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 hasA. both roots positiveB. both roots negativeC. roots of opposite signD. imaginary roots

Suppose f(x) is a function satisfying the following conditions : (i) f

1.	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 hasA. both roots positiveB. both roots negativeC. roots of opposite signD. imaginary roots
Answer» Correct Answer - D `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)`