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 $a > 11/9$