Find all values of a for which both roots of the equation `x^2-6ax+2-2a+9a^2=0` are greater then 3.

Options

A. `a gt (9)/(11)`
B. `a ge (11)/(9)`
C. `a gt (11)/(9)`
D. `a lt (11)/(9)`

Answer

C

Explanation

Let `f(x) = x^(2) - 6ax + 2 - 2a + 9a^(2)`. Clearly, coefficient of `x^(2)` is positive. So, y = f(x) represents a parabola opening upward. Therefore, f(x) = 0 will have its both roots greater than 3, if
(i) Disc `ge` 0 (ii) x-coordinate of vertex `gt` 3
(iii) 3 lies outside the roots of `f(x) = 0 i.e. f(3) gt 0`
Now,
(i) Disc `ge` 0
`rArr" "36a^(2) - 8+8a - 36a^(2) ge 0 rArr -8 + 8a ge 0 rArr a ge 1" "...(i)`
(ii) x-coordinate of vertex `gt` 3
`rArr" "3a gt 3 rArr a gt 1" "...(ii)`
(iii) `f(3) gt 0`
`rArr" "9a^(2) - 20a + 11 gt 0`
`rArr" "(9a-11) (a-1) gt 0 rArr a lt 1 or a gt (11)/(9)" "...(iii)`
From (i), (ii) and (iii), we get `a gt 11//9`.