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  • Explain solution RD Sharma class 12 Chapter 8 Continuity exercise 8.1 question 36 subquestion (i)

Explain solution RD Sharma class 12 Chapter 8 Continuity exercise 8.1 question 36 subquestion (i)

Answers (1)

Answer:

                                k=\pm 2

Hint:

 For a function to be continuous at a point, its LHL RHL and value at that point should be equal.

Given:

                f(x)=\left\{\begin{array}{c} \frac{1-\cos 2 k x}{x^{2}}, \text { if } x \neq 0 \\\\ 8, \text { if } x=0 \end{array}\right.

Solution:

                f(x)=\left\{\begin{array}{c} \frac{1-\cos 2 k x}{x^{2}}, \text { if } x \neq 0 \\\\ 8, \text { if } x=0 \end{array}\right.

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

                 \begin{aligned} &\lim _{x \rightarrow 0} f(x)=f(0) \\\\ &\lim _{x \rightarrow 0} \frac{1-\cos 2 k x}{x^{2}}=8 \end{aligned}

                \lim _{x \rightarrow 0} \frac{2 k^{2} \sin ^{2} k x}{k^{2} x^{2}}=8                                           (\cos 2x= 1-2\sin ^{2}x)                    

                \begin{aligned} &2 k^{2} \lim _{x \rightarrow 0}\left(\frac{\sin k x}{k x}\right)=8 \\\\ &2 k^{2} \times 1=8 \\\\ &k^{2}=4 \\\\ &k=\pm 2 \end{aligned}

               

 

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