2010/013) Form a quadratic equation whose coefficients are integers and in AP and it has got rational roots

Let the quadratic equation be

ax^2+bx+c = 0


as the coefficients are in ap so let common difference be y so b-a = y c-b = y

or a = b-y and c = b+ y

for the equation to have rational roots we have

discriminant is a perfect square that is say n^2


b^2 – 4 ac = n^2

or b^2-4(b-y)(b+y) = n^2

or b^2-4(b^2 – y^2) = n^2

or 4y^2-3b^2= n^2

dividing by n^2 and letting b/n = t and y/n = s we ge

4s^2-3 t^2= 1 ..1

s and t are real numbers

And from inspection we have a root s = 1/2, t =0 is a rational root and to find another rational root we draw a straight line from (0,1/2)

We get s = mt+ 1/2 is a general root …(2)

So we put in 1 to get the intersection of the curve given in 1 with this straight line to get

(2mt+1)^2 - 3t^2 = 1

Or 4m^2t^2 +4mt+1 – 3t^2 = 1

Or (4m^2-3)t^2 + 4mt = 0

ignoring the starting solution t = 0 we get

t= - 4m/(4m^2-3)

so s= mt+1/2 we get s = -(4m^2-3)/(8m^2-6)

now putting n = 8m^2-6 to get rid of denominator we get

b = -8m and y = -4m^2+3

so we get

a = 4m^2 – 8m + 3
b = - 8m
c = -4m^2 -8m -3 as roots

and taking m as different integers we get different solutions in integers

for example
m = 1 gives (-1.-8, -15) multiply by one we get x^2+ 8x + 15 = 0 giving factor (x+3)(x+5) rational
m = 2 gives (3.-16, -35) we get 3x^2 -16x -35= 0 giving (3x+5)(x-7)
m =3 gives (15,-24,-63) divide by 3 to get 5x^2-8x-21 = (5x+7)x-3)

so on

2010/012) If 0 < x < 1, then the sum of the infinite series (1/2)X^2 + (2/3)X^3 + (3/4)X^4 + ... is ...

A) log[(1+x)/(1-x)];
B) x/(1-x) + log(1+x);
C) 1/(1-x) + log(1-x);
D) x/(1-x) + log(1-x);
Explain your answer...

et f(x) = (1/2)x^2 + (2/3)x^3 + (3/4)x^4 +

df/dx = x + 2 x^2 + 3x^3 + ... ....
= (x+x^2 + x^3...) + (x^2 + x^3 ...) + (x^3+x^4 + ....)
= x/(1-x) + x^2/(1-x) +
= x/(1-x)^2
= (1-x-1)/(1-x)^2
= 1/(1-x) - 1/(1-x)^2
integrating you get
f = log(1-x) + 1/(1-x) + constant of integration log(1-x) as 1- x > 0
this can be checked to be zero so ans is C

2010/011) If cos(x)/cos(y) = a/b, then a tan(x) + b tan(y) equals

A) (a+b)cot((x+y)/2); B) (a+b)tan((x+y)/2); ...?
C) (a+b)(tan(x/2) + tan(y/2));
D) (a+b)(cot(x/2) + cot(y/2));
Kindly explain your answer...

we have (a+b) on RHS of (a)(b)(c) and (d) so let us eliminate a and b from numerator of
( a tan(x) + b tan(y))/(a+b)

b cos x = a cos y
b / a = cos y/ cos x
(b+a)/a = (cos y + cos x)/cos x

a/(a+b) = cos x/(cos y+ cos x)
and b/(a+b) = cos y /(cos y + cos x)

(a tan x + b tan y)/(a+b)
= (a/(a+b)) tan x + (b/(a+b) tan y
= (cos x tan x + cos y tan y)/(cos y + cos x)
= (sin x + sin y)/(cos x + cos y)
= (2 sin (x+y)/2 cos (x-y)/2 )/(2 cos (x+y)/2 cos (x-y)/2 )
= tan (x+y)/2

hence ans is B

2010/010) Let x and n be positive integers such that 1 + x + x^2 + x^3 + ... + x^n-1 is a prime number. Show that n? is a prime.

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

now in case n is not a prime then say n= pq

now x^n-1 = (x^(pq)-1) = (y^q-1)/(x-1) where y = x^p
= (y-1)/(x-1)(y^(q-1) + y^(q-2) + ...1)
= (x^p-1)/(x-1) (y^(q-1) + y^(q-2) + ...1)
= (x^p-1 + +1)((y^(q-1) + y^(q-2) + ...1)
has got 2 factors and hence not a prime
so n must be prime

2010/009 Evaluate : Lim(n-> infinity) {(1 + 1/2n)(1 + 3/2n)(1 + 5/2n)...(1+(2n - 1)/2n)}^(1/2n)?

this can put as sum of

((1+1/2n)(1+2/2n) .....* * (1+(2n)/2n))^(1/2n)/
(1+2/2n)(1+4/2n) ........(1+ 2n/2n)^(1/2n)

basically multiply it by the missing terms and then devide back

now numerator say
N = ((1+1/2n)(1+2/2n) .....* * (1+(2n)/2n))^(1/2n)
take log to base e

log N = 1/2n (log (1+2/n) + log (1+ 2/2n) + .... + 2)

as n-> infinite this tehds to

1/2 int (log x) for x from 1 to 2

integral of log x = x log x - x
definite integral = (2 ln 2 -1) - (-1) =2 ln 2
so N = 1/2e^(2 ln 2) = e^2

now you can evaluate the denominator as
ln d = 1/2n (ln (1+2/2n ) + ln (1+4/2n ) + ...+ ln (2))
as n->infinite this is int log x

integral of log x = x log x - x
definite integral = (2 ln 2 -1) - (-1) =2 ln 2
so D = e^(2 ln 2) = 2e^2
so N/D = value = 1/2

2010/008) Show that there is exactly one value of x which satisfies the equation,?

2 cos^2 (x^3 + x) = 2^x + 2^(-x).

proof:
the maximum value of LHS = 2

as cos^2(x^3+x) <= 1

the minimum value of RHS

let 2^x = y

so y + 1/y = (sqrt(y) - 1/(sqrt(y))^2 + 2

so minum value of RHS = 2


so both sides are same when both are 2

RHS = 2 when y = 1 or x= 0

LHS = 2 when 2 cos^2(x^3+x) = 2

or cos^2(x^3+x) = 1

x = 0 satisies LHS

so x = 0 is the only value that satisfied equality

2010/007) prove 64 {Cos^8(x) + Sin^8(x)} = cos8 x + 28cos 4x + 35

We know

(a+b)^8 = a^8 + 8 a^7b + 28 a^6b^2 + 56 a^5b^3 + 70a^4b^4 + 56 a^3b^5 + 28 a^2b^6+8ab^7+b^8

And (a-b)^8 = = a^8 - 8 a^7b + 28 a^6b^2 - 56 a^5b^3 + 70a^4b^4 - 56 a^3b^5 + 28 a^2b^6 -8ab^7+b^8

So (a+b)^8 + (a-b)^8 = 2(a^8 + b^8) + 56 (a^6b^2+a^2b^6) + 140(a^4b^4)

Putting a = e^ix and b = e^-ix

LHS = (e^ix+e^-ix)^8 + (e^ix-e^-ix)^8 = 2(e^i8x + e^-i8x)+ 56(e^4ix + e^-4ix) + 140

Or (2 cos x)^8 + (2i sin x)^8 = 2 ( 2 * cos 8x) + 56 * 2 * cos 4x + 140

Or 2^8(cos ^8x + sin ^8x) = 4 * ( cos 8x + 28 cos 4x + 35)

Or 2^6(cos ^8x + sin ^8x) = ( cos 8x + 28 cos 4x + 35)

Or 64(cos ^8x + sin ^8x) = ( cos 8x + 28 cos 4x + 35)

2010/006) equation sin(w) = (a^2 + b^2 + c^2)/(ab+bc+ca), where a,b,c are fixed nozero real numbers

has a solution for w
A) Whatever be a,b,c;
B) iff a^2 + b^2 + c^2 < 1;
C) iff a,b and c all lie in the interval (-1,1);
D) iff a = b = c;
Explain your answer.

ans

we know by GM AM enaquality a^2+b^2 >= 2ab
a^2+c^2 >= 2ac
and b^2+c^2 >= 2bc

adding all 3 we get 2(a^2+b^2+c^2) >= 2(ab+bc+ca)

or (a^2+b^2+c^2)/(ab+bc+ca)>= 1 and equal only when a= b= c

sin (w) cannot be > 1 so ans is when (a^2+b^2+c^2)/(ab+bc+ca) = 1 or a= b =c

that is d

2010/005) If 0 < x < 1, then the sum of the infinite series (1/2)X^2 + (2/3)X^3 + (3/4)X^4 + ... is ...

let f(x) = (1/2)x^2 + (2/3)x^3 + (3/4)x^4 + ...

df/dx = x + 2 x^2 + 3x^3 + ... ....
= (x+x^2 + x^3...) + (x^2 + x^3 ...) + (x^3+x^4 + ....)
= x/(1-x) + x^2/(1-x) + ...
= x/(1-x)^2
= (1-x-1)/(1-x)^2
= 1/(1-x) - 1/(1-x)^2
integrating you get
f = log(1-x) + 1/(1-x) + constant of integration log(1-x) as 1- x > 0
this can be checked to be zero
so f(x) = 1/(1-x) + log(1-x)

2010/004) Let x and n be positive integers such that 1 + x + x^2 + x^3 + ... + x^n-1 is a prime number.

Let x and n be positive integers such that 1 + x + x^2 + x^3 + ... + x^n-1 is a prime number. Show that n?
is a prime.

Proof:
1 + x + x^2 + x^3 + ... + x^n-1 = (x^n-1)/(x-1)

now in case n is not a prime then say n= pq

now x^n-1 = (x^(pq)-1) = (y^q-1)/(x-1) where y = x^p
= (y-1)/(x-1)(y^(q-1) + y^(q-2) + ...1)
= (x^p-1)/(x-1) (y^(q-1) + y^(q-2) + ...1)
= (x^p-1 + +1)((y^(q-1) + y^(q-2) + ...1)
has got 2 factors and hence not a prime
so n must be prime

2010/003) The equation x^3 + 2x^ + 2x + 1 = 0 and x^200 + x^130 + 1 = 0 have ...?

exactly one common root;
b) no common root;
c) exactly three common roots;
d) exactly two common roots;
Kindly explain...

ans:
x^3 + 2x^2 + 2x + 1 = x(x+1)^2 + (x+1) = (x+1)(x^2 + x + 1) has zeros -1, w and w^2 where w is cube root of 1

-1 is not a zero of x^200 + x^130 + 1 since the expression equals 3 when x = -1.

So we are down to 2 possible roots in common.

let f(x) = x^200 + x^130 + 1

f(w) = w^200 + w^130 + 1= w^2+ w + 1 = 0 so w is a root
f(w^2) = w^400 + w^260 + 1 = w+w^2+1 = 0 so w^2 is a root

so there are exactly 2 common roots and hence d)

2010/002) The numbers 12n + 1 and 30n + 2 are relatively prime for ...?

any positive integer n;
b) infinitely many, but not all, integers n;
c) for infinitely many integers n;
d) none of the above;
Kindly explain...

Ans:
GCD(12n + 1 ,30n + 2)
= GCD(12n + 1 ,2(15n+1))
= GCD(12n + 1 ,15n+1) as 12n+1 is odd
= GCD(12n+1,(15n+1-(12n+1)) as GCD(a,b) = GCD(a,b-a)
= GCD(12n+1,3n)
= GCD(3n, 12n+1- 4*3n) as GCD(a,b) = GCD(b,a- n * b))
= GCD(3n,1)
= 1 for all n
hence ans is a

2010/001) Prove that (2222^5555)+(5555^2222) is divisible by 7

Prove that (2222^5555)+(5555^2222) is divisible by 7

Proof:
we know

2222 mod 7 = 3

and 5555 mod 7 = 4 or -3

so (2222^5555)+(5555^2222) mod 7

= 3^5555 + (-3)^ 2222 mod 7
= 3^5555+ 3^2222 mod 7
= 3^2222(3^3333 + 1) mod 7


as 3^2222 is not divisible by 7 so

we need to show that

3^3333 + 1 mod 7 = 0

now as 7 is prime so as per format's little theorem

3^6 mod 7 = 1

3^3333 mod 7 = 3^(3333 mod 6) mod 7
= 3^ 3 mod 7 = 27 mod 7

so 3^3333 + 1 mod 7 = 28 mod 7 = 0

hence proved