#### Provide Solutio for RD Sharma Class 12 Chapter Continuity Exercise 8.1 Question 42

For any value  $\lambda, f(x)$  is Continuous at $x=1$

Hint:

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

Given:

$f(x)=\left\{\begin{array}{l} \lambda\left(x^{2}-2 x\right), \text { if } x \leq 0 \\ 4 x+1, \text { if } x>0 \end{array}\right.$

Solution:

$f(x)=\left\{\begin{array}{l} \lambda\left(x^{2}-2 x\right), \text { if } x \leq 0 \\ 4 x+1, \text { if } x>0 \end{array}\right.$

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

\begin{aligned} &\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=f(0) \\\\ &\lim _{x \rightarrow 0^{-}} \lambda\left(x^{2}-2 x\right) \end{aligned}

Putting  $x=0-h \text { as } x \rightarrow 0^{-}$

When,$h\rightarrow 0$

\begin{aligned} &\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} \lambda\left((0-h)^{2}-2(0-h)\right) \\\\ &=\lim _{h \rightarrow 0} \lambda\left(h^{2}+2 h\right) \\\\ &=0 \end{aligned}

At $x=0$ , RHL = $\lim _{x \rightarrow 0^{+}} f(x)$

$=\lim _{x \rightarrow 0^{+}} 4 x+1$

Putting $x=0+h \text { as } x \rightarrow 0^{+}$

When,  $h\rightarrow 0$

\begin{aligned} &\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0}[4(0+h)+1] \\\\ &=\lim _{h \rightarrow 0}(4 h+1) \\\\ &=1 \end{aligned}

LHL $\neq$ RHL. Thus, $f\left ( x \right )$  is discontinuous at $x=0$ for any value $\lambda$ .

At  $x=1$ , LHL=    $\lim _{x \rightarrow 1^{-}} f(x)$

$=\lim _{x \rightarrow 1^{-}} 4 x+1$

Putting $x=1-\text {has } x \rightarrow 1^{-}$

When,  $h\rightarrow 0$

\begin{aligned} &\lim _{x \rightarrow 1^{-}} f(x)=\lim _{h \rightarrow 0}[4(1+h)+1] \\\\ &=\lim _{h \rightarrow 0}(4+4 h+1) \\\\ &=5 \end{aligned}

At  $x=1$, RHL =  $\lim _{x \rightarrow 1^{+}} f(x)$

$=\lim _{x \rightarrow 1^{+}} 4 x+1$

Putting  $x=1+h$  as $x\rightarrow 1^{+}$

When,$h\rightarrow 0$

\begin{aligned} &\lim _{x \rightarrow 1^{-}} f(x)=\lim _{h \rightarrow 0}[4(1+h)+1] \\\\ &=\lim _{h \rightarrow 0}(4+4 h+1)\\ \\ &=5 \end{aligned}

LHL $=$  RHL

Therefore, for any value of  $\lambda$  for which $f\left ( x \right )$ is continuous at  $x=1$