JEE Main 2020 · 7 January · evening · Single correct · +4 / −1

The value of c in the Lagrange's mean value theorem for the function
ƒ(x) = x³ - 4x² + 8x + 11, when x ∈ [0, 1] is:

  1. (a) 2/3
  2. (b) (√7 - 2)/3
  3. (c) (4 - √5)/3
  4. (d) (4 - √7)/3
Answer and solution

Answer (d) (4 - √7)/3

Solution

Maths sir · 4 steps

Find the slope of the chord from x = 0 to x = 1, then find where the tangent has that same slope.

  1. Chord slope. f(0) = 11 and f(1) = 1 - 4 + 8 + 11 = 16. So the chord joining the end points has slope (16 - 11)/(1 - 0) = 5.
  2. Match the tangent. f'(x) = 3x² - 8x + 8. Lagrange's theorem says there is a c in (0, 1) with f'(c) = 5.

    3c² - 8c + 3 = 0 ⇒ c = (8 ± √(64 - 36))/6 = (4 ± √7)/3

  3. Pick the root. Since √7 ≈ 2.646, (4 + √7)/3 ≈ 2.22 lies outside (0, 1). (4 - √7)/3 ≈ 0.45 lies inside.
  4. So. c = (4 - √7)/3, which is option (d).

The idea to keep. For LMVT, set f'(c) equal to the chord slope (f(b) - f(a))/(b - a), solve, and keep only the root that lies strictly between a and b.

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

Study Applications of Derivatives

Asked about 160 times in JEE Main since 2020. This question is from 2020.