Question

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

(a) a = 2
(b) a = 1
(c) a = 0
(d) a = 1/2

Answer 1

(d) a = 1/2


Given: $f\left(x\right)=\left\{\begin{array}{l}a{x}^2+1,x>1\\ x+\frac{1}{2},x\le 1\end{array}\right.$
The function is derivable at x = 1, iff left hand derivative and right hand derivative of the function are equal at x = 1.

$$\begin{gathered} \left(\mathrm{LHD}\mathrm{at}x=1\right)=\underset{x\to 1^{-}}{\lim }\frac{f\left(x\right)-f\left(1\right)}{x-1} \\ \left(\mathrm{LHD}\mathrm{at}x=1\right)=\underset{h\to 0}{\lim }\frac{f\left(1-h\right)-f\left(1\right)}{1-h-1} \\ \left(\mathrm{LHD}\mathrm{at}x=1\right)=\underset{h\to 0}{\lim }\frac{f\left(1-h\right)-f\left(1\right)}{-h} \\ \left(\mathrm{LHD}\mathrm{at}x=1\right)=\underset{h\to 0}{\lim }\frac{\left(1-h+{\displaystyle \frac{1}{2}}\right)-{\displaystyle \frac{3}{2}}}{-h}=1 \\ \left(\mathrm{RHD}\mathrm{at}x=1\right)=\underset{x\to 1^{+}}{\lim }\frac{f\left(x\right)-f\left(1\right)}{x-1} \\ \left(\mathrm{RHD}\mathrm{at}x=1\right)=\underset{h\to 0}{\lim }\frac{f\left(1+h\right)-f\left(1\right)}{1+h-1} \\ \left(\mathrm{RHD}\mathrm{at}x=1\right)=\underset{h\to 0}{\lim }\frac{f\left(1+h\right)-f\left(1\right)}{h} \\ \left(\mathrm{RHD}\mathrm{at}x=1\right)=\underset{h\to 0}{\lim }\frac{a{\left(1+h\right)}^2+1-{\displaystyle \frac{3}{2}}}{h} \\ \left(\mathrm{RHD}\mathrm{at}x=1\right)=\underset{h\to 0}{\lim }\frac{a\left(1+h^2+2h\right)-{\displaystyle \frac{1}{2}}}{h} \\ \because \mathrm{LHD}=\mathrm{RHD} \\ \Rightarrow a-\frac{1}{2}=0 \\ \Rightarrow a=\frac{1}{2} \end{gathered}$$