JEE Main · 2021 · 18 Mar 2021 · Shift 2 · Definite Integrals

Let

g (x) = \int_{0}^{x} f (t) d t

, where

f

is continuous function in

[0, 3]

such that

\frac{1}{3} \leq f (t) \leq 1

for all

t \in [0, 1]

and

0 \leq f (t) \leq \frac{1}{2}

for all

t \in (1, 3]

. The largest possible interval in which

g (3)

lies is :

Answer: (B)

Solution

We have

g (3) = \int_{0}^{3} f (t) d t = \int_{0}^{1} f (t) d t + \int_{1}^{3} f (t) d t

.

For the minimum value:

On $[0, 1]$ : $f (t) \geq \frac{1}{3}$ , so $\int_{0}^{1} f (t) d t \geq \frac{1}{3}$
On $(1, 3]$ : $f (t) \geq 0$ , so $\int_{1}^{3} f (t) d t \geq 0$
Minimum: $g (3) \geq \frac{1}{3}$

For the maximum value:

On $[0, 1]$ : $f (t) \leq 1$ , so $\int_{0}^{1} f (t) d t \leq 1$
On $(1, 3]$ : $f (t) \leq \frac{1}{2}$ , so $\int_{1}^{3} f (t) d t \leq \frac{1}{2} \cdot 2 = 1$
Maximum: $g (3) \leq 2$

Therefore,

g (3) \in [\frac{1}{3}, 2]

JEE Main · 2021 · 18 Mar 2021 · Shift 2 · Single correct option