JEE Main 2023 · 24 January · evening · Single correct · +4 / −1

If f(x) = x³ - x²f'(1) + xf''(2) - f'''(3),x ∈ ℝ, then

  1. (a) 2f(0) - f(1) + f(3) = f(2)
  2. (b) f(1) + f(2) + f(3) = f(0)
  3. (c) f(3) - f(2) = f(1)
  4. (d) 3f(1) + f(2) = f(3)
Answer and solution

Answer (a) 2f(0) - f(1) + f(3) = f(2)

Solution

Maths sir · 4 steps

Those three derivative values are just constants: call them a, b, c and let f pin them down.

  1. Name the constants. f'(1), f''(2) and f'''(3) are fixed numbers, not functions of x. Call them a, b and c, so f(x) = x³ - ax² + bx - c.
  2. Differentiate. f'(x) = 3x² - 2ax + b, f''(x) = 6x - 2a and f'''(x) = 6. So c = f'''(3) = 6.

    b = f''(2) = 12 - 2a, a = f'(1) = 3 - 2a + b

  3. Solve and evaluate. Substituting b gives a = 15 - 4a, so a = 3 and b = 6. Then f(x) = x³ - 3x² + 6x - 6, which gives f(0) = -6, f(1) = -2, f(2) = 2 and f(3) = 12.
    OptionLeft sideRight sideHolds?
    (a)2(-6) - (-2) + 12 = 2f(2) = 2yes
    (b)-2 + 2 + 12 = 12f(0) = -6no
    (c)12 - 2 = 10f(1) = -2no
    (d)3(-2) + 2 = -4f(3) = 12no
  4. So. Only (a) is true: 2f(0) - f(1) + f(3) = f(2).

The idea to keep. When a function's definition contains its own derivatives at fixed points, treat each one as an unknown constant. Differentiate, then evaluate at those points to get equations for the constants.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Differentiation

Asked about 40 times in JEE Main since 2020. This question is from 2023.