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  • Provide Solutio for RD Sharma Class 12 Chapter Continuity Exercise 8.1 Question 27

Provide Solutio for RD Sharma Class 12 Chapter Continuity Exercise 8.1 Question 27

Answers (1)

Answer:

                k=\pm 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)=\left\{\begin{array}{cl} \frac{1-\cos k x}{x \sin x} & , x \neq 0 \\\\ \frac{1}{2} & , x=0 \end{array}\right.

Solution:

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

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

Consider,

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

 \Rightarrow         \lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0}\left(\frac{2 \sin ^{2} k \frac{x}{2}}{x^{2}\left(\frac{\sin x}{x}\right)}\right)

                                 =\lim _{x \rightarrow 0}\left(\frac{2 \frac{k^{2}}{4}\left(\sin k \frac{x}{2}\right)^{2}}{\left(\frac{k x}{2}\right)^{2}\left(\frac{\sin x}{x}\right)}\right)         [Multiplying and dividing by \left ( \frac{k}{2} \right )^{2} ]

                                 =\frac{2 k^{2}}{4} \lim _{x \rightarrow 0}\left(\frac{\left(\sin k \frac{x}{2}\right)^{2}}{\left(\frac{k x}{2}\right)^{2}\left(\frac{\sin x}{x}\right)}\right)

                   \lim _{x \rightarrow 0} f(x)=\frac{2 k^{2}}{4}\left(\frac{\lim _{x \rightarrow 0} \frac{\left(\sin k \frac{x}{2}\right)^{2}}{\left(\frac{k x}{2}\right)^{2}}}{\lim _{x \rightarrow 0}\left(\frac{\sin x}{x}\right)} \right)                                                      \left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right]

                    \lim _{x \rightarrow 0} f(x)=\frac{2 k^{2}}{4} \times 1=\frac{k^{2}}{2}                                                                               … (i)

From (i)

            \begin{aligned} &\frac{k^{2}}{2}=f(0) \\\\ &\frac{k^{2}}{2}=\frac{1}{2}, k=\pm 1 \end{aligned}

            

Posted by

infoexpert27

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