JEE MAIN - Mathematics (2024 - 29th January Evening Shift - No. 10)

$$\text { Let } y=\log _e\left(\frac{1-x^2}{1+x^2}\right),-1 < x<1 \text {. Then at } x=\frac{1}{2} \text {, the value of } 225\left(y^{\prime}-y^{\prime \prime}\right) \text { is equal to }$$
732
736
742
746

Explanation

$$\begin{aligned} & y=\log _e\left(\frac{1-x^2}{1+x^2}\right) \\ & \frac{d y}{d x}=y^{\prime}=\frac{-4 x}{1-x^4} \end{aligned}$$

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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