Question

Find the values of ' $a$ ' in the domain of the definition of the function, $f(a)=\sqrt{2 a^2-a}$ for which the roots of the equation, $x^2+(a+1) x+(a-1)=0$ lie between $-2 \& 1$.

Answer 1

SOLUTION : .$f(a)=\sqrt{2 a^2-a}$ for domain of $f(x) \quad ; \quad 2 a^2-a \geq 0 \quad \Rightarrow \quad a(2 a-1) \geq 0$

$\therefore \quad a \in(-\infty, 0] \cup\left[\frac{1}{2}, \infty\right)$. Let $g(x) \equiv x^2+(a+1) x+(a-1)=0$

(i) $\mathrm{D} \geq 0$

$(a+1)^2-4(a-1) \geq 0 \quad \Rightarrow \quad a \in R$

(ii) $-2<-\frac{B}{2 A}<1$

$\Rightarrow \quad-2<-\frac{(a+1)}{2}<1$

$\Rightarrow \quad a \in(-3,3)$

(iii) $g(-2)>0 \quad \Rightarrow \quad 4-2(a+1)+(a-1)>0 \quad \Rightarrow \quad a<1 \\$

$g(1)>0 \quad \Rightarrow \quad 1+a+1+a-1>0 \quad \Rightarrow \quad a>-1 / 2 \\$

(iv) Now (i) $\cap($ ii $) \cap($ iii $) \cap($ iv) we get

Ans. : $a \in\left(-\frac{1}{2}, 0\right] \cup\left[\frac{1}{2}, 1\right)$