Question

Let the values of $a$ for which the roots of the equation $(x-a)(x-a-1)=0$ lie between the roots of the equation $(x+a)(x+a^2-2)=0$ be $a\in (-\infty ,p)\cup (q,\frac{-1+\sqrt{5}}{2})$, then the value of $q-p$ is

• A. $1$
• B. $2$
• C. $-2$
• D. $-1$

Answer 1

Answer: (B)

$2$

Roots of $(x-a)(x-a-1)=0$ are $a,a+1$.
Roots of $(x+a)(x+a^2-2)=0$ are $-a,-a^2+2$.
Let $f\left(x\right)=(x+a)(x+a^2-2)$.
Since the roots of $(x-a)(x-a-1)=0$ lie between roots of $(x+a)(x+a^2-2)=0$.
$\Rightarrow f\left(a\right)=(a+a)(a^2-2+a)<0$
$\left(2a\right)(a^2-2+a)<0$$=\left(2a\right)(a-1)(a+2)<0$
$\Rightarrow a\in (-\infty ,-2)\cup (0,1)$
$f(a+1)=(2a+1)[a+1-2+a^2]<0$$=(2a+1)(a^2+a-1)<0$
$=(2a+1)(a+\frac{1+\sqrt{5}}{2})(a+\frac{1-\sqrt{5}}{2})<0$$a\in (-\infty ,-\frac{1+\sqrt{5}}{2})\cup (-\frac{1}{2},-\frac{1}{2}+\frac{\sqrt{5}}{2})$
$\Rightarrow a\in (-\infty ,-2)\cup (0,\frac{-1+\sqrt{5}}{2})$
$p=-2$ , $q=0\Rightarrow q-p=2$.