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  • Provide Solution For R.D.Sharma Maths Class 12 Chapter 9 Differentiability Exercise 9.1 Question 10 Maths Textbook Solution.

Provide Solution For  R.D.Sharma Maths Class 12 Chapter 9 Differentiability Exercise 9.1 Question 10 Maths Textbook Solution.

Answers (1)

Answer: a=\frac{-1}{2},b=\frac{-3}{2}

Hint: For differentiability, LHD = RHD where LHD is left hand derivative and RHD is right hand derivative. Also, LHD and RHD should exist in given limit.

Given: f\left ( x \right )= \left\{\begin{array}{c} a x^{2}-b, \text { if }|x|<1 \\ \frac{1}{|x|}, \text { if }|x| \geq 1 \end{array}\right.

Solution:

\Rightarrow\left\{\begin{array}{c} \frac{-1}{x}, x \leq 1 \\ a x^{2}-b,-1<x<1 \\ \frac{1}{x}, x \geq 1 \end{array}\right.

LHL\; at\: x=1:\lim_{x\rightarrow 1^{-}}f\left ( x \right )

\begin{aligned} &=\lim _{h \rightarrow 0} f(1-h) \\ &=\lim _{h \rightarrow 0} a(1-h)^{2}-b \\ &\therefore\left\{(a-b)^{2}=a^{2}+b^{2}-2 a b\right\} \\ &=\lim _{h \rightarrow 0} a\left(1+h^{2}-2 h\right)-b \\ &=a-b \end{aligned}

RHL\; at\: x=1:\lim_{x\rightarrow 1^{+}}f\left ( x \right )

=\lim_{h\rightarrow 0}f\left ( 1+h \right )

=\lim_{h\rightarrow 0}\frac{1}{1+h}

=1

Since f(x) is continuous, so LHS = RHS.

a-b=1                                                                                                        ....(1)

\begin{aligned} &\text { LHD } \Rightarrow \lim _{x \rightarrow 1^{-}} \frac{f(x)-f(1)}{x-1} \\ &\Rightarrow \lim _{h \rightarrow 0} f \frac{f(1-h)-1}{1-h-1} \end{aligned}

\begin{aligned} &\Rightarrow \lim _{h \rightarrow 0} \frac{a(1-h)^{2}-b-1}{-h} \\ &\therefore\left\{(a-b)^{2}=a^{2}+b^{2}-2 a b\right\} \\ &=\lim _{h \rightarrow 0} \frac{a h^{2}-2 a h}{-h} \\ &=\lim _{h \rightarrow 0}(2 a-a h) \\ &=2 \mathrm{a} \end{aligned}

\begin{aligned} &\mathrm{LHD} \Rightarrow \lim _{x \rightarrow 1^{+}} \frac{f(x)-f(1)}{x-1} \\ &\Rightarrow \lim _{h \rightarrow 0} f \frac{f(1+h)-1}{1+h-1} \\ &\Rightarrow \lim _{h \rightarrow 0} \frac{\frac{1}{1+h}-1}{h} \\ &=\lim _{h \rightarrow 0} \frac{-h}{(1+h) h} \\ &=\lim _{h \rightarrow 0} \frac{-1}{1+h} \\ &=-1 \end{aligned}

Since f(x) is differentiable at x = 1, (LHD at x = 1) = (RHD at x = 1)

2a=-1

a=\frac{-1}{2}

Substituting ‘a’ in (1),

a-b=1

\frac{-1}{2}-b=1

-b=1+\frac{1}{2}

b=\frac{-3}{2}

Hence,a=\frac{-1}{2} and b=\frac{-3}{2}

 

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