The set of values of 'a' for which the roots of the equation `(a-3)x^(2) - 2ax + 5a = 0` are positive, is

Options

A. `(-oo, 0)uu(3, oo)`
B. `[0, 15//4]`
C. `(3, 15//4)`
D. `(3, 15//4]`

Answer

D

Explanation

Let `f(x) = (a-3)x^(2) - 2ax + 5a`. For the roots of f(x) = 0 to be positive, we must have
(i) Discriminant `le` 0
(ii) The x-coordinate of vertex of the parabola y = f(x) is positive.
(iii) 0 is outside the roots i.e. `(a-3) f(0) gt 0`
Now,
(i) Discriminant `le` 0
`rArr" "4a^(2) - 20a (a-3) ge 0`
`rArr" "-16a^(2) + 60a ge 0`
`rArr" "4a(4a-15) le 0 rArr 0 le a le 15//4" "...(i)`
(ii) The x-coordinate of the vertex is positive
`rArr" "(a)/(a-3) gt 0 rArr a lt 0 or a gt 3" "...(ii)`
and,
(iii) `(a-3) f(0) gt 0`
`rArr" "(a-3) 5a gt 0 rArr a (a-3) gt 0 rArr a lt 0 or a gt 3" "...(iii)`
From (i), (ii) and (iii), we get
`3 lt a le 15//4 i.e. a in (3, 15//4]`