QJEEAdv18Mab32a: JEE Advanced 2018 Mathematics - Class 12 JEEAdv18Mab32a Solution: For every twice differentiable function with , which of the…

We have,

${(f(0))^2} + {(f'(0))^2} = 85$

and $f:R \to [ - 2,2]$

(a) Since, f is twice differentiable function, so f is continuous function.

$\therefore$ This is true for every continuous function.

Hence, we can always find x $\in$ (r, s), where f(x) is one-one.

$\therefore$ This statement is true.

(b) By L.M.V.T.

$f'(c) = {{f(b) - f(a)} \over {b - a}}$

$\Rightarrow |f'(c)| = \left| {{{f(b) - f(a)} \over {b - a}}} \right|$

$\Rightarrow |f'({x_0})| = \left| {{{f(0) - f( - 4)} \over {0 + 4}}} \right|$

$= \left| {{{f(0) - f( - 4)} \over 4}} \right|$

Range of f is [ $-$ 2, 2]

$\therefore$ $- 4 \le f(0) - f( - 4) \le 4$

$\Rightarrow 0 \le \left| {{{f(0) - f( - 4)} \over 4}} \right| \le 1$

Hence, |f'(x₀)| = 1.

Hence, statement is true.

(c) As no function is given, then we assume

$f(x) = 2\sin \left( {{{\sqrt {85} x} \over 2}} \right)$

$\therefore$ $f'(x) = \sqrt {85} \cos \left( {{{\sqrt {85} x} \over 2}} \right)$

Now, ${(f(0))^2} + {(f'(0))^2} = {(2\sin 0)^2} + {(\sqrt {85} \cos 0)^2}$

${(f(0))^2} + {(f'(0))^2} = 85$

and $\mathop {\lim }\limits_{x \to \infty } f(x)$ does not exists.

Hence, statement is false.

(d) From option b, $|f'({x_0})|\, \le 1$

${x_0} \in ( - 4,0)$

$\therefore$ ${(f'({x_0}))^2}\, \le 1$

Hence, $g({x_0}) = {(f({x_0}))^2} + {(f'({x_0}))^2}\, \le 4 + 1$

[ $\because$ f(x₀) $\in$ [ $-$ 2, 2]

$\Rightarrow g({x_0})\, \le 5$

Now, let p $\in$ ( $-$ 4, 0) for which g(p) = 5

Similarly, let q be smallest positive number q $\in$ (0, 4)

such that g(q) = 5

Hence, by Rolle's theorem is (p, q)

g'(c) = 0 for $\alpha$ $\in$ ( $-$ 4, 4) and since g(x) is greater than 5 as we move from x = p to x = q

and f(x))² $\le$ 4

$\Rightarrow$ (f'(x))² $\ge$ 1 in (p, q)

Thus, g'(c) = 0

$\Rightarrow$ f'f + f'f" = 0

So, f( $\alpha$ ) + f"( $\alpha$ ) = 0 and f'( $\alpha$ ) $\ne$ 0

Hence, statement is true.