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

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 $x=0$