Let $f\left(x\right)=\{\begin{array}{ccc}a{x}^2+1& \text{for}& x>1\\ x+a& \text{for}& x\le 1\end{array}$ then $f$ is derivable at $x=1$ if

$\underset{x\to 1+}{Lt}\frac{f\left(x\right)-f\left(1\right)}{x-1}=\underset{x\to 1-}{Lt}\frac{f\left(x\right)-f\left(1\right)}{x-1}$

$\underset{x\to 1+}{Lt}\frac{a{x}^2+1-\left(a+1\right)}{x-1}=\underset{x\to 1-}{Lt}\frac{x+a-\left(1+a\right)}{x-1}$

$\underset{x\to 1+}{Lt}\frac{a\left(x^2-1\right)}{x-1}=\underset{x\to 1-}{Lt}\frac{x-1}{x-1}$

$\Rightarrow \underset{x\to 1+}{Lt}a\left(x+1\right)=1$

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