Jee Main 2024 29 01 2024 Shift 2 - Question14

Answer: (736)

Solution:

$f(x)=\ln \left(1-x^{2}\right)-\ln \left(1+x^{2}\right)$

$$ \begin{alignedat} & f^{\prime}(x)=\frac{-2 x}{1-x^{2}}-\frac{2 x}{1+x^{2}} \\ & =-2 x\left[\frac{2}{1-x^{4}}\right] \end{aligned} $$

$$ f^{\prime}(x)=\frac{4 x}{x^{4}-1} $$

$f^{\prime \prime}(x)=4\left[\frac{\left(x^{4}-1\right)-4 x^{4}}{\left(x^{4}-1\right)^{2}}\right]$

$=4\left[\frac{-3 x^{4}-1}{\left(x^{4}-1\right)^{2}}\right]$

$f^{\prime}(x)-f^{\prime \prime}(x)=4\left[\frac{x}{x^{4}-1}+\frac{3 x^{4}+1}{\left(x^{4}-1\right)^{2}}\right]$

At $x=\frac{1}{2}$

$225\left[f^{\prime}(x)-f^{\prime \prime}(x)\right]=736$