Question

If $f:R\to R$ is defined by
$f\left(x\right)=\{\begin{array}{ccc}\frac{x+2}{x^2+3x+2},& if& x\in R-\{-1,-2\}\\ -1,& if& x=-2\\ 0,& if& x=-1\end{array}$
then $f$ is continuous on the set

• A. $R$
• B. $R-\{-2\}$
• C. $R-\{-1\}$
• D. $R-\{-1,-2\}$

Answer 1

Answer: (C)

$R-\{-1\}$

Since, $f\left(x\right)$ is continuous for every value of R except $\{-1,-2\}$. Now, we have to check that points.

At $x=-2$,

L.H.L $=\underset{n\to 0}{lim}\frac{(-2-n)+2}{(-2-n{)}^2+3(-2-n)+2}$

$=\underset{n\to 0}{lim}\frac{-n}{n^2+n}=-1$

R.H.L $=\underset{n\to 0}{lim}\frac{(-2+n)+2}{(-2+n{)}^2+3(-2+n)+2}$

$=\underset{n\to 0}{lim}\frac{n}{n^2-n}=-1$

$\therefore LHL=RHL=f(-2)$

$\therefore f$ is continuous at $x=-2$

Now, check for $x=-1$

L.H.L $=\underset{n\to 0}{lim}\frac{(-1-n)+2}{(-1-n{)}^2+3(-1-n)+2}$

$=\underset{n\to 0}{lim}\frac{1-n}{n^2-n}=\infty$

R.H.L $=\underset{n\to 0}{lim}\frac{(-1+n)+2}{(-1+n{)}^2+3(-1+n)+2}$

$=\underset{n\to 0}{lim}\frac{1+n}{n^2+n}=\infty$

$\therefore LHL=RHL\ne f(-1)$

$\therefore f$ is not continuous at $x=-1$

Hence, the function $f$ is continuous on the set $R-\{-1\}$