If the roots of the equation `x^2-2a x+a^2-a-3=0` are real and less than 3, then (a)`a<2` b. `2<-alt=3` c. `34``

Options

A. `a lt 2`
B. `2 le a le 3`
C. `3 lt a le 4`
D. `a gt 4`

Answer

A

Explanation

Let `f(x) = x^(2) - 2ax + a^(2) + a -3`. Clearly, y = f(x) represents a parabola opening upward.
If roots of f(x) = 0 are less than 3, we must have
(i) Discriminant `gt 0`
(ii) x-coordinate of vertex of y = f(x) is less than 3
(iii) 3 lies outside the roots of `f(x) = 0 i.e. f(3) gt 0`
Now,
(i) Discriminant `ge 0`
`rArr" "4a^(2) -4(a^(2) + a - 3) ge 0 rArr -4 (a-3) ge 0 rArr a - 3 le 0`
`rArr" "a le 3" "...(i)`
(ii) x-coordinate of vertex `lt 3`
`rArr" "a lt 3" "...(ii)`
(iii) `f(3) lt 0 rArr a^(2) - 5a + 6 gt 0 rArr (a-2) (a-3) gt 0`
`rArr" "a lt 2 or a gt 3" "...(iii)`
From (i), (ii) and (iii), we get `a lt 2`.