If the function g(x) = begin{cases} ae^x, & x | vedclass

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(B) For the function $g(x)$ to be differentiable at $x = 0$,it must first be continuous at $x = 0$.Continuity at $x = 0$ implies $\lim_{x 	o 0^-} g(x

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$2$

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(B) For the function $g(x)$ to be differentiable at $x = 0$,it must first be continuous at $x = 0$.Continuity at $x = 0$ implies $\lim_{x 	o 0^-} g(x) = \lim_{x 	o 0^+} g(x) = g(0)$.$\lim_{x 	o 0^-} ae^x = ae^0 = a$.$\lim_{x 	o 0^+} (b\cos x + x) = b\cos(0) + 0 = b$.Thus,$a = b$  .......$(1)$For differentiability,the left-hand derivative $(LHD)$ must equal the right-hand derivative $(RHD)$ at $x = 0$.$LHD = \frac{d}{dx}(ae^x) = ae^x$. At $x = 0$,$LHD = a$.$RHD = \frac{d}{dx}(b\cos x + x) = -b\sin x + 1$. At $x = 0$,$RHD = -b\sin(0) + 1 = 1$.Equating $LHD = RHD$,we get $a = 1$.Since $a = b$,we have $b = 1$.Therefore,$a^2 + b^2 = 1^2 + 1^2 = 1 + 1 = 2$.

If the function $g(x) = \begin{cases} ae^x, & x \le 0 \\ b\cos x + x, & x > 0 \end{cases}$ is differentiable,then the value of $a^2 + b^2$ is

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