If $f\left(x\right)=\{\begin{array}{cc}\frac{\tan \left[x^2\right]\pi }{a{x}^2}+a{x}^3+b,& 0<x\le 1\\ 2\cos \pi x+{\tan }^{-1}x,& 1<x\le 2\end{array}$  is differentiable in $(0,2],$  then   $a=\frac{1}{k_1}$and $b=\frac{\pi }{4}-\frac{26}{k_2}.$   Find  $\frac{k_1^2+k_2^2}{90}$ {where [.] denotes greatest integer function}.

$f\left(x\right)=\{\begin{array}{cc}a{x}^3+b,& 0\le x\le 1\\ 2\cos \pi x+{\tan }^{-1}x,& 1<x\le 2\end{array}$  as  $\tan \left[x^2\right]\pi =\tan n\pi ,n\in I$
$f\text{'}\left(x\right)=\{\begin{array}{cc}3a{x}^2,& 0<x<1\\ -2\pi \sin \pi x+\frac{1}{1+x^2},& 1<x<2\end{array}$
As the function is differentiable in [0, 2]
 function is differentiable at  $x=1$
$$\begin{gathered} \therefore f^1\left(1^{-}\right)=f^1\left(1^{+}\right) \\ \Rightarrow 3a=\frac{1}{2}\Rightarrow a=\frac{1}{6} \end{gathered}$$
Function will also be continuous at  $x=1$
$$\begin{gathered} \therefore \underset{x\to 1^{-}}{\lim }f\left(x\right)=\underset{x\to 1^{+}}{\lim }f\left(x\right) \\ \Rightarrow a+b=-2+\frac{\pi }{4} \\ \therefore b=-2-\frac{1}{6}+\frac{\pi }{4}=\frac{\pi }{4}-\frac{13}{6} \\ \Rightarrow k_1=6\&k_2=12 \\ \Rightarrow k_1^2+k_2^2=180 \end{gathered}$$