Question

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 :

• A. [1, 3]
• B. [-1, -1/2]
• C. [2, 4]
• D. [1/3, 2]

Hint:

For any integral $\int_a^b f(x) dx$, if $m \leq f(x) \leq M$, then $m(b-a) \leq \int_a^b f(x) dx \leq M(b-a)$.

Answer 1

The Correct Option is D

Step 1: $g(3) = \int_0^3 f(t) dt = \int_0^1 f(t) dt + \int_1^3 f(t) dt$.
Step 2: For the minimum value of $g(3)$: $g(3)_{min} = \int_0^1 (1/3) dt + \int_1^3 (0) dt = [t/3]_0^1 + 0 = 1/3$.
Step 3: For the maximum value of $g(3)$: $g(3)_{max} = \int_0^1 (1) dt + \int_1^3 (1/2) dt = [t]_0^1 + [t/2]_1^3 = 1 + (3/2 - 1/2) = 1 + 1 = 2$. Thus, $g(3) \in [1/3, 2]$.