#### Provide solution for RD Sharma maths class 12 chapter Continuity exercise 8.2 question 3 subquestion (viii)

f(x) is nowhere discontinuous

Hint:

$f(x)=|x|= \begin{cases}x & x>0 \\ -x & x<0\end{cases}$

Given:

$f(x)=|x|= \begin{cases}|x-3| & x \geq 1 \\ \frac{x^{2}}{4}-\frac{3 x}{2}+\frac{15}{4} & x<1\end{cases}$

Explanation:

\begin{aligned} &f(x)= \begin{cases}x-3 & x \geq 3 \\ -(x-3) & 1 \leq x<3 \\ \frac{x^{2}}{4}-\frac{3 x}{2}+\frac{13}{4} & x<1\end{cases} \end{aligned}

Now, we check the limit at x = 1 and x = 3

At x = 3

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

As

$\lim _{x \rightarrow 3} f(x)=f(3)$

Hence, f(x) is continuous at x = 3

Now, at x = 1

\begin{aligned} &\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1} \frac{x^{2}}{4}-\frac{3 x}{2}+\frac{13}{4} \\ &=\frac{1}{4}-\frac{3}{2}+\frac{13}{4} \\ &=2 \\ &\lim _{x \rightarrow 1^{+}} f(x)=-(1-3)=2 \\ &f(1)=-(1-3)=2 \end{aligned}

As

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

$\therefore$f(x) is continuous at x = 1

As polynomial function is everywhere continuous & greater integer function is continuous except its end point &

$\lim _{x \rightarrow 1} f(x) \text { \& } \lim _{x \rightarrow 3} f(x)$

is continuous.

Hence, f(x) is everywhere continuous.