Question
Let f(x) be a real valued function, then its
  • Left Hand Derivative (L.H.D.) : $\operatorname{Lf}^{\prime}(a)=\lim _{h \rightarrow 0} \frac{f(a-h)-f(a)}{-h}$
    Right Hand Derivative (R.H.D.) : $\operatorname{Rf}^{\prime}(a)=\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}$
    Also, a function f(x) is said to be differentiable at x = a if its L.H.D. and R.H.D. at x = a exist and are equal.
    For the function $\text{f}(\text{x})=\begin{cases}|\text{x}-3|,\text{x}\geq1\\\\\frac{\text{x}^2}{4}-\frac{3\text{x}}{2}+\frac{13}{4},\text{x}<1\end{cases},$ answer the following questions.
    1. R.H.D. of f(x) at x = 1 is:
    1. 1
    2. -1
    3. 0
    4. 2
    1. L.H.D. of f(x) at x = 1 is:
    1. 1
    2. -1
    3. 0
    4. 2
    1. f(x) is non-differentiable at:
    1. x = 1
    2. x = 2
    3. x = 3
    4. x = 4
    1. Find the value of f'(2).
    1. 1
    2. 2
    3. 3
    4. -1
    1. The value of f'(-1) is:
    1. 2
    2. 1
    3. -2
    4. -1

Answer

We have, $\text{f}(\text{x})=\begin{cases}\text{x}-3&,\text{x}\geq3\\3-\text{x}&,1\leq\text{x}<3\\\\\frac{\text{x}^2}{4}-\frac{3\text{x}}{2}+\frac{13}{4}&,\text{x}<1\end{cases}$
  1. (b) -1
Solution:
$\text{Rf}'(1)=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(1+\text{h})-\text{f}(1)}{\text{h}}$
$\lim\limits_{\text{h}\rightarrow0}\frac{3-(1+\text{h})-2}{\text{h}}=\lim\limits_{\text{h}\rightarrow0}-\frac{\text{h}}{\text{h}}=-1$
  1. (b) -1
Solution:
$\text{Lf}'(1)=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(1-\text{h})-\text{f}(1)}{-\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{-1}{\text{h}}\Big[\frac{(1-\text{h})^2}{4}-\frac{3(1-\text{h})}{2}+\frac{13}{4}-2\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\frac{-1}{\text{h}}\Big(\frac{1+\text{h}^2-2\text{h}-6+6\text{h}+13-8}{-4\text{h}}\Big)$
$=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{\text{h}^2+4\text{h}}{-4\text{h}}\Big)=-1$
  1. (c) x = 3
Solution:
Since, R.H.D. at x = 3 is 1
and L.H.D. at x = 3 is - 1
$\therefore$ f(x) is non-differentiable at x = 3.
  1. (d) -1
  1. (c) -2
Solution:
From above, we have
$\text{f}'(\text{x})=\frac{\text{x}}{2}-\frac{3}{2},\text{x}<1$
$\therefore\text{f}'(-1)=\frac{-1}{2}-\frac{3}{2}=-2$

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