The set of values of a for which the inequation `x^(2)+ax+a^(2)+6alt0` is satisfied for all `x epsilon(1,2)` lies in the interval

Options

A. `(4-sqrt(15),4+sqrt(15))`
B. `(5-sqrt(21),5+sqrt(21))`
C. `(5-sqrt(21),4+sqrt(15))`
D. `(4-sqrt(15),5+sqrt(21))`

Answer

C

Explanation

`Let f(x)=x^(2)-2ax+(a^(2)-6a)`.
we have,
`x^(2)-2ax+(a^(2)-6a) lt 0` for all `x in [1,2]`
`implies `1 and 2 lie between the roots of f(x)=0.
`implies f(1) lt 0,f(2) lt 0 "and Discriminant" gt 0`.
Now,
`f(1) lt 0`
`implies 1-2a+a^(2)-6a lt 0`
`implies a^(2)-8a+1 lt 0 rArr4-sqrt(5) lt a lt 4+sqrt(15) " "`......(i)
`f(2) lt 0`
`rArr 4-4a+a^(2)-6a lt 0`
`impliesa^(2)-10a+4 lt 0 rArr 5-sqrt(21) lt a lt 5+sqrt(21) " "`.......(ii)
and,
`"Discriminant" gt 0`
`rArr 4a^(2)-4(a^(2)-6a) gt 0 rArra gt 0 " "`...........(iii)
From (i), (ii)and (iii), we obtain `a in sqrt(21),4+sqrt(15)`.