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

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

Answers (1)

Answer:

No value of k exists.

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}{r} k\left(x^{2}-2 x\right), \text { if } x<0 \\\\ \cos x, \text { if } x \geq 0 \end{array}\right.

Solution:

                f(x)=\left\{\begin{array}{r} k\left(x^{2}-2 x\right), \text { if } x<0 \\\\ \cos x, \text { if } x \geq 0 \end{array}\right.

We have

(LHL 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} k\left(h^{2}+2 h\right)=0

(RHL at x=0 )

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

Thus no value of k  exists for which f\left ( x \right )  is continuous at x=0

 

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