JEE Main 2020 · 6 September · morning · Single correct · +4 / −1

The position of a moving car at time t is
given by f(t) = at² + bt + c, t > 0, where a, b and c are real numbers greater than 1. Then the average speed of the car over the time interval [t₁ , t₂ ] is attained at the point :

  1. (a) ( t₁ + t₂ )/2
  2. (b) ( t₂ - t₁ )/2
  3. (c) 2a(t₁ + t₂) + b
  4. (d) a(t₂ – t₁) + b
Answer and solution

Answer (a) ( t₁ + t₂ )/2

Solution

Maths sir · 4 steps

Find the average speed, set the instantaneous speed equal to it, and solve for the time.

  1. Average speed. Average speed over [t₁, t₂] is (f(t₂)-f(t₁))/(t₂-t₁). The constant c cancels in the subtraction.

    (a(t₂²-t₁²)+b(t₂-t₁))/(t₂-t₁) = a(t₁+t₂)+b

  2. Instantaneous speed. The speed at time t is f'(t) = 2at + b. By the Mean Value Theorem, some t in (t₁, t₂) has f'(t) equal to the average speed.
  3. Equate them. 2at + b = a(t₁+t₂) + b gives 2at = a(t₁+t₂). Since a > 1, you can divide by 2a to get t = (t₁+t₂)/2.
  4. So. The average speed is attained at the midpoint of the interval, (t₁+t₂)/2, which is option (a).

The idea to keep. For any quadratic, the Mean Value Theorem point is always the midpoint of the interval. The chord slope equals the tangent slope exactly halfway between the ends.

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.