Question:
Let f be a polynomial function such that \(f(3x) = f'(x)\cdot f''(x)\), for all \(x \in \mathbf{R}\). Then :
Let f be a polynomial function such that \(f(3x) = f'(x)\cdot f''(x)\), for all \(x \in \mathbf{R}\). Then :
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Compare degrees to get a cubic, then compare coefficients.
Updated On: Nov 11, 2026
- \(f(2) + f'(2) = 28\)
- \(f''(2) - f'(2) = 0\)
- \(f''(2) - f(2) = 4\)
- \(f(2) - f'(2) + f''(2) = 10\)
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The Correct Option is B
Solution and Explanation
Let the degree of f be n. The left side has degree n and the right side has degree \((n-1) + (n-2) = 2n - 3\), so \(n = 2n - 3 \Rightarrow n = 3\).
Let \(f(x) = ax^3 + bx^2 + cx + d\). Then \(f'(x) = 3ax^2 + 2bx + c\), \(f''(x) = 6ax + 2b\) and \(f(3x) = 27ax^3 + 9bx^2 + 3cx + d\).
Product \(f'f'' = 18a^2x^3 + 18ab\,x^2 + (4b^2 + 6ac)x + 2bc\).
Compare: \(27a = 18a^2 \Rightarrow a = \frac32\); \(9b = 18ab = 27b \Rightarrow b = 0\); \(3c = 4b^2 + 6ac = 9c \Rightarrow c = 0\); \(d = 2bc = 0\).
So \(f(x) = \frac32 x^3\), \(f'(x) = \frac92 x^2\), \(f''(x) = 9x\).
\(f(2) = 12\), \(f'(2) = 18\), \(f''(2) = 18\).
(1) \(12 + 18 = 30\); (2) \(18 - 18 = 0\) true; (3) \(18 - 12 = 6\); (4) \(12 - 18 + 18 = 12\).
Answer: (2) \(f''(2) - f'(2) = 0\)
Let \(f(x) = ax^3 + bx^2 + cx + d\). Then \(f'(x) = 3ax^2 + 2bx + c\), \(f''(x) = 6ax + 2b\) and \(f(3x) = 27ax^3 + 9bx^2 + 3cx + d\).
Product \(f'f'' = 18a^2x^3 + 18ab\,x^2 + (4b^2 + 6ac)x + 2bc\).
Compare: \(27a = 18a^2 \Rightarrow a = \frac32\); \(9b = 18ab = 27b \Rightarrow b = 0\); \(3c = 4b^2 + 6ac = 9c \Rightarrow c = 0\); \(d = 2bc = 0\).
So \(f(x) = \frac32 x^3\), \(f'(x) = \frac92 x^2\), \(f''(x) = 9x\).
\(f(2) = 12\), \(f'(2) = 18\), \(f''(2) = 18\).
(1) \(12 + 18 = 30\); (2) \(18 - 18 = 0\) true; (3) \(18 - 12 = 6\); (4) \(12 - 18 + 18 = 12\).
Answer: (2) \(f''(2) - f'(2) = 0\)
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