# Paper

• Q(2)

• @vikram-ghotra

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)(1-x22)2(1-x)-2

Now we have to calculate coefficient of x21 in this expression so (1-x22)2 can be omitted

General term of (1-x)-2 is (r+1)xr

So required coefficient = 22+20+18+16+14+12+10+8+6+4+2 = 132

• @yash-kapoor i said Q(2)

• @vikram-ghotra

let g(y) = 1 / [y-f(1)] + 2/[y-f(2)] +3/[y-f(3)]

g' (y) = -(1/[y-f(1)]2 + 2/[ y-f(2)]2 + 3/ [y-f(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