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  • Provide Solution For R.D.Sharma Maths Class 12 Chapter 9 Differentiability Exercise 9.2 Question 7 Maths Textbook Solution.

Provide Solution For R.D.Sharma Maths Class 12 Chapter 9 Differentiability Exercise 9.2 Question 7 Maths Textbook Solution.

Answers (1)

Answer: Function is differentiable at x = 0 & not differentiable at x = -2

Hint: Function is differentiable on x \epsilon  [-3, 2], x \epsilon  (-2, 0) and x \epsilon  [0, 1]. So, we need to check differentiability of f(x) at x = - 2 and x = 0.

Given: f\left ( x \right )=\left\{\begin{array}{c} 2 x+3, \text { if }-3 \leq \mathrm{x}<-2 \\ x+1, \text { if }-2<x<0 \\ x+2, \text { if } 0 \leq \mathrm{x} \leq 1 \end{array}\right.

Solution:

To check differentiability at x = -2,

(\mathrm{LHD} \text { at } \mathrm{x}=-2)=\lim _{x \rightarrow-2^{-}} \frac{f(x)-f(-2)}{x-(-2)}

                                     =\lim _{x \rightarrow-2^{-}} \frac{2x+3-\left ( 2\left ( -2 \right ) +3\right )}{x+2}                                                    \left [ \because f\left ( x \right )=2x+3,x< -2 \right ]

                                      \begin{aligned} &=\lim _{x \rightarrow-2^{-}} \frac{2 x+3-(-4+3)}{x+2} \\ &=\lim _{x \rightarrow-2^{-}} \frac{2 x+3-(-1)}{x+2} \\ &=\lim _{x \rightarrow-2^{-}} \frac{2 x+4}{x+2} \\ &=\lim _{x \rightarrow-2^{-}} \frac{2(x+2)}{x+2} \\ &=2 \end{aligned}

(\mathrm{RHD} \text { at } \mathrm{x}=-2)=\lim _{x \rightarrow-2^{+}} \frac{f(x)-f(-2)}{x-(-2)}

                                    =\lim _{x \rightarrow-2^{+}} \frac{x+1-\left ( -2+1 \right )}{x-\left ( -2 \right )}                                                        \left [ \because f\left ( x \right )=x+1,x> -2 \right ]                 

                                   =\lim _{x \rightarrow-2^{+}} \frac{x+1+1}{x+2}

                                    =\lim _{x \rightarrow-2^{+}} \frac{x+2}{x+2}

                                    =1

\because  (LHD at x = -2) \neq (RHD at x = -2)

Thus, f(x) is not differentiable at x = -2.

Now, we need to check differentiability at x = 0.

\begin{aligned} (\mathrm{LHD} \text { at } \mathrm{x}=0) &=\lim _{x \rightarrow 0^{-}} \frac{f(x)-f(0)}{x-0} \\ &=\lim _{x \rightarrow 0^{-}} \frac{x+1-(0+1)}{x-0} \end{aligned}                                                           \left [ \because f\left ( x \right ) =x+1,x< 0\right ]

                                =\lim _{x \rightarrow 0^{-}} \frac{x}{x}

                                =1       

\begin{aligned} (\mathrm{RHD} \text { at } \mathrm{x}=0) &=\lim _{x \rightarrow 0^{+}} \frac{f(x)-f(0)}{x-0} \\ &=\lim _{x \rightarrow 0^{+}} \frac{x+2-(0+2)}{x-0} \end{aligned}                                                     \left [ \because f\left ( x \right ) =x+2,x< 0\right ]

                                 =\lim _{x \rightarrow 0^{+}} \frac{x}{x}

                                 =1

\because  (LHD at x = 0) = (RHD at x = 0)

Thus, f(x) is differentiable at x = 0.

                               

Posted by

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