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  • Explain solution RD Sharma class 12 Chapter 8 Continuity exercise 8.1 question 45

Explain solution RD Sharma class 12 Chapter 8 Continuity exercise 8.1 question 45

Answers (1)

Answer:

                k=1

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)= \begin{cases}\frac{1-\cos 2 x}{2 x^{2}} & , \text { if } x<0 \\\\ k & , \text { if } x=0 \\\\ \frac{x}{|x|} & , \text { if } x>0\end{cases}

Solution:

                f(x)= \begin{cases}\frac{1-\cos 2 x}{2 x^{2}} & , \text { if } x<0 \\\\ k & , \text { if } x=0 \\\\ \frac{x}{|x|} & , \text { if } x>0\end{cases}

                f(x)= \begin{cases}\frac{1-\cos 2 x}{2 x^{2}} & , \text { if } x<0 \\\\ k & \text { , if } x=0 \\\\ 1 & , \text { if } x>0\end{cases}

We have

(LHL at x=0  )

                \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0}\left(\frac{1-\cos 2(-h)}{2(-h)^{2}}\right)=\lim _{h \rightarrow 0}\left(\frac{1-\cos 2 h}{2 h^{2}}\right)=\frac{1}{2} \lim _{h \rightarrow 0}\left(\frac{2 \sin ^{2} h}{h^{2}}\right)

                                                                                                            \left[\because 1-\cos 2 x=2 \sin ^{2} x\right] 

                =\frac{2}{2} \lim _{h \rightarrow 0}\left(\frac{\sin ^{2} h}{h^{2}}\right)=\lim _{h \rightarrow 0}\left(\frac{\sin h}{h}\right)^{2}=1                                                      \left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right]

(RHL at  x=0  )

                \lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h)=\lim _{h \rightarrow 0} f(h)=\lim _{h \rightarrow 0}(1)=1        

Also

f(0)=k

If f\left ( x \right )  is continuous at x=0 .

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

Hence, the required value of  k  is 1  .

 

Posted by

infoexpert27

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