QJEEAdv15Ma48be: JEE Advanced 2015 Mathematics - Class 12 JEEAdv15Ma48be Solution: Let be a continuous odd function, which vanishes exactly at one …

Here, $\mathop {\lim }\limits_{x \to 1} {{F(x)} \over {G(x)}} = {1 \over {14}}$

$\Rightarrow \mathop {\lim }\limits_{x \to 1} {{F'(x)} \over {G'(x)}} = {1 \over {14}}$ [using L' Hospital's rule] ....... (i)

As $F(x) = \int_{ - 1}^x {f(t)dt}$

$\Rightarrow F'(x) = f(x)$ ...... (ii)

and $G(x) = \int_{ - 1}^x {t|f\{ f(t)\} |dt}$

$\Rightarrow G'(x) = x|f\{ f(x)\} |$ ...... (iii)

$\therefore$ $\mathop {\lim }\limits_{x \to 1} {{F(x)} \over {G(x)}} = \mathop {\lim }\limits_{x \to 1} {{F'(x)} \over {G'(x)}} = \mathop {\lim }\limits_{x \to 1} {{f(x)} \over {x|f\{ f(x)\} |}}$

$= {{f(1)} \over {1|f\{ f(1)\} |}} = {{1/2} \over {|f(1/2)|}}$ ....... (iv)

Given, $\mathop {\lim }\limits_{x \to 1} {{F(x)} \over {G(x)}} = {1 \over {14}}$

$\therefore$ ${{{1 \over 2}} \over {\left| {f\left( {{1 \over 2}} \right)} \right|}} = {1 \over {14}} \Rightarrow \left| {f\left( {{1 \over 2}} \right)} \right| = 7$