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Provide Solutio for RD Sharma Class 12 Chapter Continuity Exercise 8.1 Question 36 Subquestion (iv)

Answers (1)

Answer:

                k=\frac{-2}{\pi}

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}{l} k x+1, \text { if } x \leq \pi \\\\ \cos x, \text { if } x>\pi \end{array}\right.

               

Solution:

                f(x)=\left\{\begin{array}{l} k x+1, \text { if } x \leq \pi \\\\ \cos x, \text { if } x>\pi \end{array}\right.

We have

(LHL at x=\pi )

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

(RHL at x=\pi )

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

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

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

               

 

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