If `6,8 and 12` are `l^(th),m^(th) and n^(th)` terms of an `A.P. and f(x)=nx^2+2lx-2m`, then the equation `f(x)= 0` has -

Options

A. both roots negative
B. both roots greater than 2
C. one root negative other greater than 1
D. exactly one root in (0, 1).

Answer

D

Explanation

Let a be the first term term and d be the common difference of the A.P. It is given that
`a_(r) = 6, a_(s) = 8 and a_(t) = 12`
`rArr a + (r-1)d = 6, a + (s-1)d = 8 and a + (t-1)d = 12` Solving first two equation, we get
`a=(6s-8r+2)/(s-r) and d = (2)/(s-r)`
Substituting these values in a+(t-1) d = 12, we obtain `2r + t = 3s rArr t + 2r - 2s = s rArr f(1) = s`
Clearly, f(0) = - 2s
`therefore f(0) f(1) = - 2s^(2) lt 0`
Hence, f(x) has exactly one root in (0, 1).