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Q.
The function $g (x) = \bigg[\begin{matrix}x+b, &x<\,0\\ cos\,x,&x \ge0\end{matrix}$ can be made differentiable at x = 0,
Continuity and Differentiability
Solution:
First $f(x)$ must be continuous at $x=0,$ for which $0+b=\cos 0$ or $b=1$
Also $g'(x) = \bigg[\begin{matrix}1, & x<\,0\\ -sin\,x,&x >\,0\end{matrix}$
$\Rightarrow g'\left(0^{+}\right)=0$ and $g'\left(0^{-}\right)=1$
Hence, $f(x)$ is non-differentiable at $x=0$.
Thus $g$ cannot be made differentiable for any value of $b$