MCQ
The value of $\int_{\pi /4}^{3\pi /4} {\frac{\phi }{{1 + \sin \phi }}\,d\phi ,} $ is
- ✓$\pi \tan \frac{\pi }{8}$
- B$\log \tan \frac{\pi }{8}$
- C$\tan \frac{\pi }{8}$
- DNone of these
$\left\{ \because \frac{\pi }{4}+\frac{3\pi }{4}=\pi \right\}$
==> $2I = \int_{\pi /4}^{3\pi /4} {\frac{\pi }{{1 + \sin \phi }}d\phi } $
On simplification, we get
$I = \pi (\sqrt 2 - 1) = \pi \tan \frac{\pi }{8}.$
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$f(x)= \begin{cases}n(1-2 n x) \text { if } 0 \leq x \leq \frac{1}{2 n}2 n(2 n x-1) \text { if } \frac{1}{2 n} \leq x \leq \frac{3}{4 n}4 n(1-n x) \text { if } \frac{3}{4 n} \leq x \leq \frac{1}{n} \\ \frac{n}{n-1}(n x-1) \text { if } \frac{1}{n} \leq x \leq 1\end{cases}$
If $n$ is such that the area of the region bounded by the curves $x=0, x=1, y=0$ and $y=f(x)$ is $4$ , then the maximum value of the function $f$ is