The function g(x) = begin{cases} x + b, & x

Question

(D) For $g(x)$ to be differentiable at $x = 0$,it must first be continuous at $x = 0$. For continuity: $\lim_{x 	o 0^-} g(x) = \lim_{x 	o 0^+} g(x)

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for no value of $b$

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(D) For $g(x)$ to be differentiable at $x = 0$,it must first be continuous at $x = 0$. For continuity: $\lim_{x 	o 0^-} g(x) = \lim_{x 	o 0^+} g(x) = g(0)$. $\lim_{x 	o 0^-} (x + b) = b$ and $\lim_{x 	o 0^+} \cos x = \cos(0) = 1$. Thus,$b = 1$. Now,check for differentiability using the left-hand derivative $(LHD)$ and right-hand derivative $(RHD)$ at $x = 0$: $RHD = g'(0^+) = \lim_{h 	o 0^+} \frac{\cos(0 + h) - \cos(0)}{h} = \lim_{h 	o 0^+} \frac{\cos h - 1}{h} = 0$. $LHD = g'(0^-) = \lim_{h 	o 0^-} \frac{(0 + h + b) - (0 + b)}{h} = \lim_{h 	o 0^-} \frac{h}{h} = 1$. Since $LHD 
eq RHD$ $(1 
eq 0)$,the function $g(x)$ cannot be made differentiable at $x = 0$ for any value of $b$.

The function $g(x) = \begin{cases} x + b, & x < 0 \\ \cos x, & x \geqslant 0 \end{cases}$ can be made differentiable at $x = 0$.

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