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Periodic f: ∫f(2x) from 3 to 3+3T equals KI | JEE

JEE Maths question with a full step-by-step solution.

Question

Let

T>0

be a fixed real number. Suppose

f

is continuous with

f(x+T)=f(x)

for all

x\in\mathbb{R}

. If

I=\displaystyle\int_{0}^{T}f(x)\,dx

, then

\displaystyle\int_{3}^{3+3T}f(2x)\,dx=KI

, where

K

1

2

3

correct

4

Solution

Step 1: Substitute

t=2x

, so

dt=2\,dx

and

dx=\dfrac{dt}{2}

. The limits map as

x=3\Rightarrow t=6

and

x=3+3T\Rightarrow t=6+6T

\int_{3}^{3+3T}f(2x)\,dx=\dfrac12\int_{6}^{6+6T}f(t)\,dt.

Step 2: Use periodicity. For a

T

-periodic function, the integral over any interval of length

T

equals

I

, and the integral over a length that is an integer multiple of

T

scales accordingly. Here the interval

[6,\,6+6T]

has length

6T=6

periods, and the lower endpoint

6

may be shifted to

0

by periodicity without changing the value:

\int_{6}^{6+6T}f(t)\,dt=\int_{0}^{6T}f(t)\,dt=6\int_{0}^{T}f(t)\,dt=6I.

Step 3: Therefore

\int_{3}^{3+3T}f(2x)\,dx=\dfrac12\cdot 6I=3I.

Step 4: Comparing with

KI

gives

K=3

. Correct answer: (3)

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