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

The Correct Option is A

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 \(\underset{x\rightarrow 2}{\lim}f(x)=\underset{x\rightarrow 2}{\lim}\frac{x^3+x^2-16x+20}{(x-2)^2}\)
\(=\underset{x\rightarrow 2}{\lim}(x+5)=7\)
\(\therefore f(2)=7\)