If roots of `x^2-(a-3)x+a=0` are such that at least one of them is greater than 2, then `a in [7,9]` b. `a in [7,oo]` c. `a in [9,oo]` d. `a in [7,9]`

Options

A. `a in [7,9]`
B. `a in [7, infty)`
C. `a in [ 9 , infty)`
D. ` a in [ 7,9)`

Answer

3

Explanation

` x^(2) - (a - 3) x + a = 0`
`rArr D = (a - 3)^(2) - 4a `
` = a^(2) - 10 a + 9`
`= (a - 1) (a - 9)`
Case I:
Both the roots are greater then 2.
`D ge 0,f(2) gt 0 , (B)/(2A) gt 2`
`rArr (a - 1) (a - 9) ge 0, 4 - (a - 3) 2 + a gt 0, (a - 3)/(2) gt 2`
`rArr a in (-infty, 1] cup [ 9 , infty), a lt 10, a gt 7`
Case II:
One root is greater then 2 and the other root is less then or
equal to 2 .Hence ,
`f(2) le 0 `
`rArr 4 - (a - 3) 2 + a le 0 `
`rArr a ge 10`
From (1) and (2) , (2)
` a in [ 9 , 10 )cup [10, infty)`
`rArr a in [ 9, infty)`