Question

If $f\left(x\right)=\frac{2x-4}{x^2-1}$ and $f^{\prime }\left(x\right)=\frac{p}{{(x^2-1)}^2}$ then p equals-

• A. $x^2-8x-2$
• B. $-2x^2+8x+2$
• C. $4x+2$
• D. $-2x^2+8x-2$

Answer 1

The correct option is D $-2x^2+8x-2$

$f\left(x\right)=\frac{2x-4}{x^2-1}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{(x^2-1)\left(2\right)-(2x-4)\left(2x\right)}{{(x^2-1)}^2}$
$=\frac{{2x}^2-2-{4x}^2+8x}{{(x^2-1)}^2}=\frac{-{2x}^2+8x-2}{{(x^2-1)}^2}$
$\therefore p=-{2x}^2+8x-2$