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  • Explain Solution R.D.Sharma Class 12 Chapter 9 Differentiability Exercise 9.1 Question 9 Maths Textbook Solution.

Explain Solution R.D.Sharma Class 12 Chapter 9 Differentiability Exercise 9.1 Question 9 Maths Textbook Solution.

Answers (1)

Answer: f(x) is continuous but not differentiable at x = 1.

Hint: The function y = f(x) is said to be differentiable in the closed interval [a, b] if R f ` (a) and L f ` (b) exist and f `(x) exist for every point of (a, b).

If the left hand limit, right hand limit and the value of the function at x = 3 exist and are equal to each other, then f is said to be continuous at x = 3.

Given: f\left ( x \right )=\left\{\begin{array}{c} |2 x-3|[x], \quad x \geq 1 \\ \sin \left(\frac{n x}{2}\right), \mathrm{x}<1 \end{array}\right.

Solution:     

f(x)=\left\{\begin{array}{r} |2 x-3|[x], \quad x \geq \frac{3}{2} \\ -(2 x-3), 1 \leq x \leq \frac{3}{2} \\ \sin \left(\frac{n x}{2}\right), x<1 \end{array}\right.

For continuity at x = 1,

\begin{aligned} &\mathrm{LHL}=\lim _{x \rightarrow 1^{-}} \mathrm{f}(\mathrm{x}) \\ &=\lim _{h \rightarrow 0} \mathrm{f}(1-\mathrm{h}) \\ &=\lim _{h \rightarrow 0} \sin \left(\frac{n(1-h)}{2}\right) \\ &=\sin \left(\frac{n}{2}\right) \end{aligned}                                                                                        \left [ \frac{n}{2}=90^{o} ,\sin 90^{o}=1\right ]

=1

Now consider,

\begin{aligned} &\mathrm{RHL}=\lim _{x \rightarrow 1^{+}} \mathrm{f}(\mathrm{x}) \\ &=\lim _{h \rightarrow 0} \mathrm{f}(1+\mathrm{h}) \\ &=\lim _{h \rightarrow 0}-(2(1+\mathrm{h})-3) \\ &=\lim _{h \rightarrow 0}-(2+2 \mathrm{~h}-3) \\ &=\lim _{h \rightarrow 0}-(-1+2 \mathrm{~h}) \\ &=1 \end{aligned}

LHL=RHL=f\left ( 1 \right )

Since f(x) is continuous at x = 1, we have to find its differentiability using the formula,

\begin{aligned} &f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\ &\mathrm{LHD} \text { at } \mathrm{x}=1: \lim _{x \rightarrow 1^{-}} \frac{f(x)-f(1)}{x-1} \\ &=\lim _{h \rightarrow 0} \frac{f(1-h)-f(1)}{1-h-1} \\ &=\lim _{h \rightarrow 0} \frac{\sin \left(n\left(\frac{1-h}{2}\right)\right)-1}{-h} \\ &=\lim _{h \rightarrow 0} \frac{\sin \left(\frac{n}{2}-\frac{n h}{2}\right)-1}{-h} \end{aligned}                                                                \left [ \therefore \sin \left ( \frac{n}{2}-\theta \right )=\cos \theta \right ]

=-1

\begin{aligned} &\mathrm{RHD} \text { at } \mathrm{x}=1: \lim _{x \rightarrow 1^{+}} \frac{f(x)-f(1)}{x-1} \\ &=\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{1+h-1} \\ &=\lim _{h \rightarrow 0} \frac{-[2+2 h-3]-1}{h} \\ &=\lim _{h \rightarrow 0} \frac{-2 h}{h} \\ &=-2 \end{aligned}

LHD at x = 1 \neq RHD at x = 1. Hence, f(x) is continuous but not differentiable.

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