Let g(x) = ∫₀ˣ f(t)dt, where f is continuous function in [ 0, 3 ] such that 1/3 ≤ f(t) ≤ 1 for all t∈ [0, 1] and 0 ≤ f(t) ≤ 1/2 for all t∈ (1, 3]. The largest possible interval in which g(3) lies is :

Solution

Maths sir · 4 steps

Split the integral at 1 and bound each piece by its height times its width; then add the bounds.

1. Split at 1. The bounds on f change at t = 1, so write g(3) = ∫₀¹ f(t) dt + ∫₁³ f(t) dt.

   m ≤ f(t) ≤ M on [a,b] ⇒ m(b-a) ≤ ∫ₐᵇ f(t) dt ≤ M(b-a)

2. First piece. On [0, 1] the width is 1 and 1/3 ≤ f ≤ 1. So 1/3 ≤ ∫₀¹ f(t) dt ≤ 1.
3. Second piece. On [1, 3] the width is 2 and 0 ≤ f ≤ 1/2. So 0 ≤ ∫₁³ f(t) dt ≤ 2 × 1/2 = 1.
4. So. Adding the two, 1/3 + 0 ≤ g(3) ≤ 1 + 1, that is g(3) ∈ [1/3, 2]. The answer is (d).

The idea to keep. When the bounds on f change partway, split the integral there. Each piece lies between (least value × width) and (greatest value × width); add the lower bounds and the upper bounds separately.

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