JEE Main 2024 — Differentiation Question with Solution

Let, $f\left(x\right)=\log \left(\frac{1-x^2}{1+x^2}\right)$

$\Rightarrow f\left(x\right)=\log \left(1-x^2\right)-\log \left(1+x^2\right)$

$\Rightarrow f^{\text{'}}\left(x\right)=\frac{-2x}{\left(1-x^2\right)}-\frac{2x}{\left(1+x^2\right)}$

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

$\Rightarrow f^{\text{'}}\left(x\right)=\left(\frac{4x}{x^4-1}\right)$

$\Rightarrow f^{"}\left(x\right)=\frac{\left(x^4-1\right)\left(4\right)-\left(4x\right)\left(4x^3\right)}{{\left(x^4-1\right)}^2}$

$\Rightarrow f^{"}\left(x\right)=\frac{4\left(-3x^4-1\right)}{{\left(x^4-1\right)}^2}$

$\Rightarrow 225\left[f^{\text{'}}\left(x\right)-f^{"}\left(x\right)\right]=225\left[\left(\frac{4x}{x^4-1}\right)-\frac{4\left(-3x^4-1\right)}{{\left(x^4-1\right)}^2}\right]$

Putting, $x=\frac{1}{2}$

$\Rightarrow 225\left[f^{\text{'}}\left(\frac{1}{2}\right)-f^{"}\left(\frac{1}{2}\right)\right]=225\left[\left(\frac{2}{-{\displaystyle \frac{15}{16}}}\right)-\frac{4\left(-{\displaystyle \frac{3}{16}}-1\right)}{{\left({\displaystyle \frac{1}{16}}-1\right)}^2}\right]$

$\Rightarrow 225\left[f^{\text{'}}\left(\frac{1}{2}\right)-f^{"}\left(\frac{1}{2}\right)\right]=225\left[\left(\frac{-32}{15}\right)-\frac{4\left(\frac{-19}{16}\right)}{{\left({\displaystyle \frac{-15}{16}}\right)}^2}\right]$

$\Rightarrow 225\left[f^{\text{'}}\left(\frac{1}{2}\right)-f^{"}\left(\frac{1}{2}\right)\right]=225\left[\left(\frac{-32}{15}\right)+\frac{4\times 19\times 16}{225}\right]$

$\Rightarrow 225\left[f^{\text{'}}\left(\frac{1}{2}\right)-f^{"}\left(\frac{1}{2}\right)\right]=\left[-480+1216\right]=736$

$\Rightarrow 225\left[f^{\text{'}}\left(\frac{1}{2}\right)-f^{"}\left(\frac{1}{2}\right)\right]=736$

$\Rightarrow 225\left[y^{\text{'}}\left(\frac{1}{2}\right)-y^{"}\left(\frac{1}{2}\right)\right]=736$