Consider a polynomial f(x), which satisfies the following conditions:
The value of f ′(0) is
(A) 0
(B) 1/4
(C) 1/2
(D) 1
Answer is (D) 1
Let f(x) = ax2 + bx + c. Then f′(0) = b > 0.
Also, f(x) = {f′(x)}2
⇒ ax2 + bx + c = 4a2x2 + 4abx + b2
Thus, a = 4a2, b = 4ab and c = b2.
From which, we get
since b > 0 and so c = 1
Hence, f′(0) = b = 1.
