Question

If $f:R\to [-1,2],f(x)=\frac{x^2+9bx+17}{a{x}^3+x^2+bx+33}$ is onto function and $f^{\prime }(d)=f^{\prime }(e)=0$. Then

• A. eccentricity of curve $\frac{x^2}{(d+e{)}^2}+\frac{y^2}{b+1}=1$ can be $\frac{1}{2}$
• B. eccentricity of curve $\frac{x^2}{(d+e{)}^2}+\frac{y^2}{b+1}=1$ can be $\frac{\sqrt{5}}{2}$
• C. eccentricity of curve $\frac{x^2}{(d+e{)}^2}+\frac{y^2}{a+1}=1$ can be $\frac{\sqrt{3}}{2}$
• D. eccentricity of curve $\frac{x^2}{(d+e{)}^2}+\frac{y^2}{a+1}=1$ can be $\frac{\sqrt{7}}{2}$.

Answer 1

Answer: (A), (B), (C)

$\because$ Domain of function is real number.
$\therefore a=0$
Now, $y=\frac{x^2+9bx+17}{x^2+bx+33}\Rightarrow (y-1)x^2+(by-9b)x+(33y-17)=0$
$\because x$ is real
$\therefore (by-9b{)}^2-4(y-1)(33y-17)\ge 0$
$\because$ Range is $[-1,2]$
$\therefore -1$ and 2 should be root
$\therefore b=+2$ and -2
$\therefore$ for $b=2\Rightarrow d=-5,e=7$
and $b=-2\Rightarrow d=-7,e=+5$