#### Provide solution for RD Sharma maths class 12 chapter 8 Continuity exercise multiple choice question 38

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}2 \sqrt{x} &, 0 \leq x \leq 1 \\ 4-2 x &, 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} f(4-2(1+h)) \\ &=2 \end{aligned}

Again,

\begin{aligned} &\lim _{x \rightarrow 1^{-}} f(x)=\lim _{h \rightarrow 0} f(1-h) \\ &=\lim _{h \rightarrow 0}(2 \sqrt{1-h}) \\ &=2 \\ &=\lim _{x^{+} \rightarrow 0} f(x)=f(0)=\lim _{x \rightarrow 0^{-}} f(x) \end{aligned}

Therefore, function is continuous at x = 0

Now we will check the continuity of f(x) at

\begin{aligned} x = \frac{5}{2} \end{aligned}

\begin{aligned} &\text { So } \lim _{x \rightarrow \frac{5}{2}} f(x)=\lim _{h \rightarrow 0} f\left(\frac{5}{2}-h\right) \\ &=\lim _{h \rightarrow 0}\left(4-2\left(\frac{5}{2}-h\right)\right) \quad\left[\because f(x)=4-2 x, x<\frac{5}{2}\right] \\ &=4-2\left(\frac{5}{2}-0\right) \end{aligned}

\begin{aligned} &=4-2 \times \frac{5}{2} \\ &=4-5 \\ &=-1 \end{aligned}

\begin{aligned} \text { And } \lim _{x \rightarrow \frac{5}{2}^{+}} f(x) &=\lim _{h \rightarrow 0} f\left(\frac{5}{2}+h\right) \\ &=\lim _{h \rightarrow 0} 2\left(\frac{5}{2}+h\right)-7 \quad\left[\because f(x)=2 x-7, x>\frac{5}{2}\right] \\ &=2\left(\frac{5}{2}+0\right)-7 \end{aligned}

\begin{aligned} &=2 \times \frac{5}{2}-7 \\ &=5-7 \\ &=-2 \\ \lim _{x \rightarrow \frac{5}{2}} f(x) & \neq \lim _{x \rightarrow \frac{5}{2}} f(x) \\ \text { le }-1 \neq-2 & \end{aligned}

Therefore the function f(x) is discontinuous at

\begin{aligned} x = \frac{5}{2} \end{aligned}

So, correct option is (b)