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  • Provide Solutio for RD Sharma Class 12 Chapter Continuity Exercise 8.1 Question 26

Provide Solutio for RD Sharma Class 12 Chapter Continuity Exercise 8.1 Question 26

Answers (1)

Answer:

                a=\frac{-3}{2}, b \neq 0, c=\frac{1}{2} ; b \in R-\{0\}

Hint:

f\left ( x \right )  must be defined. The limit of the f\left ( x \right ) approaches the value x must exist.

Given:

                f(x)=\left\{\begin{array}{c} \frac{\sin (a+1) x+\sin x}{x}, \text { for } x<0 \\\\ c \quad, \text { for } x=0 \\\\ \frac{\sqrt{x+b x^{2}}-\sqrt{x}}{b x^{\frac{3}{2}}}, \text { for } x>0 \end{array}\right.

Solution:

                f(x)=\left\{\begin{array}{c} \frac{\sin (a+1) x+\sin x}{x}, \text { for } x<0 \\\\ c \quad, \text { for } x=0 \\\\ \frac{\sqrt{x+b x^{2}}-\sqrt{x}}{b x^{\frac{3}{2}}}, \text { for } x>0 \end{array}\right.

 Since f\left ( x \right ) is continuous at x=0 , we have

\lim _{x \rightarrow 0^{-}} f(x)=f(0)=\lim _{x \rightarrow 0^{+}} f(x) 

 (LHL at x=0 )

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

\begin{aligned} &=\lim _{x \rightarrow 0} \frac{(\sin a x \times \cos x)}{x}+\lim _{x \rightarrow 0} \frac{(\cos a x \times \sin x)}{x}+\lim _{x \rightarrow 0} \frac{\sin x}{x} \\ &=\lim _{x \rightarrow 0} \frac{\sin a x}{x} \lim _{x \rightarrow 0} \cos x+\lim _{x \rightarrow 0} \cos a x \lim _{x \rightarrow 0} \frac{\sin x}{x}+\lim _{x \rightarrow 0} \frac{\sin x}{x} \end{aligned}

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

\begin{aligned} &=a \lim _{x \rightarrow 0} \frac{\sin a x}{a x}+\lim _{x \rightarrow 0} \frac{\sin x}{x}+\lim _{x \rightarrow 0} \frac{\sin x}{x} \\ &=a(1)+1+1 \\ &=a+2 \end{aligned}                                                                              \left[\begin{array}{l} \because \lim _{x \rightarrow 0} \cos x=1 \\ \because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1 \end{array}\right]                             

RHL at  x=0

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

\begin{aligned} &=\lim _{x \rightarrow 0} \frac{x+b x^{2}-x}{b x \sqrt{x}\left(\sqrt{x+b x^{2}}+\sqrt{x}\right)} \\\\ &=\lim _{x \rightarrow 0} \frac{b x^{2}}{b x \sqrt{x}\left(\sqrt{x+b x^{2}}+\sqrt{x}\right)} \end{aligned}

\begin{aligned} &=\lim _{x \rightarrow 0} \frac{x}{\sqrt{x}\left(\sqrt{x+b x^{2}}+\sqrt{x}\right)} \\\\ &=\lim _{x \rightarrow 0} \frac{\sqrt{x}}{\left(\sqrt{x+b x^{2}}+\sqrt{x}\right)} \end{aligned}

=\lim _{x \rightarrow 0} \frac{\left(\frac{\sqrt{x}}{\sqrt{x}}\right)}{\left(\frac{\sqrt{x+b x^{2}}+\sqrt{x}}{\sqrt{x}}\right)}

\begin{aligned} &=\lim _{x \rightarrow 0} \frac{1}{(\sqrt{1+b x}+\sqrt{1})} \\\\ &=\frac{1}{2} \\\\ &f(0)=\lim _{x \rightarrow 0} f(x)=c \end{aligned}

Since f\left ( x \right ) is continuous at x=0 , then

 \begin{aligned} &\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=f(0) \\\\ &a+2=c=\frac{1}{2} \end{aligned}

a=\frac{-3}{2}, c=\frac{1}{2}, b  is any real number except  0

 

Posted by

infoexpert27

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