Let \(F(x)\) be an indefinite integral of \(\sin ^{2} x\). STATEMENT-1 : The function \(F(x)\) satisfies \(F(x+\pi)=F(x)\) for all real \(x\). because STATEMENT-2 : \(\sin ^{2}(x+\pi)=\sin ^{2} x\) for all real \(x\).

(D) Statement-1 is False, Statement-2 is True

(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1

(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1

(C) Statement-1 is True, Statement-2 is False

JEE Advanced · 2007 · 08 Apr 2007 · Paper 1 · Indefinite Integrals

Let

F (x)

be an indefinite integral of

sin^{2} x

.

STATEMENT-1 : The function

F (x)

satisfies

F (x + π) = F (x)

for all real

x

.

because

STATEMENT-2 :

sin^{2} (x + π) = sin^{2} x

for all real

x

(A)Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1
(B)Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
(C)Statement-1 is True, Statement-2 is False
(D)Statement-1 is False, Statement-2 is True✓ Correct

Answer: (D)

Statement-2 is clearly true since

sin^{2} (x + π) = sin^{2} x

(period of

sin^{2} x

π

).

For Statement-1:

F (x) = \int sin^{2} x d x = \int \frac{1 - c o s 2 x}{2} d x = \frac{x}{2} - \frac{s i n 2 x}{4} + C

F (x + π) = \frac{x + π}{2} - \frac{s i n 2 ( x + π )}{4} + C = \frac{x + π}{2} - \frac{s i n 2 x}{4} + C

Since

F (x + π) - F (x) = \frac{π}{2} \neq = 0

, Statement-1 is False.

Therefore, Statement-1 is False and Statement-2 is True.

JEE Advanced · 2007 · 08 Apr 2007 · Paper 1 · Single correct option