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If $f(x)\left\{\begin{array}{cc}\frac{x-1}{2 x^2-7 x+5}, & \text { for } x \neq 1 \\ -\frac{1}{3} & , \text { for } x=1\end{array}\right.$, then $f^{\prime}(1)$ is equal to :
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The correct answer is:
$-\frac{2}{9}$
We have,
$f(x)=\frac{x-1}{2 x^2-7 x+5}, x \neq 1$
$=\frac{(x-1)}{2 x^2-2 x-5 x+5}$
$=\frac{x-1}{2 x(x-1)-5(x-1)}=\frac{1}{2 x-5}$
$f(x)=\left\{\begin{array}{cc}\frac{1}{2 x-5}, & x \neq 1 \\ -\frac{1}{3}, & x=1\end{array}\right.$
Now, $\quad f^{\prime}(1)=\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{h}$
$=\lim _{h \rightarrow 0} \frac{\frac{1}{2(1+h)-5}-\left(-\frac{1}{3}\right)}{h}$
$=\lim _{h \rightarrow 0} \frac{\frac{1}{2 h-3}+\frac{1}{3}}{h}$
$=\lim _{h \rightarrow 0} \frac{3+2 h-3}{3 h(2 h-3)}$
$=\lim _{h \rightarrow 0} \frac{2}{3(2 h-3)}=-\frac{2}{9}$
$f(x)=\frac{x-1}{2 x^2-7 x+5}, x \neq 1$
$=\frac{(x-1)}{2 x^2-2 x-5 x+5}$
$=\frac{x-1}{2 x(x-1)-5(x-1)}=\frac{1}{2 x-5}$
$f(x)=\left\{\begin{array}{cc}\frac{1}{2 x-5}, & x \neq 1 \\ -\frac{1}{3}, & x=1\end{array}\right.$
Now, $\quad f^{\prime}(1)=\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{h}$
$=\lim _{h \rightarrow 0} \frac{\frac{1}{2(1+h)-5}-\left(-\frac{1}{3}\right)}{h}$
$=\lim _{h \rightarrow 0} \frac{\frac{1}{2 h-3}+\frac{1}{3}}{h}$
$=\lim _{h \rightarrow 0} \frac{3+2 h-3}{3 h(2 h-3)}$
$=\lim _{h \rightarrow 0} \frac{2}{3(2 h-3)}=-\frac{2}{9}$
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