(c) = 2ac$

Now, Consider

$\rm \: f'(a) + f'(c)$

$\rm \: = \: {2a}^{2} + 2ac$

$\rm \: = \: 2a(a + c)$

$\rm \: = \: 2a(2b)$

$\rm \: = \: 2(2ab)$

$\rm \: = \: 2f'(b)$

$\bf\implies \:f'(a), \: f'(b), \: f'(c) \: are \: in \: AP$

So, Option (1) is correct.