Let F(x) be an indefinite integral of ^2 x . Statement I The function F(x) satisfies F(x+π)=F(x) for all real x . Statement II ^2(x+π)= ^2 x for all real x .

Correct answer: D. Statement I is false, Statement II is true aligned & F(x)=∫ ^2 x d x=∫ (1- 2 x)/(2)d x & F(x)=(1)/(4)(2 x- 2 x)+c aligned Since, F(x+π) ≠ F(x) . Hence, Statement I is false. But Statement II is true as ^2 x is periodic with period π .

JEE Advanced 2007Paper 1

Exam: JEE Advanced
Year: 2007
Paper: Paper 1
Subject: Mathematics
Chapter: Indefinite Integration

JEE Advanced 2007 · Paper 1 · Mathematics · Indefinite Integration

Let F(x) be an indefinite integral of ^2 x . Statement I The...

Let

F(x)

be an indefinite integral of

\sin ^2 x

. Statement I The function

F(x)

satisfies

F(x+\pi)=F(x)

for all real

x

. Statement II

\sin ^2(x+\pi)=\sin ^2 x

for all real

x

Options

AStatement I is true, Statement II is true; Statement II is a correct explanation for Statement I
BStatement I is true, Statement II is true; Statement II is not a correct explanation for Statement I.
CStatement I is true, Statement II is false
DStatement I is false, Statement II is true

Correct answer

D. Statement I is false, Statement II is true

Step-by-step solution

\begin{aligned}& F(x)=\int \sin ^2 x d x=\int \frac{1-\cos 2 x}{2}d x \\ & \Rightarrow F(x)=\frac{1}{4}(2 x-\sin 2 x)+c \end{aligned}

Since,

F(x+\pi) \neq F(x)

. Hence, Statement I is false. But Statement II is true as

\sin ^2 x

is periodic with period

\pi

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