Paper

Q(2)

You can solve this question by the following method
Possile algebric expression for the given equation is
(x0+x2+x4+........+x20)(x0+x1+x2+..................+x21)(x0+x1+x2+x3+....................x21) (i)
Here three brackets are for x , y and z, and powers are based on the range of least to greatest possile values of x, y, z
on solving eq (i) will become (x0+x2+x4+...........+x20)(1x22)2(1x)2
Now we have to calculate coefficient of x21 in this expression so (1x22)2 can be omitted
General term of (1x)2 is (r+1)xr
So required coefficient = 22+20+18+16+14+12+10+8+6+4+2 = 132


@yashkapoor i said Q(2)

let g(y) = 1 / [yf(1)] + 2/[yf(2)] +3/[yf(3)]
g' (y) = (1/[yf(1)]2 + 2/[ yf(2)]2 + 3/ [yf(3)]2)
since g '(y) is negative summ of squares , its always negative
so, g '(y) < 0
g(y) is strictly decreasing function
here f(1) = 102 + 7 sin1
and f(2) = 212 + 7 sin2
clearly f(2) > f(1)
g( f(1) + 0 ) > infinity
g( f(2)  0 ) >  infinity
So, the mean value theorem
between ( infinity,infinity) as 0 will lie btn y=f(1) & y=f(2)
hence proved