The second degree polynomial f(x), satisfying f(0)=o,
`f(1)=1,f'(x)gt0AAx in (0,1)`

Options

A. `f(x)=phi`
B. `f(x)=ax+(1-a)x^(2),a in (0,oo)`
C. `f(x)=ax+(1-a)x^(2),x in (0,2)`
D. non-existent

Answer

C

Explanation

Let `f(x)=ax^(2)+bx+c`, Then f(0)=0 and f(1)=1
`Rightarrow c=0 and a+b+c=1 Rightarrow a+b=1`
`therefore f(x)=ax^(2)bx Rightarrow f'(x)=2ax+b and f''(x)=2a`
CASE-II: When `a gt 0`
`f''(x)=2a gt 0 "for all "x`
`Rightarrow f'(x)` is increasing for all x
`therefore f'(x) gt 0"for all "x in (0,1),"if "f'(0) gt 0, i.e. b gt 0`
CASE-II: When `a lt 0`
`f''(x)=2a lt 0"for all x"`
`Rightarrow f'(x)` is decreasing for all x
`therefore f'(x) gt 0"for all "x in (0,1)`
`"if "f'(1) gt 0 i.e. 2a+b gt 0`
`Rightarrow 2(1-b)+b gt 0" "[therefore a+b=1]`
`Rightarrow b lt 2`
Hence, `f'(x) gt 0` for all `x in (0,1)` only when `b in (0,2)`
`therefore f(x)=ax^(2)+bx=(1-b)^(2)x^(2)+bx,"where", b in(0,2)`