The values of `a` for which the roots of the equation `(a+1)x^(2)-3ax+4a=0 (a ne -1)` are real and greater than `1` are

Options

A. `a in (-oo, -1)uu(2, oo)`
B. `a in (-16//7, -0]`
C. `a in - [16//7, -1)`
D. `a in (-1//2, oo)`

Answer

C

Explanation

Let `f(x) = (a+1) x^(2) - 3ax + 4a`. The equation f(x) = 0 will have roots greater than 1, iff
(i) Discriminant `ge` 0
(ii) `1 lt` x-coordinate of the vertex of the parabola y = f(x)
(iii) 1 lies outside the roots i.e. `(a+1) f(1) gt 0`
Now,
(i) Discriminant `ge` 0
`rArr" "9a^(2) - 16 a (a+1) ge 0`
`rArr" "-7a^(2) - 16a ge 0 rArr a (7a + 16) le 0 rArr -(16)/(7) le a le 0" "...(i)`
(ii) `1 lt` x-coordinate of the vertex
`rArr" "1 lt (3a)/(2(a+1))`
`rArr" "(a-2)/(a+1) gt 0 rArr a lt -1 or a gt 2" "...(ii)`
and, `(a+1) f(1) gt 0`
`rArr" "(a+1) (a+1-3a + 4a) gt 0`
`rArr" "(a+1) (2a + 1) gt 0`
`rArr" "a lt - 1 or a gt -1//2" "...(iii)`
From (i), (ii) and (iii), we get
`(-16)/(7) le a lt -1 i.e. a in [-16//7, -1)`