Question

If $f\left(x\right)=\{\begin{array}{c}b{x}^2-a;x<-1\\ a{x}^2-bx-2;x\ge -1\end{array}$
If $f$ and $f^{\prime }$ are continuous everywhere, then the equation whose roots are $a$ and $b$ is:

• A. $x^2+3x-2=0$
• B. $x^2-3x+2=0$
• C. $x^2+3x+2=0$
• D. $x^2-3x-2=0$

Answer 1

The correct option is B $x^2-3x+2=0$

$f\left(x\right)=\{\begin{array}{c}b{x}^2-a;x<-1\\ a{x}^2-bx-2;x\ge -1\end{array}$
$\because f\left(x\right)$ is a continuous function
$\therefore f(-1^{-})=f(-1^{+})=f(-1)$
$\Rightarrow b-a=a+b-2$
$\Rightarrow a=1$
$f^{\prime }\left(x\right)=\{\begin{array}{c}2bx;x<-1\\ 2ax-b;x>-1\end{array}$
$\because f^{\prime }\left(x\right)$ is continuous everywhere,
$\therefore f^{\prime }(-1^{-})=f^{\prime }(-1^{+})=f^{\prime }(-1)$
$\Rightarrow -2b=-2a-b$
$\Rightarrow b=2a$
$\therefore a=1,b=2$
Equation whose roots are $1$ and $2$ is
$(x-2)(x-1)=0\Rightarrow x^2-3x+2=0$