Q13/090) if M={x,xy,log(xy)}, N={0,∣x∣ ,y} given: M=N find : ( x + 1/y) + (x^2 + 1/y^2) + (x^3 + 1/y^3) …. + (x^2001 + 1/y^2001)

x and y cannot be zero as we have log(xy) and log zero is undefined .

so we have xy = 1 as log(xy) = 0

as x and y both are positive or –ve.

We have |x| in N . now x = |x| and xy = y => x = y = 1

but x and xy have to be different and |x| and y has to be different

so y = x = - 1 giving M={- 1,1,0 } and N={0, 1 ,0-1}

So x = y =-1 and sum = x^n + 1/y^n = -2 for n = odd and 2 for n = even

So sum = -2 as 1st 1000 pairs cancel leaving with x^2001 + 1/y^2001 = - 2

Q13/089) Find the real solution(s) to the equation 36/x^(1/2) +9/y^(1/4) = 42−9x^(1/2) −y^(1/2)

We have

36/x^(1/2) + 9x^(1/2) +     9/y^(1/4)+ y^(1/2) = 42

Complete the square for x and y terms giving

36/x^(1/2) + 9x^(1/2) - 36  +    9/y^(1/4)+ y^(1/2) = 42

Hence 9( 2/x^1/4 – x^(1/4))^2 + (3/y^(1/4) – y^(1/4))^2 = 0
so  2/x^(1/4) – x^(1/4) = 0 or x = 2^2 = 4
And 3/y^(1/4) – y^(1/4) = 0 or y = 9

Q13/ 088) Determine the smallest integer that is square and starts with the first four digits 3005.

we can have even number of digits or odd number of digits and they have to be treated separately

if even then we have

sqrt (3005 * 10^2n) = 54.8177 * 10^n
sqrt (3006* 10^2n) = 54.8270 * 10^n

if odd digits then

sqrt (30050 * 10^2n) = 173.34 * 10^n

sqrt (30060 * 10^2n)= 173.37 * 10 ^n

from the 1st set we get square root  5482 as smallest where a digit is different and from the second at least

17335

so it is 5482^2 = 30052324

Q13/087) Prove that if x = log_a(bc),y = log_b(ca),z = log_c(ab) then the value of xyz - x - y - z is 2?

we get

bc = a^x ..1
ca = b^y...2
ab = c ^z ...3

from (3)

c^(xyz) = (ab)^(xy)
= a^(xy) * b^(xy)
= (a^x)^y) * (b^y)^x)
= (bc)^y ( ac)^x ( from 1 and 2)
= c^y c^x ( b^y)(a^x)
= c^y c^x (ac) (bc)
= c^y c^x c^2(ab)
= c^ x c^y c^2 c^z
= c^(x+y+z+2)

hence xyz = x + y + z + 2 or
xyz - x - y - z = 2

Alternatively we can solve as below

\

multiply (1) by a

abc = a^(x+1) => a = (abc)^(1/(x+1) ..4

similarly

b = (abc)^(1/(y+1) ..5

c = (abc)^(1/(z+1) .. 6

multiply to get

abc = (abc)^(1/(x+1) + 1/(y+1) + 1(z+1))

or (1/(x+1) + 1/(y+1) + 1(z+1)) = 1

or (y+1)(z+1) + (x+1)(z+1) + (x+1)(y+1) = ( x+1)(y+1) (z+1)

or yz + y + z + 1 + xz + z + x + 1 + xy + x + y + 1 = xyz + xy + yz + xz + x + y+ z + 1

or x+y + z + 2 = xyz

Q13/086) Solve x- 2/(x-1) = 1 – 2/(x-1)

X cannot be 1 as x-1 is in denominator

Canceling 2/(x-1) from both sides x = 1

As x cannot be 1 so no solution

Q13/085) The sum of all real root of the equation |x-2|^2 + |x-2| - 2= 0 is

Let |x -2 | = t

So t > = 0

t^2 + t – 2 = 0 or (t-1)(t+2) = 0

so t = 1 as – 2 is not possible

|x-2| = 1 or x -2 = 1or -1

So x = 3 or 1 so sum of real roots = 4

Q13/084) The sum of the rational terms in the expansion of (2^(1/2) +3^(1/5))^10 is

For it the power of 2 and of 3 both should be integer

So the terms should be (10C 2p)  (2^(1/2)^2p * (3^(1/5)^ 5q) where 2p + 5q = 10

Solution of 2p + 5q =1 0 are p = 0 q =2 or q = 0 p = 5

P = 0 q = 2 give (10c0) 3^2 = 9

P = 5 q = 0 give (10c10) 2^5 = 32

So the rational terms are 32 and 9 and sum is 41