Question

If the function $f\left(x\right)=\{\begin{array}{ccc}2a^{2}x-1& ;& x\le 1\\ x^2+x+b& ;& x>1\end{array}$
is differentiable everywhere, where $a,b\in R$, then the value of $2a^2+b$ is

Answer 1

Given : $f\left(x\right)=\{\begin{array}{ccc}2a^{2}x-1& ;& x\le 1\\ x^2+x+b& ;& x>1\end{array}$
For the function to be continuous at $x=1$
$f\left(1^{-}\right)=f\left(1^{+}\right)=f\left(1\right)\Rightarrow 2a^2-1=1+1+b\Rightarrow 2a^2-b=3\cdots \left(1\right)$
Differentiating the function w.r.t. $x$, we get
$f^{\prime }\left(x\right)=\{\begin{array}{ccc}2a^2& ;& x\le 1\\ 2x+1& ;& x>1\end{array}$
If the function is differentiable at $x=1$, then it is differentiable everywhere, so
$f^{\prime }\left(1^{-}\right)=f^{\prime }\left(1^{+}\right)\Rightarrow 2a^2=3$
From equation $\left(1\right)$, we get
$b=0$
Hence, $2a^2+b=3$