Question

The values of ${}^{\prime }{a}^{\prime }$ for which the roots of the equation $(a+1)x^2-3ax+4a=0(a\ne -1)$ to be greater than unity are

• A. $-\frac{16}{7}\le a<-1$
• B. $-2<a<-1$
• C. $0<a<1$
• D. none of these

Answer 1

Answer: (A)

$-\frac{16}{7}\le a<-1$

Let $f\left(x\right)=(a+1)x^2-3ax+4a$ and let $\alpha ,\beta$ be the roots of the equation $f\left(x\right)=0$

The equation will have roots greater than $1$ iff

(1) $D\ge 0$

(2) $\alpha +\beta >2$

(3) $(a+1)f\left(1\right)>0$

Now $D\ge 0\Rightarrow 9a^2-16a(a+1)\ge 0\Rightarrow -7a^2-16a\ge 0$

$\Rightarrow a(7a+16)\le 0\Rightarrow -\frac{16}{7}\le a\le -0$ ...(1)

$\alpha +\beta >2\Rightarrow \frac{3a}{a+1}>0\Rightarrow \frac{3a}{a+1}-2>0\Rightarrow \frac{3a-2a-2}{a+1}>0$

$\frac{a-2}{a+1}>0\Rightarrow a<-1$ or $a>2$ ...(ii)

$(a+1)f\left(1\right)>0\Rightarrow (a+1)(a+1-3a+4a)>0$

$\Rightarrow (a+1)(2a+1)>0\Rightarrow a<-1$ or

$a>-\frac{1}{2}$ ...(iii)

From (i),(ii) and (iii) we get

$-\frac{16}{7}\le a<-1$