2011/102) Prove z/ (1 + z^2) is real if and only if z is real or |z| =1.

if z is zero it is true so let us take for the case z not zero

so z/(1+z^2) = 1/(z + 1/z) is real

let z = r cos t + ir sin t

so 1/z = 1/r cos t - 1/r i sin t

z + 1/z imaginary part is zero

r sin t = - 1/r sin t
sin t = 0 or r = -1/r or |r| = 1 and hence |z| = 1


sin t = 0 means real

hence z is real or |z| = 1

2011/101) Prove this inequality? 4x^10+x^8+4x^2+1 ≥ 2x^9+4x^5+2x (x is a positive number)

we shall use the AM GM enaquality to prove it

we know (a+b)/2 > = sqrt(ab) or (a+b) > = 2 sqrt(ab)

a = 4x^10, b = x^8 gives 4x^10 + x^ 8 > = 4 x^9
a = 4x^10, b = 1 gives 4x^10 + 1 > = 4 x^5
a = x^8, b = 4x^2 gives x^ 8 +4x^2 > = 4x^5
a = 4x^2, b = 1 gives 4x^2 + 1 > = 4 x

adding we get

8x^10 + 2x^2+ 8x + 2 >=4x^9 + 8x^5 + 2

dividing by 2 we get 4x^10+x^8+4x^2+1 ≥ 2x^9+4x^5+2x

2011/100) Solve for x: (2x + 14) / (5x - 1) ≤ x + 3.

we need to get rid of denominator so multiplying by (5x-1) is problematic as we do not know whether it is positive or negative

so multiply by (5x-1)^2 which is positive to get

(2x+ 14) (5x -1 ) <= (5x-1)^2(x+3)

or (2x+14) (5x-1) - (5x-1)^2 (x+ 3) <= 0

or (5x-1) ((2x+14 - (5x- 1 )(x+3)) < = 0

or (5x-1)(2x+14 - (5x^2 + 14x - 3)) <= 0
or (5x-1)(-5x^2 - 12x + 17) < = 0
or (5x-1)(5x^2+ 12x - 17) >= 0
or (5x-1) (5x+17)(x-1) > = 0

there are 4 ranges - infinite to -17/5
- 17/5 to 1/5 (in this it is positive but x cannot be 1/5 as 5x-1 is in denominator)
1/5 to 1
and 1 to infinite (in this it is positive

x belongs to [- 17/5, 1/5) U [1, ∞).

2011/099) Find cube root of 9 sqrt 3+11 sqrt 2

because there is sqrt(3) and sqrt(2) in the number the cube root is a combination of them
let it be x sqrt (3) + y sqrt (2)
so (x sqrt (3) + y sqrt (2))^3 = 9 sqrt 3+11 sqrt 2
or 3 x^3 sqrt (3) + 9x^2y sqrt 2 + 6xy^2 sqrt (2) + 2y^3 sqrt 2 = 9 sqrt 3+11 sqrt 2
comparing coefficient of sqrt (3) and sqrt(2) we get
3x^3 + 6x y^2 = 9 ..1
9x^2 y + 2y^3 = 11 .. 2
multiply (1) by 11 and 2nd one by 9 to get
33 x^3 + 66 xy^2 = 81 x^2y + 18y^3
divide both sides by 311 x^3 + 22xy^2 = 27x^2y + 6y^3
or 11x^3 - 27 x^2y + 22 xy^2 - 6y^3 = 0
putting x/y = t we get
11t^3-27 t^2 + 22t- 6 = 0
f(1) = 0 gives t = 1 is a sqrt ( as 11 + 22 = 27+ 6)
so t = 1 or 11t^2-16t + 6 = 0
or x = y is a solution
and
from (1) 9x^3 = 1 or x = 1 and y = 1
so result = sqrt(3) + sqrt (2)

2011/098) if a=0.5(5-root 21) then find a^3 + a^-3- 5a^2- 5a^-2+ a+ a^-1.?

a = .5(5-sqrt(21)

so 1/a = 2/(5- sqrt(21) = 2(5+ sqrt(21)/(5^2-21) = .5(5 + sqrt(21))

so a + 1/a = 5

a^3+ 1/a^3- 5 ( a^2+ 1/a^2) + (a+ 1/a)
= (a+1/a)^3 - 3 ( a + 1/a) - 5((a+1/a2^2) - 2) + (a+ 1/a)
= 5^3 - 3 * 5 - 5(5^2 - 2) + 5
= 125 - 15 - 125 + 10 + 5 = 0

2011/097) Prove the identity: (1+cosx+sinx)/(1+cosx-sinx) = secx+tanx?

regular solution
(1+ cos x + sin x)/(1+ cos x - sin x)
= (1+ cos x + sin x)^2/ (( 1+ cos x + sin x)(1+ cosx - sin x))
= (1+ cos x + sin x)^2/ ((1+ cos x) ^2 - sin ^2 x)
= (1+ cos x + sin x)^2/ ((1+ 2cos x + cos ^2 x - sin ^2 x)
= (1+ cos x + sin x)^2/ ((cos ^2 x+ 2cos x + cos ^2 x)
= (1+ cos x + sin x)^2/ (2(cos^2 x + 2 cos x))
= (1+ cos x + sin x)^2/ (2 cos x( 1+ cos x))
= ( 1 + 2 cos x + 2 sin x + 2 sin x cos x + cos ^2 x + sin ^2 x)/ (2 cos x( 1+ cos x))
= ( 2 + 2 cos x + 2 sin x + 2 cos x sin x)/(2 cos x( 1+ cos x))
= 2(1+ cos x)(1+ sin x)/ (2 cos x (1+ cos x))
= (1+ sin x)/cos x
= sec x + tan x


alternatively ( simpler solution)

1+ cos x/ sin x = ( 2 cos ^2 x/2)/(2 cos x/2 sin x/2)
= (cos x/2)/ sin x/2

using compnondo dividendo
we get
(1+cosx+sinx)/(1+cosx-sinx) = (cos x/2 + sin x/2)/( cos x/2 - sin x/2)
= ( cos x/2 + sin x/2)^2 / (( cos x/2 + sin x/2)(cos x/2 - sin x/2)
= ( cos ^2 x/ 2 + sin ^2 x/2 + 2 cos x/2 sin x/2)/(cos^2 x/2 - sin ^2 x/2)
= (1 + sin x)/ cos x
= sec x + tan x

2011/096) The equation P(x) = x^4 – 16x^3 + 94x^2 +px + q = 0 has two double roots. Solve

Because it has 2 double roots so it is square of a quadratic polynomial

Let it be (x^2+ax+b)^2

= x^4 + 2ax^3 + x^2(2b+a^2) + 2abx+ b^2

Comparing coefficients

2x = -16 or a = - 8

2b+a^2 = 94 or b = 15

p = 2ab = - 240

q = b^2 = 225

so equation

= x^4-16x^2 + 94x^2 – 240x + 225

It is (x^2 -8 x + 15) ^2 = (x-3)^2(x-5)^2 and roots are x = 3,3,5,5

2011/095) Prove: Let p and q be distinct primes, k be a natural number, and W be a natural number less than pq. Then W^(1+k(p-1)(q-1)) is congruent to W (mod pq

because p is a prime as per Formats Little theorem
now there are 3 cases and we deal as below
1) when w is coprime to p and q

w^(p-1) = 1 mod pwhen w is coprime to p

so w^(p-1)(q-1) = 1 mod p

similarly w^(p-1)(q-1) = 1 mod q

so w^(p-1)(q-1) = 1 mod pq

so w^k(p-1)(q-1) = 1 mod pq
or w^(1+k(p-1)(q-1)) = w mod pq

case 2
let w be a miultiple of p and not q

w^ t = w mod p as both are zero

so w ^(1+(k(p-1)(q-1)) = w mod p
as per argument in 1
w ^(1+(k(p-1)(q-1)) = w mod q

so w ^(1+(k(p-1)(q-1)) = w mod pq

3)
w is coprime to p and not q
same arguments as in 2

hence proved for all cases

2011/094) Show that the equation 2x+cosx=0 has exactly one real root?

we have
cos x + 2x = 0
so cos x = - 2x

as cos x is between -1 and + 1 so x has to be between -1/2 and + 1/2

now between -1/2 and 1/2

d/dx (cos x + 2x) = 2 - sin x > 0 so increasing

at 1/2 we have cos x + 2x > 0 (both cos x and x > 0)
at -1/2 we have cos x + 2x < 0 cos x < 1 and 2x = - 1

so it is increasing from -1 to 1 and hence exactly one real root between (-1/2 and 1/2)

2011/093) If a=4 root 6 /(root 2+ root 3) then find value of (a+2 root 2)/(a-2 root 2) + (a+ 2 root 3)/(a - 2 root 3)?

this is a perfect example of using componendo dividendo method

a=4 root 6 /(root 2+ root 3)

so a/(2 root 2) = 2 root3/(root 2 + root 3)

by componendo dividendo

( a+ 2 root 2)/(a- 2 root 2) = (2 root 3 + root 2 + root 3)/(root 3 - root 2)

= (3 root 3 + root 2) /(root 3 - root2) ...1

again for the second term

a=4 root 6 /(root 2+ root 3)
so a/(2 root 3) = 2 root2/(root 2 + root 3)

by componendo dividendo

( a+ 2 root 3)/(a- 2 root 3) = (2 root 2 + root 2 + root 3)/(root 2 - root 3)
= ( 3 root 2 + root 3)/ (root 2 - root 3) = - (- root 3 - 3 root 2)/(root 3 - root 2) .. 2

add (1) and (2) to get
( a+ 2 root 2)/(a- 2 root 2) + ( a+ 2 root 3)/(a- 2 root 3)
= 2(root 3 – root 2)/(root 3 – root 2) = 2


(comonendo and dividendo: for the persons who are not familiar

if a/b = c/d then (a+b)/(a-b) = (c+d)/(c-d))

for a proof of it (a+b)/(a-b) = (a/b +1)/(a/b-1)= (c/d + 1)/(c/d - 1) = (c+d)/(c-d))

2011/092) When is one of the roots of a cubic equal to the product of the other two roots?

Find, in terms of the coefficients, a necessary and sufficient condition for one of the roots of ax³ + bx² + cx + d = 0 (a ≠ 0) to be equal to the product of the other two roots.

Let the roots be p,q,pq

So a(x-p)(x-q)(x-pq)

= a x^3 – ax^2(p + q + pq) + ax(pq + pq^2 + p^2q) – ap^q^2

Comparing coefficients

b/a = -(p+q+pq) .. 1

c/a = pq + pq^2+ p^2q = pq(1+p+q) … 2

d/a = -p^2 q^2 ..3

subtract 1 from both sides of (1)

b/a - 1 = - (p+1)(q+1)

from (2 ) and (3)

c/a - d/a = pq(1+p+q+pq) = -pq(b/a- 1)

or c-d = - pq(b-a)

square both sides

(c-d)^2 = p^2q^2(b-a)^2 = -d(b-a)^2

Or (c-d)^2 + d(b-a)^2= 0

2011/91) What is the graph of xy = 3y

(A) one point
(B) two points
(C) one line
(D) two lines
(E) a right angle

this is a tricky question and ans is simple

xy = 3y => y(x-3) = 0 so y = 0 or x = 3

y= 0 is a line and x= 3 is another line and hence ans is D

2011/090) Prove that there do not exist four positive real numbers, all distinct from each other, such that a^3 + b ^3 = c^3 + d^3 and a+b = c+d

proof:
without loss of generality assume a>=b and c >= d

now as a+b = c+ d
cube both sides a^3+b^3 + 3ab(a+b)= c^3+d^3 + 3cd(c+d)

and a a^3+b^3 = c^3+d ^3 so 3ab(a+b) =3cd(c+d)
and hence ab = cd
(a-b)^2 = (a+b)^2 - 4ab = (c+d)^2 - 4cd = (c-d)^2 and hence a- b = c- d

as
a + b = c+ d
a- b = c - d
adding 2a = 2c = a = c and hence b = d
hence proved

2011/089) Evaluate the infinite product Π(n=2..∞)n^2/(n^2-1).

First we try to get a closed form for the product

We have t(n) = n^2/((n+1)(n-1))

Π(n=2) n^2/(n^2-1)= 2^2/(3)

Π(n=2..3)n^2/(n^2-1)= 2^2/(3) * 3^2 /(2*4) = 2 * 3 / 4

Π(n=2..4)n^2/(n^2-1)= 2*3/(4) * 4^2 /(3*5) = 2 * 4 / 5

From the pattern we feel that

Π(n=2..k)n^2/(n^2-1) = 2 * k/(k+1)

It is true for n = 2

If it is true for n =k then

Π(n=2..k+1)n^2/(n^2-1) = 2 * k/(k+1) * (k+1)^2/(k*(k+2) = 2 *(k+1)/(k+2)

So it is true for k+ 1

Π(n=2..k)n^2/(n^2-1) = 2k/(k+1)

Now that we have found the closed form as k goes to inifinite above product goes to 2 (converges to 2)

2011/088) Find a right angled triangle with integer sides whose all three sides are fibonacci numbers.

No solution

Proof:

Let the legs be x and y

X and y cannot be same then hypotenuse shall be x qrt(2) and it is not integer.

Now let x < y and so hypotenuse >= (x+y)

as next Fibonacci number = (x+y) if x and y are consecutive Fibonacci numbers

and > (x+y) if they are not consecutive

So x^2+y^2 >= (x+y)^2 which is not possible unless x = 0

Hence no solution

2011/087) Prove Cos^4x =(1/8)(3+4cos2x+cos4x)

realizing that RHS has got cos of multiple of x

Cos^4x = 1/4(2 cos ^2 x )^2
= 1/4( cos 2x + 1)^2 knowing cos 2x = 2 cos^2 x - 1
= 1/4( cos^2 2x + 2 cos 2x + 1)
= 1/8( 2 cos ^2 2x + 4 cos 2x + 2)
= 1/8( cos 4x +1 + 4 cos 2x + 2) as 2 cos^2 2x - 1 = cos 4x)]
= 1/8( cos 4x + 4 cos 2x + 3)

2011/086) prove that the expression 2x+3y and 9x + 5y are divisible by 17 for same integral values of x and y

if 9x + 5y is divisible by 17 then
4(9x + 5y) is divisible by 17 (note below why multiply by 4)
or 36x + 20 y is divisible by 17
or 36x-34 x + 20y - 17 y or 2x + 3y is divisible by 17

multiplication by 4 was not magic

as 2 = 9m mod 17 gives m = 4

to find it

let us find inverse 0f 9 mod 17

we have 17 = 9 + 8
8 = 17-9
now 9= 8+1 = (17-9) + 1 or
2 * 9 = 17 + 1 so 2 is inverse of 9
so m = 4 as multiplying by 2 gives 1 so multiply by 4 to give 2

2011/085) The quadratic function f(x) is negative for x > 9/2 and x < -1, but for no other value of x. If f(1) = 28 find f(x)

As it is –ve for x > 9/2 and x < -1 it is of the form –A (x-9/2)(x+1) where A > 0

Or – B(2x-9)(x+1)(where B = A/2)

F(1) = -B(-7)(2) = 28A = 28 or B = 2

So we get

F(x) = – 2(2x-9)(x+1)

2011/084) If p q r are roots of equation x^3 – 3px^2 + 3q^2x – r^3 = 0 then prove that p = q = r

proof:

as sum of roots is 3p so p+ q + r = 3p

so there exists t such that

q = p- t, r = p+ t

coefficient of x = pq + pr + qr = 3q

or p(q+r) + qr = 3q^2

or p* 2p + (p – t)(p+t) = 3(p-t)^2

or 3p^2 – t^2 = 3p^2-6pt + 3t^2

or 4t^2-6pt = 0 or

t(2t-3p) = 0 ..1

constant = p(p+t)(p-t) = (p+t)^3

p+t = 0 or p(p-t) = (p+t)^2

or p^2-pt = p^2 +2 pt + t^2

or t^2 + 3pt =0

t(t+3p) = 0 ….2

fom (1) and (2) t = 0

or

2t-3p = 0 and t+ 3p = 0 which again gives t = 0 (also p = 0 this is redundant as a sub set of 1^st solution)

So t = 0

So p = q =r

hence proved

2011/083) Show that for p and q odd x^2 + 2px + 2q = 0 does not have rational solution

proof

the discriminant is

b^2-4ac = (2p)^2 – 8q = 4(p^2-2q)

as p and q are odd

p^2-2q = (2s+1)^2 – 2(2t+1) = (4s^2+ 4s+1) – (4t+2) =(4s^2 + 4s – 4t -1) which is 3 mod 4

so discriminant cannot be a perfect square so there is no rational root

2011/082) The equation x³ + px² + qx + r = 0 (where p, q, r are non zero) has roots α, β, γ such that 1/α , 1/β, 1/γ...?

are consecutive terms in an arithmetic sequence, show that β = -3r / q

f(x) = x^3+px^2+qx+r = 0 has the roots α, β, γ

so f(1/x) has the roots 1/α , 1/β, 1/γ. Which are in AP

f(1/x) = 1/x^3+p/x^2+q/x+ r = 0

or rx^3+qx^2+px+1 = 0

sum of roots = - q/r = 3/ β or β = -3r/q (sum of 3 terms of AP = 3 * middle term)

proved