Question
If the function $g\left( x \right) = \left\{ \begin{array}{l} a{e^x},\,\,\,\,\,x \le 0\\ b\cos x + x,\,\,x > 0 \end{array} \right.$ is differentiable, then the value of $a^2 + b^2$ is

Answer

$\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = f\left( 0 \right)$
$\Rightarrow b=a=a......(1)$
$\mathop {\lim }\limits_{h \to {0^ + }} \frac{{f(0 + h) - f(0)}}{h} $
$= \mathop {\lim }\limits_{h \to {0^ - }} \frac{{f(0 - h) - f(0)}}{h}$
$1=a......(2)$
$\therefore a=1=b$

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