Question

Let $\alpha +1$ and $\frac{1}{\alpha }+1$ be the roots of $x^2-2(p+1)x+5p-p^2=0$ for non-zero $\alpha .$ If $f:R\to [0,\infty )$ defined by $f\left(x\right)=x^2-2(p+1)x+5p-p^2$ is surjective function, then $p$ can be

• A. $1$
• B. $\frac{1}{2}$
• C. $2$
• D. $4$

Answer 1

The correct option is A $1$

For $f$ to be surjective,
$f_{min}=\frac{-D}{4a}=0\Rightarrow (-2(p+1){)}^2-4(5p-p^2)=0\Rightarrow 2p^2-3p+1=0\Rightarrow (p-1\left)\right(2p-1)=0$
$\Rightarrow p=1$ or $p=\frac{1}{2}$

Sum of roots $:\alpha +\frac{1}{\alpha }+2=2(p+1)\dots \left(1\right)$
Product of roots $:(\alpha +1)(\frac{1}{\alpha }+1)=5p-p^2$
$\Rightarrow \alpha +\frac{1}{\alpha }+2=5p-p^2\dots \left(2\right)$

From $\left(1\right)$ and $\left(2\right),$
$2(p+1)=5p-p^2$
$\Rightarrow p^2-3p+2=0$
$\Rightarrow (p-2)(p-1)=0$
$\Rightarrow p=1$ or $2$

Common value of $p$ is $1$