#### Provide Solutio for RD Sharma Class 12 Chapter Continuity Exercise 8.1 Question 41

$k$  can be any real number.

Hint:

$f\left ( x \right )$ must be defined. The limit of the $f\left ( x \right )$  approaches the value $x$ must exist.

Given:

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

Solution:

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

We have

(LHL at $x=0$)

$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0} f(-h)=\lim _{h \rightarrow 0}- 2-(h)^{2}+k=k$

(RHL at $x=0$ )

$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h)=\lim _{h \rightarrow 0} f(h)=\lim _{h \rightarrow 0} 2(h)^{2}+k=k$

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

\begin{aligned} &\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{-}} f(x)=f(0) \\ &\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{-}} f(x)=k \end{aligned}

$k$ can be any real number.