Provide solution for RD Sharma maths class 12 chapter 8 Continuity exercise multiple choice question 30

The correct option is (a)

Hint:

If a function f  is continuous at x = a, then

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

Given:

the function

$f(x)=\frac{x^{3}+x^{2}-16 x+20}{x-2}$

is not defined for x = 2. In order to make f(x) continuous at x = 2

Step 1: Take

$f(x)={x^{3}+x^{2}-16 x+20}$

and calculate further

$\therefore f(x)={x^{3}+x^{2}-16 x+20}$

\begin{aligned} &=x^{2}(x-2)+3 x(x-2)-10(x-2)\\ \end{aligned}

Simplify further,

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

Therefore, the function can be rewritten as

\begin{aligned} &\therefore f(x)=\frac{(x-2)^{2}(x+5)}{(x-2)} \\ &f(x)=(x-2)(x+5) \end{aligned}

Step 2: Understand that if f(x) is continuous at x = 2

Therefore,

\begin{aligned} &\therefore \lim _{h \rightarrow 0} f(x)=f(x) \\ &=\lim _{h \rightarrow 0}(x-2)(x+5)=f(x) \\ &f(2)=0 \end{aligned}

Hence, the correct answer is option (a)