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  • Provide Solution For R.D.Sharma Maths Class 12 Chapter 9 Differentiability Exercise 9.1 Question 1 Maths Textbook Solution.

Provide Solution For R.D.Sharma Maths Class 12 Chapter 9 Differentiability Exercise 9.1 Question 1 Maths Textbook Solution.

Answers (1)

Answer: f(x) is continuous but not differentiable at x = 3.

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 )=|x-3|

Solution:

Therefore, we can write given function as,

\mathrm{f}(\mathrm{x})=\left\{\begin{array}{c} -(x-3), x<3 \\ x-3, x \geq 3 \end{array}\right.

But f\left ( 3 \right )=3-3=0

LHL=\lim_{x\rightarrow 3^{-}}f\left ( x \right )

=\lim_{h\rightarrow 0}f\left ( 3-h \right )

=\lim_{h\rightarrow 0}\: 3-\left ( 3-h \right )

=3-\left ( 3-0 \right )

=0

Now Consider,

RHL=\lim_{x\rightarrow 3^{+}}f\left ( x \right )

=\lim_{h\rightarrow 0} f\left ( 3+h \right )

=\lim_{h\rightarrow 0} 3+h-3

=3+0-3

=0

LHL=RHL=f \left ( 3 \right )

Since f(x) is continuous at x = 3, we have to find its differentiability using the formula,

f'\left ( x \right )=\lim_{h\rightarrow 0}\frac{f\left ( x+h \right )-f\left ( x \right )}{h}

Left Hand Derivate (LHD) ==\lim_{h\rightarrow 0}\frac{f\left ( c+h \right )-f\left ( c \right )}{h}, denoted by L f `(c)

Right Hand Derivate (RHD ==\lim_{h\rightarrow 0}\frac{f\left ( c+h \right )-f\left ( c \right )}{h}, denoted by R f `(c)

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

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

LHD at x = 3 \neq RHD at x = 3

Hence, f(x) is continuous but not differentiable at x = 3.

 

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