Question

Let $f(x) = \begin{cases} \frac{x^3+x^2-16x+20}{(x-2)^2} & if x \neq 2 \\\ b & \text{if x = 2} \end{cases}$ If $f(x)$ is continuous for all $x$ , then $b$ is equal to

• A. 7
• B. 3
• C. 2
• D. 5

Answer 1

Answer: (A)

7

Answer 2

The correct answer is A:7
Given that;
$f(x)=\frac{x^3+x^2-16x+20}{(x-2)^2},x\neq2$
The given function is continuous at $x=2$
$\therefore \underset{x\rightarrow 2^-}{\lim}f(x)=\underset{x\rightarrow 2^+}{\lim}f(x)=f(2)$
$\underset{x\rightarrow 2}{\lim}=f(2)$
Now,consider $\underset{x\rightarrow 2}{\lim}f(x)=\underset{x\rightarrow 2}{\lim}\frac{x^3+x^2-16x+20}{(x-2)^2}$
$=\underset{x\rightarrow 2}{\lim}(x+5)=7$
$\therefore f(2)=7$