Question

$\frac{d}{dx}\left[a{\tan }^{-1}x+b\log \left(\frac{x-1}{x1}\right)\right]=\frac{1}{x^4-1}\Rightarrow a-2b=$

• A. 1
• B. -1
• C. 0
• D. 2

Answer 1

Answer: (B)

$\frac{d}{dx}\left[a{\tan }^{-1}x+b\log \left(\frac{x-1}{x+1}\right)\right]=\frac{1}{x^4-1}$
$\Rightarrow \frac{a}{1+x^2}+b\left[\frac{1}{x-1}-\frac{1}{x+1}\right]=\frac{1}{x^4-1}$
$\Rightarrow \frac{a}{1+x^2}+b.\frac{2}{x^2-1}=\frac{1}{x^4-1}$
$\Rightarrow \frac{a\left(x^2-1\right)+2b\left(1+x^2\right)}{x^4-1}=\frac{1}{x^4-1}$
$\Rightarrow a\left(x^2-1\right)+2b\left(1+x^2\right)=1$
$\Rightarrow \left(a+2b\right)x^2+\left(2b-a\right)=1$
$\therefore 2b-a=1$ and $a+2b=0$
$\therefore a-2b=-1$