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

Q3/083) solve

Given a,b,c integers solve for a,b,c

(a−b)(b−c)(c+a)=−90 ..1
(a−b)(b+c)(c−a)=42.. 2
(a+b)(b−c)(c−a)=−60 ..3

  

Solution

from the (3)

as 7 divides RHS and not the other 2 equations we have

b+ c = 7 ( as b and c > 0)

so we get (a-b)(c-a) = 6

so combinations a-b = 6 , c- a = 1
a- b= 3 , c - a = 2
a - b = 2 c - a = 3
a- b = 1 , c- a = 6
and (-6, -1),(-3,-2), (-2,-3) and (-1,6) each can be tried with b+ c to give
a= 3, b = 1 and c = 6 then they are seen to satisfy other 2 equations

Q13/082) if iz^3 + z^2 – z + i = 0 ( i is square root of -1) show that |z| = 1

iz^3 + z^2 – z + i

= iz^2( z-i) – (z-i)

= (iz^2 – 1)(z –i)

So z – i = 0 => |z| = 1

or iz^2 = 1 => z^2 = i => |z|^2 = |i| = 1 => |z| = 1

Q13/081) If α and β are non real complex roots of unity prove that α^4 + β^4 + 1/(αβ) = 0

 α^4 = α ,  β^4 = β, αβ = 1

so α^4 +  β^4 + 1/(αβ) = α +  β + 1 = 0

Q13/080) Let α +i β : α and β are real , be a root of the equation x^3 + qx +r = 0 , q and r are real. Find a real cubic equation, independent of α and β whose one root is 2 α.

As q and r are real and  α +i β is a root so  α - i β is a root.
Now sum of roots = (coefficient of x^2  =) 0)

Ler 3^rd root be γ, so γ + (α +i β) + (α- i β) = γ + 2 α = 0

Or γ = -  2 α

So -  2 α is root of f(x) = x^3 + qx +r  =0

So   2 α is root of f(-x) = - x^3 – q x + r = 0 or x^3 + qx – r = 0

equation  x^3 + qx  - r = 0

Q13/079) Find all pairs (p,q) of integers such that 1+1996p+1998q=pq.

pq – 1996p – 1998q = 1

Or (p-1998)(q-1996) – 1998 * 1996 = 1

Or (p-1998)(q-1996) = 1998 * 1996 + 1 = 1997^2

We get all the solution set for (p-1998, q- 1996) to be ( 1, 1997^2), (1997,1997), ( 1997^2, 1)
(-1, - 1997^2), (-1997,- 1997), (- 1997^2, - 1) as 1997 is prime

Q13/078) Let x,y, and z be distinct non-zero real numbers such that x+1/y=y+1/z=z+1/x. What is the value of |xyz|?

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

now putting 1/a for z , 1/b for x and 1/c for y we get

1/b + c = 1/c + a = 1/a + b

the above equation is same as (1) with c for x b for y and a for z and hence

|xyz| = |abc| as the value is unique

= | 1/z 1/x 1/y| = | 1/xyz|

so |xyz| = 1

Q13/077) X=3^10000 Given log3=0.4771 All the digits sum of X =A All the digits sum of A =B All the digits sum of B =C find C

This is 4772 digits so A < 4772 * 9 or A < 50000

B < 41 so C <= 12 ( if b were 39)

X mod 9 = C mod 9

ax x mod 9 = 0 so C= 0 or 9 as it cannot be 18

as C cannot be zero it is 9

Q13/076) prove that (1+x) * ( 1 + x^2) * (1+x^3) * (1+ x^4) … (1+x^n) >= (1+x^(n+1)/2)^n

we have x^a + x^b >= 2x^(a+b)/2 by AM GM inequality

adding 1 + x^(a+b) on both sides we get

(1+x^a)(1+x^b) >= 1 +  2x^(a+b)/2 + x^(a+b) = ( 1+ x^(a+b)/2)^2

Putting b = n+ 1 -a we get

(1+x^a)(1+x^(n+1- a) >= (1+ x^(n+1)/2)^2

For n even taking a from 1 to n/2 we get n/2 expressions and multiplying them out we get

(1+x)(1+x^2)( 1 + x^3) .. (1+x^n) >= (1+ x^(n+1)/2)^n as

 For n odd we have n-1 ( running a from 1 to (n-1)/2 we get (n-1)/2 pairs and

As (1+x^(n+1)/2)= (1+x^(n+1)/2) we get 1 term and multiplying we get the result