Q230 / 270 · NDA 2018 (Phase 1) · Mathematics
If $f(x)=\frac{x^2-9}{x^2-2 x-3}, x \neq 3$ is continuous at $x=3$, then which one of the following is correct?
- A$f(3)=0$
- Correct answerB$f(3)=1.5$
- C$f(3)=1.5$
- D$f(3)=-1.5$
Solution
$f(x)=\frac{x^2-9}{x^2-2 x-3}$
Given, the function is continuous at $\mathrm{x}=3$.
$\begin{aligned}
& f(3)=\lim _{x \rightarrow 3} \frac{x^2-9}{x^2-2 x-3} \\
& =\lim _{x \rightarrow 3} \frac{(x+3)(x-3)}{(x-3)(x+1)} \\
& =\frac{3+3}{3+1}=\frac{6}{4}=\frac{3}{2}=1.5
\end{aligned}$
Hence, the correct answer is option: 'B'.