#### Need solution for RD Sharma Maths Class 12 Chapter 8 Continuity Excercise 8.1 Question 43

$k=2$

Hint:

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

Given:

$f(x)=\left\{\begin{array}{l} 2 x+1, \text { if } x<2 \\ k, \text { if } x=2 \\ 3 x-1, \text { if } x>2 \end{array}\right.$

Solution:

$f(x)=\left\{\begin{array}{l} 2 x+1, \text { if } x<2 \\ k, \text { if } x=2 \\ 3 x-1, \text { if } x>2 \end{array}\right.$

We have

(LHL at  $x=2$ )

$\lim _{x \rightarrow 2^{-}} f(x)=\lim _{h \rightarrow 0} f(2-h)=\lim _{h \rightarrow 0}[2(2-h)+1]=[4+1]=5$

(RHL at $x=2$  )

$\lim _{x \rightarrow 2^{+}} f(x)=\lim _{h \rightarrow 0} f(2+h)=\lim _{h \rightarrow 0}[3(2+h)-1]=[6-1]=5$

Also  $f\left ( 2 \right )=k$

If $f\left ( x \right )$  is continuous at  $x=2$ .

\begin{aligned} &\lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2^{-}} f(x)=f(2) \\\\ &5=5=k \\\\ &k=5 \end{aligned}

Since  $LHS =RHS ,f\left ( x \right )$   is continuous

Thus,$f\left ( x \right )$  is continuous at $x=2$ .