Question

Find 'a' and 'b', if the function given by $f\left(x\right)=\{\begin{array}{c}a{x}^2+b,ifx<1\\ 2x+1,ifx\ge 1\end{array}$ is differentiable at x = 1.

Answer 1

As f is differentiable at x = 1 so, it is continuous at x = 1 as well.
So, $\underset{x\to 1^{+}}{\text{lim}}f\left(x\right)=\underset{x\to 1^{-}}{\text{lim}}f\left(x\right)=f\left(1\right)i.e.,\underset{x\to 1^{-}}{\text{lim}}a{x}^2+b=2\left(1\right)+1$ $\Rightarrow a+b=\mathrm{3...}\left(i\right)$
Also $L{f}^{\prime }\left(1\right)=R{f}^{\prime }\left(1\right)i.e.,\underset{x\to 1^{-}}{\text{lim}}\frac{a{x}^2+b-3}{x-1}=\underset{x\to 1^{+}}{\text{lim}}\frac{2x+1-3}{x-1}$
$\Rightarrow \underset{x\to 1^{-}}{\text{lim}}\frac{a{x}^2-a}{x-1}=\underset{x\to 1^{+}}{\text{lim}}\frac{2(x-1)}{x-1}\Rightarrow \underset{x\to 1^{-}}{\text{lim}}\frac{a(x^2-1)}{x-1}=\underset{x\to 1^{+}}{\text{lim}}2\left[By\right(i),b-3=-a]$
$\Rightarrow \underset{x\to 1^{-}}{\text{lim}}a(x+1)=2$ $\Rightarrow a(1+1)=2$ $\therefore a=1,b=2$ [By (i).]