Question

STATEMENT-1 : $\underset{0}{\overset{2\pi }{\int }}{\tan }^{2}xdx=4\underset{0}{\overset{\pi /2}{\int }}{\tan }^{2}xdx$
STATEMENT-2 : $\underset{0}{\overset{nT}{\int }}f(x)dx=n\underset{0}{\overset{T}{\int }}f(x)dx$, where $n$ is an integer and $T$ is a period of $f(x)$

• A. Both the statements are true.
• B. Statement-I is true, but Statement-II is false.
• C. Statement-I is false, but Statement-II is true.
• D. Both the statements are false.

Answer 1

Answer: (A)

$\underset{0}{\overset{2\pi }{\int }}{\tan }^{2}xdx=2\underset{0}{\overset{\pi }{\int }}{\tan }^{2}xdx$
$=2\left[\underset{0}{\overset{\pi /2}{\int }}{\tan }^{2}xdx+\underset{0}{\overset{\pi /2}{\int }}{\tan }^2(\pi -x)dx\right]=4\underset{0}{\overset{\pi /2}{\int }}{\tan }^{2}xdx$
$\therefore$ Statement 1 is true
statement-2 $\underset{0}{\overset{nT}{\int }}f(x)dx=\underset{0}{\overset{T}{\int }}f(x)dx+\underset{T}{\overset{2T}{\int }}f(x)dx+\dots ..+\underset{(n-1)T}{\overset{nT}{\int }}f(x)dx$
$=\underset{0}{\overset{T}{\int }}f(x)dx+\underset{0}{\overset{T}{\int }}f(x+T)dx+\dots \dots +\underset{0}{\overset{T}{\int }}f(x+(n-1)T)dx$
$=\underset{0}{\overset{T}{\int }}f(x)dx+\underset{0}{\overset{T}{\int }}f(x)dx+\dots ..+\underset{0}{\overset{T}{\int }}f(x)dx$
$(\because f$ has a period $T)$
$=n\underset{0}{\overset{T}{\int }}f(x)dx$