#### Please solve RD Sharma class 12 chapter 8 Continuity exercise multiple choice question 13 maths textbook solution

The correct option is (d)

Hint:

Use the given formula:

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

$\lim _{x \rightarrow a^{+}} f(a+h)=\lim _{x \rightarrow a^{-}} f(a-h)=f(a)$

Given:

$f(x)=\left\{\begin{array}{l} \frac{x^{4}-5 x^{2}+4}{|(x-1)(x-2)|} \\ 6 \\ 12 \end{array}\right.$    \begin{aligned} ,\: &x \neq 1,2 \\ ,\: &x=1 \\ ,\: &x=2 \end{aligned}

Solution:

we write

\begin{aligned} x^{4}-5 x^{2}+4 &=x^{4}-x^{2}-4 x^{2}+4=x^{2}\left(x^{2}-1\right)-4\left(x^{2}-1\right) \\ &=\left(x^{2}-4\right)\left(x^{2}-1\right) \\ &=(x+2)(x-2)(x+1)(x-1) \end{aligned}

$\therefore$ we have

$f(x)=\left\{\begin{array}{c} \frac{(x+2)(x-2)(x+1)(x-1)}{|(x-1)(x-2)|} \\ 6 \\ 12 \end{array}\right.$    \begin{aligned} ,\: &x \neq 1,2 \\ ,\: &x=1 \\ ,\: &x=2 \end{aligned}

$f(x)=\left\{\begin{array}{c} \frac{x^{4}-5 x^{2}+4}{(x-1)(x-2)} \\ 6 \\ 12 \end{array}\right.$    \begin{aligned} ,\: &x \neq 1,2 \\ ,\: &x=1 \\ ,\: &x=2 \end{aligned}

$=\left\{\begin{array}{c} \int(x+1)(x+2), x<1 \\ -(x+1)(x+2), 12 \\ 6, x=1 \\ 12, x=2 \end{array}\right.$

Using RHL at x = 1

\begin{aligned} &\lim _{x \rightarrow 1^{+}} f(x)=\lim _{h \rightarrow 0} f(1+h) \\ &=-(2) \cdot(3)=-6 \end{aligned}

Using LHL at x = 1

\begin{aligned} &\lim _{x \rightarrow 1^{-}} f(x)=\lim _{h \rightarrow 0} f(1-h) \\ &=(2)(3)=6 \end{aligned}

And using RHL at x = 2

$\lim _{x \rightarrow 2^{+}} f(x)=\lim _{h \rightarrow 0} f(1+h)=12$

Using LHL at x = 2

$\lim _{x \rightarrow 2^{-}} f(x)=\lim _{h \rightarrow 0} f(1-h)=-12$

f(x) is discontinuous at x=1 and x=2 for f(x) continuous at R[1, 2]

So, the correct option is (d)