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  • Provide solution for RD Sharma maths class 12 chapter 8 Continuity exercise multiple choice question 14

Provide solution for RD Sharma maths class 12 chapter 8 Continuity exercise multiple choice question 14

Answers (1)

Answer:

 The correct option is (c)

Hint:

 

Use the given formula:

(i) A function f(x) is said to be continuous at a point x = a of its domain, if

\begin{aligned} &\lim _{x \rightarrow a^{+}} f(a+h)=\lim _{x \rightarrow a^{-}} f(a-h)=f(a)\\ &\text { (ii) } \lim _{x \rightarrow 0} \frac{\sin x}{x}=1 \end{aligned}

Given:

 f(x) is continuous at x = 0 and

f(x)=\left\{\begin{array}{cc} \frac{\sin (a+1) x+\sin x}{x} &, x<0 \\ C & , x=0 \\ \frac{\sqrt{x+b x^{2}}-\sqrt{x}}{b x \sqrt{x}} & , x>0 \end{array}\right.

Solution:

Using RHL

\begin{gathered} \lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h) \\ =\lim _{h \rightarrow 0} f(h) \\ =\lim _{h \rightarrow 0} \frac{\sqrt{h+b h^{2}}-\sqrt{h}}{b h \sqrt{h}} \end{gathered}

\begin{aligned} &=\lim _{h \rightarrow 0} \frac{\sqrt{h}(\sqrt{1+b h})-\sqrt{n}}{b h \sqrt{h}} \\ &=\lim _{h \rightarrow 0} \frac{\sqrt{1+b n}-1}{b h} \\ &=\lim _{h \rightarrow 0} \frac{\sqrt{1+b h}-1}{b h} \times \frac{\sqrt{1+b h}+1}{\sqrt{1+b h}+1} \end{aligned}

\begin{aligned} &=\lim _{h \rightarrow 0} \frac{(1+b h)-1}{b h(\sqrt{1+b h}+1)} \\ &=\lim _{h \rightarrow 0} \frac{b h}{b h(\sqrt{1+b h}+1)} \\ &=\lim _{h \rightarrow 0} \frac{1}{\sqrt{1+b h}+1} \end{aligned}

\begin{aligned} &=\frac{1}{\sqrt{1+6 \cdot 0}+1} \\ &=\frac{1}{1+1} \\ &=\frac{1}{2} \end{aligned}

Using L.H.L

\begin{aligned} \lim _{x \rightarrow 0^{-}} f(x) &=\lim _{h \rightarrow 0} f(0-h) \\ &=\lim _{h \rightarrow 0} f(-h) \\ &=\lim _{n \rightarrow 0} \frac{\sin (a+1)(-h)+\sin (-h)}{(-h)} \end{aligned}

\begin{aligned} &=\lim _{h \rightarrow 0} \frac{-\sin (a+1) \cdot h-\sinh }{-h} \\ &=\lim _{h \rightarrow 0}\left\{-\frac{\sin (a+1) \cdot h}{-h}+\frac{\sin h}{h}\right\} \quad[\because \sin (-\theta)=-\sin \theta] \end{aligned}

\begin{aligned} &=\lim _{h \rightarrow 0}\left\{\frac{\sin (a+1)}{(a+1)} \times(a+1)+\frac{\sinh }{h}\right\} \\ &=(a+1) \lim _{h \rightarrow 0} \frac{\sin (a+1)}{a+1}+\lim _{h \rightarrow 0} \frac{\sin h}{n} \end{aligned}

\begin{aligned} &=(a+1) \cdot 1+1 \quad\left[: \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right] \\ &=a+1+1 \\ &=a+2 \end{aligned}

Since function f(x) is continuous at x = 0
\begin{aligned} &\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x) \\ &=f(0) \\ &\Rightarrow a+2=\frac{1}{2} \\ &\Rightarrow a+2=\frac{1}{2} \\ &\text { or } c=\frac{1}{2} \end{aligned}

\begin{aligned} &\begin{aligned} \Rightarrow a &=\frac{1}{2}-2 \\ &=-\frac{3}{2} \text { or } \\ c=\frac{1}{2} \end{aligned}\\ &\text { and from } f(x) \text { , }\\ &b \in \mathbb{R}-\{0\} \end{aligned}
So, the correct option is (c).

Posted by

Gurleen Kaur

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