Question

If both the roots of the equation $x^2-6ax+2-2a+9a^2$ = 0 exceed 3, then

• A. $a<\frac{1}{2}$
• B. $a>\frac{1}{2}$
• C. $a<1$
• D. $a>\frac{11}{9}$

Answer 1

Answer: (D)

Condition for both the roots greater than $3$ be
$b^2-4ac>0,1\cdot f(3)0,\frac{-b}{2a}>3$
Now, $b^2-4ac>0$
$\Rightarrow (-6a{)}^2-4\times 1\times (2-2a+9a^2)>0$
$\Rightarrow 36a^2-4(2-2a+9a^2)>0$
$\Rightarrow 4(2a-2)>0$
$\Rightarrow a>2...(i)$
$f(3)>0\Rightarrow (3{)}^2-6a(3)+2-2a+9a^2=0$
$\Rightarrow 9-18a+2-2a+9a^2>0$
$\Rightarrow 9a^2-20a+11>0$
$\Rightarrow (9a-11)(a-1)>0$
$a<1$ and $a>\frac{11}{9}...(ii)$
and $\frac{-b}{2a}>3$,
$\therefore \frac{+6a}{2(1)}>3$
$\Rightarrow a>1...(iii)$
From Eqs. $(i),(ii)$ and $(iii)$, we get $a>11/9$