NIMCET 2021 Question Paper with solutions

Given,
F(x)= $\{\begin{array}{cc}\frac{\mathit{x}+2}{{\mathit{x}}^2+3\mathit{x}+2}& \mathit{n}\in \mathit{R}\\ -1& \mathit{n}=-2\\ 0& \mathit{n}=-1\end{array}$
For $\mathit{n}\in \mathit{R}$ we have $\mathit{f}\left(\mathit{x}\right)=\frac{\mathit{x}+2}{\left(\mathit{x}+2\right)\left(\mathit{x}+1\right)}\Rightarrow \frac{1}{\mathit{x}+1}$
F(x)= $\{\begin{array}{cc}\frac{\mathit{x}+2}{{\mathit{x}}^2+3\mathit{x}+2}& \mathit{n}\in \mathit{R}\\ -1& \mathit{n}=-2\\ 0& \mathit{n}=-1\end{array}$
$\therefore$ Checking continuity at $\mathit{x}=-2$
$\mathit{L}\mathit{H}\mathit{L}=\underset{\mathit{x}\to -2}{\lim }\mathit{f}\left(\mathit{x}\right)=\underset{\mathit{h}\to 0}{\lim }\frac{1}{2-\mathit{h}+1}$ $=\underset{\mathit{h}\to 0}{\lim }\mathit{f}\left(\mathit{x}\right)$ for $\mathit{x}\in \mathit{R}$
$=\frac{1}{-2-0+1}=-1$
$\mathit{R}\mathit{H}\mathit{L}=\underset{\mathit{x}\to -2\prime }{\lim }\mathit{f}\left(\mathit{x}\right)=\underset{\mathit{h}\to 0}{\lim }\frac{1}{-2+\mathit{n}+1}$ $=\underset{\mathit{h}\to 0}{\lim }\mathit{f}\left(\mathit{x}\right)$ for $\mathit{x}\in \mathit{R}$
$=\frac{1}{-2+0+1}=-1$
Hence $\mathit{L}\mathit{H}\mathit{L}=\mathit{R}\mathit{H}\mathit{L}=\mathit{f}\left(-2\right)=-1$
It is continuous at $\mathit{x}=-2$
$\therefore$ Checking continuity at $\mathit{x}=-1$
$\mathit{L}\mathit{H}\mathit{L}=\underset{\mathit{x}\to -1^{-}}{\lim }\mathit{f}\left(\mathit{x}\right)=\underset{\mathit{n}\to 0}{\lim }\frac{1}{-1-\mathit{h}+1}=\frac{1}{0}=\mathrm{\infty }$
and $\mathit{f}\left(-1\right)=0$
Hence It is not continuous at $\mathit{x}=-1$ from the given option. It is all real number except $-1$ i.e., $\mathit{R}-\left\{-1\right\}$