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

The Correct Option is 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 $ $\therefore a -2b = - 1$