Question

If $\frac{d}{dx}\left[a{\tan }^{-1}x+b\log \left(\frac{x-1}{x+1}\right)\right]=\frac{1}{x^4-1}$, then $a-2b=$

• A. $1$
• B. $-1$
• C. $0$
• D. $2$

Answer 1

The correct option is B

$-1$

Explanation for the correct option:

Step 1. Find the value of $a-2b$:

Given, $\frac{d}{dx}\left[a{\tan }^{-1}x+b\log \left(\frac{x-1}{x+1}\right)\right]=\frac{1}{x^4-1}$

$\Rightarrow$$\frac{a}{a+x^2}+b\left(\frac{x+1}{x-1}\right)\frac{(x+1)-(x-1)}{{(x+1)}^2}=\frac{1}{x^4-1}$

$\Rightarrow$ $\frac{a}{1+x^2}+\frac{2b}{x^2-1}=\frac{1}{x^4-1}$

$\Rightarrow$ $a(x^2-1)+2b(x^2+1)=1$

$\Rightarrow$ $x^2\left(a+2b\right)+\left(ab-1\right)=1$

Step 2. By comparing L.H.S and R.H.S of the above equation, we get

$a+2b=1$

$\Rightarrow$ $2b-a=1$

$\mathbf{\therefore }\mathit{a}\mathbf{}\mathbf{-}\mathbf{}\mathbf{2}\mathit{b}\mathbf{}\mathbf{=}\mathbf{}\mathbf{-}\mathbf{1}$

Hence, Option ‘B’ is Correct.