Verified Solution • 2025–26 Session moderate

QJEEM24Ma36a0: JEE Main 2024 Mathematics - Class 12 JEEM24Ma36a0 Solution

\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 }

Multiple Choice Options

  • A 732
  • B 736
  • C 742
  • D 746

Step-by-Step Verified Solution

amp;y=log⁡e(1−x21+x2)amp;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}

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}

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}

at x=12\mathrm{x}=\frac{1}{2} ,

y′−y′′=736225y^{\prime}-y^{\prime \prime}=\frac{736}{225}

Thus 225(y′−y′′)=225×736225=736225\left(y^{\prime}-y^{\prime \prime}\right)=225 \times \frac{736}{225}=736

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