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JEE Main Mathematics question on Application of Derivatives

Let f(x)f(x) be a polynomial of degree four having extreme values at x=1x=1 and x=2x=2. If lim⁡x→0[1+f(x)x2]=3\lim_{x\to 0}\left[1+\dfrac{f(x)}{x^2}\right]=3, then f(2)f(2) is equal to
  1. A.
    (a) −8-8
  2. B.
    (b) −4-4
  3. C.
    (c) 00
  4. D.
    (d) 44
Show answer and solution

Correct option: C

Solution

Since f(x) has degree 4 and the limit as x→0 of [1 + f(x)/x²] = 3, f(x)/x² → 2 as x→0. This means f(x) must have no constant term or x term (otherwise f(x)/x² would blow up), so f(x) = ax⁴ + bx³ + cx² with c = 2.

Thus f(x) = ax⁴ + bx³ + 2x², and f'(x) = 4ax³ + 3bx² + 4x.

Since x=1 and x=2 are extreme values, f'(1) = 0 and f'(2) = 0:
- f'(1): 4a + 3b + 4 = 0
- f'(2): 32a + 12b + 8 = 0 → 8a + 3b + 2 = 0

Subtracting: 4a - 2 = 0 → a = 1/2, then 3b = -2 - 4(1/2) = -4 → b = -4/3...

Let me redo: 4a + 3b = -4 and 8a + 3b = -2. Subtracting first from second: 4a = 2 → a = 1/2, b = (-4 - 2)/3 = -2.

So f(x) = (1/2)x⁴ - 2x³ + 2x².

f(2) = (1/2)(16) -

Previous-year question, JEE Main.

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