Question

If $x^2$ + 2(a - 1)x + a + 5 =0 has real roots belonging to the interval (1, 3) then a$ϵ$

• A. $(-\infty ,\frac{-8}{7}]$
• B. $(4,\infty )$
• C. $(-\infty ,-\frac{48}{3})$
• D. $(\frac{-8}{7},-1)$

Answer 1

The correct option is D

$(\frac{-8}{7},-1)$

$x^2$ + 2 (a - 1)x + a + 5 = 0 has roots $ϵ$ (1, 3)
$\therefore \Delta \ge 0,f\left(1\right)>0,f\left(3\right)>0,1<-\frac{B}{2A}<3\Rightarrow 4(a-1{)}^2-4(a+5)\ge 0\Rightarrow (1+a\left)\right(a-4)\ge 0\Rightarrow aϵ(-\infty ,-1]\cup [4,\infty )...(i)1+2(a-1)+a+5>0\Rightarrow a>-\frac{4}{3}...(ii)9+2(9-1)3+a+5>0\Rightarrow a>-\frac{8}{7}...(iii)1<\frac{2(1-a)}{2}<3\Rightarrow -2<a<0...(iv\left)From\right(i),(ii),(iii\left)and\right(iv)...\therefore aϵ(-\frac{8}{7},-1]$