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:

Integral Calculus: JEE Main 2021 Mathematics question

JEE Main 2021 · 18 Mar, Evening shift · Q3404

JEE Main 202118 Mar, Evening shiftMathematicsMediumIntegral Calculus369× since 2002

Let g(x) =

\int_0^x {f(t)dt}

, where f is continuous function in [ 0, 3 ] such that

{1 \over 3}

\le

f(t)

\le

1 for all t

\in

[0, 1] and 0

\le

f(t)

\le

{1 \over 2}

for all t

\in

(1, 3]. The largest possible interval in which g(3) lies is :

JEE Main 2021 · 18 Mar, Evening shift · Q3404

JEE Main 202118 Mar, Evening shiftMathematicsMediumIntegral Calculus369× since 2002

Let g(x) =

\int_0^x {f(t)dt}

, where f is continuous function in [ 0, 3 ] such that

{1 \over 3}

\le

f(t)

\le

1 for all t

\in

[0, 1] and 0

\le

f(t)

\le

{1 \over 2}

for all t

\in

(1, 3]. The largest possible interval in which g(3) lies is :