Question with Solution

$f(x)=\{\begin{array}{llc}a{x}^3+b& ,& 0\le x\le 1\\ 2\cos \pi x+{\tan }^{-1}x& ,& 1<x\le 2\end{array}$

$$f^{\prime }(x)=\{\begin{array}{llc}3a{x}^2& ,& 0<x<1\\ -2\pi \sin \pi x+\frac{1}{1+x^2}& ,& 1<x<2\end{array}$$

As the function is differentiable in $[0,2]⟹$ function is differentiable at $x=1$

$$\begin{array}{rl}& \therefore f^{\prime }\left(1^{-}\right)=f^{\prime }\left(1^{+}\right)\\ & \Rightarrow 3a=\frac{1}{2}\Rightarrow a=\frac{1}{6}\end{array}$$

Function will also be continuous at $x=1$

$$\begin{array}{ll}\therefore & {\lim }_{x\to 1^{-}}f(x)={\lim }_{x\to 1^{-}}f(x)\\ \Rightarrow & a+b=-2+\frac{\pi }{4}\\ \therefore & b=-2-\frac{1}{6}+\frac{\pi }{4}=\frac{\pi }{4}-\frac{13}{6}\\ \Rightarrow k_1=6& \\ & k_2=12\Rightarrow k_1^2+k_2^2=180\text{ Ans. }\end{array}$$