# Solve this

Question:

Let $f(x)\left\{\begin{array}{ll}a x^{2}+1, & x>1 \\ x+1 / 2, & x \leq 1\end{array}\right.$. Then, $f(x)$ is derivable at $x=1$, if

(a) $a=2$

(b) $a=1$

(c) $a=0$

(d) $a=1 / 2$

Solution:

(d) $a=1 / 2$

Given: $f(x)= \begin{cases}a x^{2}+1, & x>1 \\ x+\frac{1}{2}, & x \leq 1\end{cases}$

The function is derivable at $x=1$, iff left hand derivative and right hand derivative of the function are equal at $x=1$.

$(\mathrm{LHD}$ at $x=1)=\lim _{x \rightarrow 1^{-}} \frac{f(x)-f(1)}{x-1}$

$(\mathrm{LHD}$ at $x=1)=\lim _{h \rightarrow 0} \frac{f(1-h)-f(1)}{1-h-1}$

$(\mathrm{LHD}$ at $x=1)=\lim _{h \rightarrow 0} \frac{f(1-h)-f(1)}{-h}$

$(\mathrm{LHD}$ at $x=1)=\lim _{h \rightarrow 0} \frac{\left(1-h+\frac{1}{2}\right)-\frac{3}{2}}{-h}=1$

$(\mathrm{RHD}$ at $x=1)=\lim _{x \rightarrow 1^{+}} \frac{f(x)-f(1)}{x-1}$

$(\mathrm{RHD}$ at $x=1)=\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{1+h-1}$

$(\mathrm{RHD}$ at $x=1)=\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{h}$

$(\mathrm{RHD}$ at $x=1)=\lim _{h \rightarrow 0} \frac{a(1+h)^{2}+1-\frac{3}{2}}{h}$

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

$\because \mathrm{LHD}=\mathrm{RHD}$

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

$\Rightarrow a=\frac{1}{2}$