JEE Main 2024DifferentiationDifferentiationHardQuestion Let y=loge(1−x21+x2),−1<x<1. Then at x=12, the value of 225(y′−y′′) is equal to \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 } Let y=loge(1+x21−x2),−1<x<1. Then at x=21, the value of 225(y′−y′′) is equal to OptionsA732B736C742D746Check AnswerHide SolutionSolutiony=loge(1−x21+x2)dydx=y′=−4x1−x4\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}y=loge(1+x21−x2)dxdy=y′=1−x4−4x Again, d2ydx2=y′′=−4(1+3x4)(1−x4)2\frac{d^2 y}{d x^2}=y^{\prime \prime}=\frac{-4\left(1+3 x^4\right)}{\left(1-x^4\right)^2}dx2d2y=y′′=(1−x4)2−4(1+3x4) Again y′−y′′=−4x1−x4+4(1+3x4)(1−x4)2y^{\prime}-y^{\prime \prime}=\frac{-4 x}{1-x^4}+\frac{4\left(1+3 x^4\right)}{\left(1-x^4\right)^2}y′−y′′=1−x4−4x+(1−x4)24(1+3x4) at x=12\mathrm{x}=\frac{1}{2}x=21, y′−y′′=736225y^{\prime}-y^{\prime \prime}=\frac{736}{225}y′−y′′=225736 Thus 225(y′−y′′)=225×736225=736225\left(y^{\prime}-y^{\prime \prime}\right)=225 \times \frac{736}{225}=736225(y′−y′′)=225×225736=736