Question

If $f\left(x\right)=\{\begin{array}{c}ax,x<2\\ a{x}^2-bx+3,x\ge 2\end{array}$ is differentiable for all $x,$ then the value of $a+b$ is:

Answer 1

Given : $f\left(x\right)=\{\begin{array}{c}ax,x<2\\ a{x}^2-bx+3,x\ge 2\end{array}$
Since, $f$ is differentiable for all $x,$ so it will be also continuous.
Therefore,
$f\left(2\right)=f\left(2^{-}\right)\Rightarrow 4a-2b+3=2a$
$\Rightarrow 2a-2b+3=0\cdots \left(1\right)$

Now,
$f^{\prime }\left(x\right)=\{\begin{array}{c}ax<2\\ 2ax-bx>2\end{array}$
For differentiablity at $x=2$,
$L.H.D.=R.H.D.\Rightarrow a=4a-b\Rightarrow 3a=b\cdots \left(2\right)$
From equations $\left(1\right)$ and $\left(2\right),$ we get
$a=\frac{3}{4},b=\frac{9}{4}\therefore a+b=\frac{12}{4}=3$