Suppose f(x) is a function satisfying the following conditions :
(i) f(0)=2,f(1)=1
(ii) f has a minimum value at `x=5//2`
(iii) for all `x,f (x) = |{:(2ax,,2ax-1,,2ax+b+1),(b,,b+1,,-1),(2(ax+b),,2ax+2b+1,,2ax+b):}|`
f(x)=0 has

Options

A. both roots positive
B. both roots negative
C. roots of opposite sign
D. imaginary roots

Answer

D

Explanation

`f(x) = |{:(2ax,,2ax-1,,2ax+b+1),(b,,b+1,,-1),(2(ax+b),,2ax+2b+1,,2ax+b):}|`
Applying `(C_(1) to C_(1)-C_(3) ,C_(2) to C_(2) -C_(3)`
`f(x) =|{:(-(b+1),,-(b+2),,2ax+b+1),((b+1),,(b+2),,-1),(b,,b+1,,2ax+b):}|`
Applying `R_(1) to R_(1)+R_(2)" and " R_(3) to R_(3)-R_(2)` we get
`f(x) = |{:(0,,0,,2ax+b),(b+1,,b+2,,-1),(-1,,-1,,2ax+b+1):}|`
`=(2ax +b) [-b-1+b+2]`
`:. f(x) =2ax+b`
`:. f(x) = ax^(2) +bx +c`
`f(0) =2 rArr c=2`
`f(1) =1rArr a+b+2=1rArr a+b =1`
`f(5//2) =0 rArr 5a+ b=0`
`rArr a=1//4 ,b=-5//4`
hence `f(x) =(1)/(4) x^(2) -(5)/(4)x+2`
Clearly . discriminant (D) of the equation f(x) =0 is less than 0.
hence f(x)=0 has imaginary roots .Also f(2) `=1//2` .and minimum value of f(x) is
`-((25)/(16)-4.(1)/(4) (2))/(4.(1)/(4)) =(7)/(16)`
Hence range of the f(x) is `[(7)/(16),oo)`