MCQ
$\int\limits_0^{\frac{\pi }{4}} {} (tan^n x + tan^{n -2} x) d (x - [x])$ is : ( $[. ]$ denotes greatest integer function)
- ✓$\frac{1}{{n\,\, - \,\,1}}$
- B$\frac{1}{{n\,\, + \,\,2}}$
- C$\frac{2}{{n\,\, - \,\,1}}$
- Dnone of these
Where $[. ]$ denotes greatest integer function
Using property, $x=[x]+\{x\}$ where $\{x\}$ denotes fractional part of $x$. $I=\int_{0}^{H / 4}\left(\tan ^{n} x+\tan ^{n-2} x\right) d(\{x\})$
$\frac{\pi}{4}=0.78<1$
$\therefore$ For $x \in\left[0, \frac{\pi}{4}\right], \quad\{x\}=x$
$\therefore I=\int_{0}^{\pi / 4}\left(\tan ^{n} x+\tan ^{-2} x\right) d x$
$\therefore I=\int_{0}^{\pi / 4} \tan ^{n-2} x\left[1+\tan ^{2} n\right] d x$
[ using trigonomotric identity $, 2+\tan ^{2} x=\sec ^{2} x$] $\therefore I=\int_{0}^{\pi / 4}\left(\tan ^{n-2} x\right)\left(\sec ^{2} x d x\right)$
Let $\tan x=t$
$\therefore \sec ^{2} x d x=d t$
$\tan 0=0, \tan \frac{\pi}{4}=1$
$\therefore$ Limit of $t$ laries from 0 to 1 .
$\therefore I=\int_{0}^{1}(t)^{n-2} d t$
$\left.\therefore I=\frac{E^{n-1}}{m-1}\right]_{0}^{1}=\left[\frac{(1)^{n-1}}{n-1}-\frac{(0)^{n-1}}{n-1}\right]$
$\therefore I=\frac{1}{n-1}$
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