The second degree polynomial f(x) satisfying $f(0)=0, f(1)=1, f'(x)>0$ for all $x \in(0,1)$, is
Answer & explanation
Correct answer: option 3
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)