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  • Please Solve R.D.Sharma class 12 Chapter 9 Differentiability Exercise 9.2 Question 4 Maths textbook Solution.

Please Solve R.D.Sharma class 12 Chapter 9 Differentiability Exercise 9.2 Question 4 Maths textbook Solution.

Answers (1)

Answer: f(x) is continuous but not differentiable at x = 2.

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 = c exist and are equal to each other, then f is said to be continuous at x = c.

Given: f(x)=\left\{\begin{array}{cc} 3 x-2, & 0 \leq \mathrm{x} \leq 1 \\ 2 x^{2}-x, & 1<x \leq 2 \\ 5 x-4, & x>2 \end{array}\right.

Solution:

Now we have to check for continuity at x = 2.For continuity,

\left ( LHL\: at\: x=2 \right )=\left ( RHL\: at\: x=2 \right )

f\left ( 2 \right )=2\left ( 2 \right )^{2}-2

=8-2

=6

LHL=\lim_{x\rightarrow 2^{-}}f\left ( x \right )

=\lim_{h\rightarrow 0}f\left ( 2-h \right )

=\lim_{h\rightarrow 0}\left [ 2\left ( 2-h^{2} \right )-\left ( 2-h \right ) \right ]

=8-2

=6

Now consider,

RHL=\lim_{x\rightarrow 2^{+}}f\left ( x \right )

=\lim_{h\rightarrow 0}f\left ( 2+h \right )

=\lim_{h\rightarrow 0}5\: \left ( 2+h \right )-4

=10-4

=6

LHL=RHL=f\left ( 2 \right )

Since f(x) is continuous at x = 2, 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}=2: \lim _{x \rightarrow 2^{-}} \frac{f(x)-f(2)}{x-2} \\ &=\lim _{h \rightarrow 0} \frac{f(2-h)-f(2)}{2-h-2} \\ &=\lim _{h \rightarrow 0} \frac{\left[2(2-h)^{2}-(2-h)\right]-[8-2]}{-h} \end{aligned}

\begin{aligned} &=\lim _{h \rightarrow 0} \frac{8-8 h+2 h^{2}-2+h-6}{-h} \\ &=\lim _{h \rightarrow 0} \frac{2 h^{2}-7 h}{-h} \\ &=\lim _{h \rightarrow 0} \frac{h(2 h-7)}{-h} \\ &=\lim _{h \rightarrow 0}(7-2 h) \\ &=7 \end{aligned}

\begin{aligned} &\mathrm{RHD} \text { at } \mathrm{x}=2: \lim _{x \rightarrow 2^{+}} \frac{f(x)-f(2)}{x-2} \\ &=\lim _{h \rightarrow 0} \frac{f(2+h)-f(2)}{2+h-2} \\ &=\lim _{h \rightarrow 0} \frac{[5(2+h)-4]-[10-4]}{h} \\ &=\lim _{h \rightarrow 0} \frac{10+5 h-4-6}{h} \\ &=5 \end{aligned}

LHD at x = 2 \neq RHD at x = 2

Hence, f(x) is continuous but not differentiable at x = 2.

Posted by

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