Mathematics - Continuity - Differentiability Question with Solution | TestHub

$f(x)=\{\begin{array}{ccccc}a{x}^3+b& ,& 0\le x\le 12\cos \pi x+{\tan }^{-1}& ,& 1<x\le 2\end{array}$ $f^{\prime }(x)=\{\begin{array}{ccc}3a{x}^2& ,0<x<1-2\pi \sin \pi x+\frac{1}{1+x^2}& ,1<x<2\end{array}$ As the function is differentiable in $[0,2]\Rightarrow$ function is differentiable at $x=1$

$\begin{array}{llc}\therefore & f^{\prime }\left(1^{-}\right)=f^{\prime }\left(1^{+}\right)\Rightarrow & 3a=\frac{1}{2}\Rightarrow a=\frac{1}{6}\end{array}$ Function will also be continuous at $\mathrm{x}=1$

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