MCQ
$\int_0^{\pi /2} {\frac{{\sqrt {\cos x} }}{{\sqrt {\sin x} + \sqrt {\cos x} }}\,dx = } $
- A$0$
- B$\frac{\pi }{2}$
- ✓$\frac{\pi }{4}$
- DNone of these
and $I = \int_0^{\pi /2} {\frac{{\sqrt {\cos \left( {\frac{\pi }{2} - x} \right)} }}{{\sqrt {\sin \left( {\frac{\pi }{2} - x} \right)} + \sqrt {\cos \left( {\frac{\pi }{2} - x} \right)} }}dx} $
$I = \int_0^{\pi /2} {\frac{{\sqrt {\sin x} }}{{\sqrt {\cos x + } \sqrt {\sin x} }}} \,dx$....$(ii)$
Adding $(i)$ and $(ii),$ we get
$2I = \int_0^{\pi /2} {(1)dx = \frac{\pi }{2} \Rightarrow I = \frac{\pi }{4}} $.
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$I$. If $9$ divides $a^3+b^3+c^3$, then $3$ divides $a b c$.
$II$. If $9$ divides $a^3+b^3+c^3+d^3$, then $3$ divides abcd. Then