MCQ
$\int_{\pi /6}^{\pi /3} {\frac{{dx}}{{1 + \sqrt {\tan x} }} = } $
- ✓$\pi /12$
- B$\pi /2$
- C$\pi /6$
- D$\pi /4$
$ = \int_{\pi /6}^{\pi /3} {\frac{{\sqrt {\cos x} }}{{\sqrt {\cos x} + \sqrt {\sin x} }}\;dx} $ ..$(i)$
$I = \int_{\pi /6}^{\pi /3} {\frac{{\sqrt {\sin x} }}{{\sqrt {\cos x} + \sqrt {\sin x} }}\;} $ ..$(ii)$
(Since $\int_a^b {f(x)dx} = \int_a^b {f(a + b - x)\,dx} $)
Adding $(i)$ and $(ii),$ we get,
$2I = \int_{\pi /6}^{\pi /3} {\;dx} $
==> $I = \frac{1}{2}\left( {\frac{\pi }{3} - \frac{\pi }{6}} \right) = \frac{\pi }{{12}}$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$[A]$ $x=-1$ $[B]$ $x=0$ $[C]$ $x=2$ $[D] x=1$
$I$. Domain of $f\left((g(x))^2\right)=$ Domain of $f(g(x))$
$II$. Domain of $f(g(x))+g(f(x))=$ Domain of $g(f(x))$
$III$. Domain of $f(g(x))=$ Domain of $g(f(x))$
$IV.$ Domain of $g\left((f(x))^3\right)=$ Domain of $f(g(x))$