AIEEE Test...(2)

co-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

i 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

for 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

if question is similar to archana den putting x = cos φ , cos2φ

can also be tried

for 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

yes question (1) is surely a typo ...
question shuld have been as archana said [1]