The second degree polynomial f(x) satisfying $f(0)=0, f(1)=1, f'(x)>0$ for all $x \in(0,1)$, is

$f(x)=a x+(1-a) x^2, a \in(0,2)$

Let $f(x) = a x^2+b x+c$. Then,

f(0) = 0  and  f(1) = 1 

⇒ c = 0 and a + b + c = 1 ⇒ c = 0 and a + b = 1

∴  f(x) = ax² + bx ⇒ f'(x) = 2ax + b and f''(x) = 2a

Case I: When a > 0

f''(x) = 2a > 0 for all x

⇒ f'(x) is increasing for all x

∴  f'(x) > 0 for all x ∈ (0, 1), if f'(0) > 0 i.e. b > 0

Case II: When a < 0

f''(x) = 2a < 0 for all x

⇒ f'(x) is decreasing for all x

∴ f'(x) > 0 for all x ∈ (0, 1)

if f'(1) > 0 i.e. 2a + b > 0

⇒ 2(1 - b) + b > 0                        [∵ a + b = 1]

⇒ b < 2

Hence, f'(x) > 0 for all x ∈ (0, 1) only when b ∈ (0, 2)

∴ $ f(x) = ax^2+b x = (1-b) x^2+b x$, where b ∈ (0, 2)