Limit Continuity And Differentiability Ques 57
- Find $f^{\prime}(1)$, if $f(x)=\left\{\begin{array}{cl}\frac{x-1}{2 x^2-7 x+5}, & \text { when } x \neq 1 \\ -\frac{1}{3}, & \text { when } x=1 \end{array}\right.$
(1979,3M)
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Answer:
Correct Answer: 57.$\left(-\frac{2}{9}\right)$
Solution: Given that, $f(x)= \begin{cases}\frac{x-1}{2 x^2-7 x+5} & , \text { when } x \neq 1 \\ -\frac{1}{3} & \text {, when } x=1\end{cases}$
$ \begin{aligned} \mathrm{RHD} & =\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{h} \\ & =\lim _{h \rightarrow 0} \frac{\left[\frac{1+h-1}{2(1+h)^2-7(1+h)+5}-\left(-\frac{1}{3}\right)\right]}{h} \\ & =\lim _{h \rightarrow 0}\left[\frac{3 h+2(1+h)^2-7(1+h)+5}{3 h\left\{2(1+h)^2-7(1+h)+5\right\}}\right] \\ & =\lim _{h \rightarrow 0}\left(\frac{2 h^2}{3 h\left(-3 h+2 h^2\right)}\right)=-\frac{2}{9} \\ \mathrm{LHD} & =\lim _{h \rightarrow 0} \frac{f(1-h)-f(1)}{-h} \\ & =\lim _{h \rightarrow 0} \frac{\left[\frac{1-h-1}{2(1-h)^2-7(1-h)+5}-\left(-\frac{1}{3}\right)\right]}{-h} \\ & =\lim _{h \rightarrow 0} \frac{-3 h+2\left(1+h^2-2 h\right)-7(1-h)+5}{-3 h\left[2(1-h)^2-7(1-h)+5\right]} \\ & =\lim _{h \rightarrow 0} \frac{2 h^2}{-3 h\left(2 h^2+3 h\right)}=-\frac{2}{9} \therefore \mathrm{LHD}=\mathrm{RHD} \end{aligned} $
Hence, required value of $f^{\prime}(1)=-\frac{2}{9}$.
BETA
