Question

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

• A. $a=1,b=1$
• B. $a=1,b=0$
• C. $a=2,b=0$
• D. $a=2,b=1$

Answer 1

Answer: (C)

$a=2,b=0$

$f\left(x\right)=\{\begin{array}{cc}a{x}^2+1,& x\le 1\\ x^2+ax+b,& x>1\end{array}$ is differentiable at $x=1$.
Then $f\left(x\right)$ is continuous at $x=1$. Therefore,
$f\left(1^{-}\right)=f\left(1^{+}\right)$ or $a+1=1+a+b$ or $b=0$
Also, $f^{\prime }\left(x\right)=\{\begin{array}{cc}2ax,& x<1\\ 2x+a,& x>1\end{array}$
We must have $f^{\prime }\left(1^{-}\right)=f^{\prime }\left(1^{+}\right)$ or $2a=2+a$ or $a=2$.