Question

If the derivative function $f\left(x\right)=\{\begin{array}{c}b{x}^2+ax+4,x\ge -1\\ a{x}^2+b,x<-1\end{array}$ is everywhere continuous, then

• A. $a=2,b=3$
• B. $a=3,b=2$
• C. $a=-2,b=-3$
• D. $a=-3,b=-2$

Answer 1

Answer: (A)

$a=2,b=3$

$b(1{)}^2+a(-1)+4=a(-1{)}^2+b$

$b-a+4=+a+b\Rightarrow 2a=4\Rightarrow a=2$

Derivative is continuous

$2b(-1)+a=2a(-1)$

$-2b+a=-2a$

$+2b=+3a$

$b=3\frac{a}{2}$

$b=\frac{3\times 2}{2}\Rightarrow b=3$