Question
Let f(x) be a real valued function, then its
  • Left Hand Derivative (L.H.D.) : $\text{Lf}'(\text{a})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{a}-\text{h})-\text{f}(\text{a})}{-\text{h}}$
  • Right Hand Derivative (R.H.D.) : $\text{Rf}'(\text{a})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{a}+\text{h})-\text{f}(\text{a})}{\text{h}}$
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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