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Jee Main 2024
29-01-2024 S2
Question
Let $\mathrm{y}=\log _{\mathrm{e}}\left(\frac{1-\mathrm{x}^2}{1+\mathrm{x}^2}\right),-1<\mathrm{x}<1$. Then at $\mathrm{x}=\frac{1}{2}$, the value of $225\left(y^{\prime}-y^{\prime \prime}\right)$ is equal to
Select the correct option:
A
732
B
746
C
742
D
736
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
Sol. $y=\log _e\left(\frac{1-\mathrm{x}^2}{1+\mathrm{x}^2}\right)$ $$ \frac{d y}{d x}=y^{\prime}=\frac{-4 x}{1-x^4} $$ Again, $$ \frac{d^2 y}{d x^2}=y^{\prime \prime}=\frac{-4\left(1+3 x^4\right)}{\left(1-x^4\right)^2} $$ Again $$ y^{\prime}-y^{\prime \prime}=\frac{-4 x}{1-x^4}+\frac{4\left(1+3 x^4\right)}{\left(1-x^4\right)^2} $$ at $\mathrm{x}=\frac{1}{2}$, $$ y^{\prime}-y^{\prime \prime}=\frac{736}{225} $$ Thus $225\left(y^{\prime}-y^{\prime \prime}\right)=225 \times \frac{736}{225}=736$
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