Question

If $f\left(x\right)=\{\begin{array}{c}\begin{array}{cc}\frac{x-1}{2x^2-7x+5},& x\ne 1\end{array}\\ \begin{array}{cc}-\frac{1}{3},& x=1\end{array}\end{array}$ then $f^{\prime }\left(1\right)$ is equal to

• A. $-\left(\frac{1}{9}\right)$
• B. $-\left(\frac{2}{9}\right)$
• C. $-13$
• D. $\frac{1}{3}$

Answer 1

Answer: (B)

$-\left(\frac{2}{9}\right)$

$f^{\prime }\left(1\right)=\frac{d}{dx}\left(\frac{x-1}{2x^2-7x+5}\right)$

$=\frac{d}{dx}\left[\frac{x-1}{(x-1)(2x-3)}\right]$

$=\frac{d}{dx}\left[\frac{1}{2x-5}\right]$

$\Rightarrow f^{\prime }\left(1\right)=-\frac{2}{{(2x-5)}^2}$ at $x=1$

$=-\frac{2}{{(2-5)}^2}=-\frac{2}{9}$