1i think the first integral is
\int_{0}^{1}{\sqrt\frac{1-x}{1+x}}=\int_{0}^{1}\frac{1-x}{\sqrt{1-x^2}}= sin^-^1x(0 - 1) +\frac{1}{2}\int_{0}^{1}{\frac{-2x}{\sqrt{1-x^2}}}dx=\frac{\ \pi }{2}-1
0∫1√1-x1+x dx
a)1 b) (pi)/2 + 1 c) (pi) d) (pi)/2 - 1
the ans is d
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6 Answers
11co-efficient of x24 in (1+x2)12(1+x12)(1+x24)
a)3+ 12C6 b)2 + 12C6 c)1+ 12C6 d)12C6
i am geting c but ans given b
11for 2nd one
coefficien of x24 in (i+x2)12 (1+x12+x24)
=coefficient of x24 in (i+x2)12 + coefficient of x12 in (1+x2)12 + constant term in (1+x2)12
=1 + 12C6 + 1
=2 + 12C6.......and so b
1if question is similar to archana den putting x = cos φ , cos2φ
can also be tried
1for 2>
sandipan ans is" b" only
i not posting the soln...but
see u will get 12C0+12C12x24+...... WHICH WILL BE RESPONSIBLE FOR "2" IN THE ANSWER
1yes question (1) is surely a typo ...
question shuld have been as archana said [1]