Question

If $f\left(x\right)=\{\begin{array}{cc}ax,& x<1\\ a{x}^2+bx+2,& x\ge 1\end{array}$ is a differentiable function, then the value of $a-b$ is

Answer 1

Given: $f\left(x\right)=\{\begin{array}{cc}ax,& x<1\\ a{x}^2+bx+2,& x\ge 1\end{array}$
The function $f\left(x\right)$ is differentiable, then it is continuous and differentiable at $x=1$.
Applying condition of continuity
$L.H.L.=R.H.L.\Rightarrow a=a+b+2\Rightarrow b=-2$
Now using differentiability condition at $x=1$
$f^{\prime }\left(x\right)=\{\begin{array}{cc}a,& x<1\\ 2ax+b,& x>1\end{array}L.H.D.=R.H.D.\Rightarrow a=2a+b\Rightarrow a=-b=2\therefore a-b=4$