Question

Let $f\left(x\right)=\{\begin{array}{cc}x,& x\le 1\\ x^2+ax+b,& x>1\end{array},$
If $f\left(x\right)$ is differentiable at $x=1,$ then

• A. $a=1,b=-1$
• B. $a=1,b=1$
• C. $a=-1,b=-1$
• D. $a=-1,b=1$

Answer 1

The correct option is D $a=-1,b=1$

$f\left(x\right)=\{\begin{array}{cc}x,& x\le 1\\ x^2+ax+b,& x>1\end{array}$
$\Rightarrow f^{\prime }\left(x\right)=\{\begin{array}{cc}1,& x<1\\ 2x+a,& x>1\end{array}$

$f\left(x\right)$ is differentiable at $x=1$
Then, it must be continuous at $x=1$
$\Rightarrow \underset{x\to 1^{+}}{lim}f\left(x\right)=\underset{x\to 1^{-}}{lim}f\left(x\right)$
$\Rightarrow 1+a+b=1$
$\Rightarrow a+b=0$

Also, $f^{\prime }\left(1^{+}\right)=f^{\prime }\left(1^{-}\right)$
$\Rightarrow \underset{x\to 1^{+}}{lim}{f}^{\prime }\left(x\right)=\underset{x\to 1^{-}}{lim}{f}^{\prime }\left(x\right)$
$\Rightarrow 2+a=1$
$\Rightarrow a=-1$
$\Rightarrow b=1$