Solve this

The given function $f(x)=\left\{\begin{array}{cl}a x^{2}+3, & x>1 \\ x+\frac{5}{2}, & x \leq 1\end{array}\right.$ is differentiable at $x=1$

$\therefore L f^{\prime}(1)=R f^{\prime}(1)$

$\Rightarrow \lim _{h \rightarrow 0} \frac{f(1-h)-f(1)}{-h}=\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{h}$

$\Rightarrow \lim _{h \rightarrow 0} \frac{\left(1-h+\frac{5}{2}\right)-\left(1+\frac{5}{2}\right)}{-h}=\lim _{h \rightarrow 0} \frac{\left[a(1+h)^{2}+3\right]-\left(1+\frac{5}{2}\right)}{h}$

$\Rightarrow \lim _{h \rightarrow 0} \frac{-h}{-h}=\lim _{h \rightarrow 0} \frac{a(1+h)^{2}-\frac{1}{2}}{h}$

$\Rightarrow 1=\lim _{h \rightarrow 0} \frac{\left(a-\frac{1}{2}\right)+a\left(2 h+h^{2}\right)}{h}$

$\Rightarrow 1=\lim _{h \rightarrow 0} \frac{\left(a-\frac{1}{2}\right)+a h(2+h)}{h}$

$\Rightarrow 1=\lim _{h \rightarrow 0} \frac{\left(a-\frac{1}{2}\right)}{h}+\lim _{h \rightarrow 0} a(2+h)$

Here, $\mathrm{LHS}$ is finite.

So, for RHS to be finite, we must have

$a-\frac{1}{2}=0$

Thus, the value of $a$ is $\frac{1}{2}$.

Let $f(x)=\left\{\begin{array}{cc}a x^{2}+3, & x>1 \\ x+\frac{5}{2}, & x \leq 1\end{array} .\right.$ If $f(x)$ is differentiable at $x=1$, then $a=\underline{\frac{1}{2}}$.