NIMCET 2019 Question Paper with solutions

Given:

and $\mathit{f}$ is continuous at $\mathit{x}=0$

Let's find $\mathit{f}\left(0\right)$

$\Rightarrow \mathit{f}\left(0\right)=\cos \left[0\right]=1$
$\mathit{L}\mathit{H}\mathit{L}=\underset{\mathit{x}\to 0^{-}}{\lim }\left(\left|\mathit{x}\right|+\mathit{a}\right)=\underset{\mathit{h}\to 0}{\lim }\left(\left|0-\mathit{h}\right|+\mathit{a}\right)=\mathit{a}$

As we know that, if any function is continuous if

$\underset{\mathit{\delta }\mathit{x}\to 0^{-}}{\lim }\mathit{f}\left(\mathit{x}\right)=\mathit{f}\left(0\right)=\underset{\mathit{\delta }\mathit{x}\to 0^{+}}{\lim }\mathit{f}\left(\mathit{x}\right)$

i.e left hand limit $=$ function value $=$ right hand limit

The given function $\mathit{f}$ is continuous at $\mathit{x}=0$

$\Rightarrow \mathit{L}\mathit{H}\mathit{L}=\mathit{f}\left(0\right)$

$\Rightarrow \mathit{a}=1$