Let y = loge(1−x^2 / 1+x^2), −1 < x < 1. Then at x = 1/2. the value of

Question

Let \(y=\log _{e}\left(\frac{1-x^{2}}{1+x^{2}}\right),-1<x<1\). Then at \(x=\frac{1}{2}\), the value of \(225\left(y^{\prime}-y^{\prime \prime}\right)\) is equal to

(1) 732

(2) 746

(3) 742

(4) 736

Answer 1

Correct option is (4) 736

\(y=\log _{e}\left(\frac{1-x^{2}}{1+x^{2}}\right)\)

\(\frac{d y}{d x}=y^{\prime}=\frac{-4 x}{1-x^{4}}\)

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 \(225\left(y^{\prime}-y^{\prime \prime}\right)=225 \times \frac{736}{225}=736\)