MCQ
$\int\limits_0^{\frac{\pi }{4}} {} (cos 2x)^{3/2}. \cos\, x \,dx =$
- A$\frac{{3\,\pi }}{{16}}$
- B$\frac{{3\,\pi }}{{32}}$
- ✓$\frac{{3\,\pi }}{{16\,\sqrt 2 }}$
- D$\frac{{3\,\pi \,\sqrt 2 }}{{16}}$
Put $\sqrt 2\, \sin\, x = \sin \,\theta$
$\Rightarrow I = \frac{1}{{\sqrt 2 }}\, \int\limits_0^{\frac{\pi }{2}} \,cos^4\, \theta \,d\theta = \frac{{3\,\pi }}{{16\,\sqrt 2 }}$
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$f(x)=\min \{x-[x], 1-x+[x]\}$
$g(x)=\max \{x-[x], 1-x+[x]\}$
where $[x]$ denotes the largest integer not exceeding $x$ :
The positive integer $n$ for which
$\int_0^n(g(x)-f(x)) d x=100$ is
$f(x)=\left\{\begin{array}{ll}-55 x, & \text { if } x<-5 \\ 2 x^{3}-3 x^{2}-120 x, & \text { if }-5 \leq x \leq 4 \\ 2 x^{3}-3 x^{2}-36 x-336, & \text { if } x>4\end{array}\right.$
Let $A=\{ x \in R : f$ is increasing $\} .$ Then $A$ is equal to :