Mathematics - Application of Derivative Question with Solution | TestHub

$\begin{array}{rl}& {\mathrm{Lf}}^{\prime }(1)=\underset{h\to 0}{\lim }\frac{f(1-h)-f(1)}{-h}\\ & =\underset{h\to 0}{\lim }\frac{(1-h+a)-(1+a)}{-h}=\underset{h\to 0}{\lim }\frac{-h}{-h}=1\\ & {\mathrm{Rf}}^{\prime }(1)=\underset{h\to 0}{\lim }\frac{f(1+h)-f(1)}{h}\\ & =\underset{h\to 0}{\lim }\frac{\left[a(1+h{)}^2+1\right]-(1+a)}{h}\\ & =\underset{h\to 0}{\lim }(ah+2a)=2a\end{array}$ Since f' (1) exists, $\therefore L{f}^{\prime }(1)=R{f}^{\prime }(1)\Rightarrow a=1/2$