#### Explain solution RD Sharma class 12 chapter 8 Continuity exercise multiple choice question 40 maths

The correct option is (b)

Hint:

(i) A function f(x) is said to be continuous at a point x = a of its domain, if

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

Given:

$f(x)= \begin{cases}\frac{1}{5}\left(2 x^{2}+3\right) &, \quad x \leq 1 \\ 6-5 x & , \quad 1

Solution:

Now at x = 1

\begin{aligned} &\lim _{x \rightarrow 1^{+}} f(x)=\lim _{h \rightarrow 0} f(1+h) \\ &\lim _{h \rightarrow 0}[6-5(1+h)]=1 \\ &\lim _{h \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{-}} f(x)=f(1) \end{aligned}

Therefore continuous at x = 1

Now, at x = 3

\begin{aligned} &\lim _{x \rightarrow 3^{+}} f(x)=\lim _{h \rightarrow 0} f(3+h) \\ &\lim _{h \rightarrow 0}[(3+h)-3] \\ &=0 \end{aligned}

For left hand limit,

\begin{aligned} &=\lim _{x \rightarrow 3^{-}} f(x)=\lim _{h \rightarrow 0} f(3-h) \\ &=\lim _{h \rightarrow 0}[6-5(3-h)]=-9 \end{aligned}

We have

$=\lim _{x \rightarrow 3^{+}} f(x) \neq \lim _{x \rightarrow 3^{-}} f(x)$

Therefore discontinuous at x = 3

So, the correct option is (b).