2011/053) find the relation between a and b when a^3+b^3 = 8 - 6 ab

take 8 to the left

a^3+b^3-8 = -6 ab
or a^3+b^3 + (-2)^3 = 3(-2) a b

as x^3 +y^3 + z^3 = 3xyz then x = y = z or x + y+ z = 0

comparing we get a = b = -2 or a + b- 2 = 0

that is a= b = -2 or a + b = 2

2011/052) prove that x^2+3x+3 is a factor of (x+1)^(n+1) + (x+2)^(2n-1)

for integer n > = 1


This can be proved by induction

n =1 gives (x+1)^(n+1) + (x+2)^(2n-1) = (x+1)^2 + (x+2) = x^2 + 2x + 1 + x + 2 = x^2 + 3x + 3 divisible

so the 1st step is proved

let it be true for n = k.
we need to prove for n= k+ 1

let f(k) = (x+1)^(k+1) + (x+ 2)^(2k- 1)

so f(k+1) = (x+1)^(k+2) + (x+2)^(2k+ 1)
= (x+1)(x+1)^k + (x+2)^2 *(x+2)^(2k-1)
= (x+1)(x+1)^k + (x^2+4x + 4) *(x+2)^(2k-1)
= (x+1)(x+1)^k + (x + 1 + x^2+3x + 3) *(x+2)^(2k-1)
= (x+1)(x+1)^k + (x + 1)(x+2)^(2k-1) + (x^2+3x + 3) *(x+2)^(2k-1)
= (x+1)[(x+1)^k + (x+2)^(2k-1)] + (x^2+3x + 3) *(x+2)^(2k-1)
= (x+1) f(k) + (x^2+3x + 3) *(x+2)^(2k-1)
as both terms are divisible so f(k+1) is divisible

so step of induction is proved

hence proved

2011/051) if a+ib is root of equation x^3+px + q = 0 the form an equation whose root is 2a

and p and q are real

a + ib is a root and coefficients are real so another root = a - ib

sum of roots = 0 (coefficient of x^2 =0) so 3rd root = -2a

now -2a is root of f(x) = x^3+ px + q = 0

so 2a is root of f(-x) = (-x)^3 - px + q = 0 or x^3+px -q = 0

2011/050) If a and b are roots of the equation x^2 – 2x cos t +1 = 0 form the equation whose roots are a^n and b^n

As a and b are roots so

a + b = 2 cos t … 1
ab = 1 …2

we have (a-b)^2 = (a+b)^2 – 4ab = 4 cos^2 t – 4 = - 4 sin ^2 t

or a-b = 2 i sin t … 3

from 1 and 3 adding 2a = 2 (cos t + i sin t) = 2 e^it or a= e^it
subtracting 2b = 2 (cos t - i sin t) = 2 e^it or b = e^-it

now a^n + b^n = (e^int + e^-int) = 2 cos nt

and a^nb^n = (ab)^n = 1

so the equation = (x^2- 2x cos nt + 1) = 0

2011/049) the value of a for which the equation

(a^2+4a+3) x^2 + (a^2-a-2) x + (a+1)a = 0 has more than 2 solutions is


ans:
if above is quadratic equation then it has 2 solutions and hence it must be identity so
a^2+4a + 3 = a(a+1) = 0
(a^2-a-2) = (a+1)(a-2) = 0
a(a+1) = 0

hence a+ 1 = 0 or a = -1

2011/048) show that (a+bw+cw^2) ^3 + (a + bw^2_+cw)^3 = (2a-b-c)(2b-a-c)(2c-a-b) when w is cube root of unity

let x = (a+bw+cw^2) ...1
and y = (a + bw^2+cw) ...2

we need to factor x^3 + y^3

x^3+y^3 = (x+y) ((x+y)^2 - 3 xy) ...3

add (2) and (1) to get

x+y = (2a+b(w+w^2)+ c(w+w^2))
= (2a -b - c) ... 4

now xy = (a+bw+cw^2)(a + bw^2+cw)
= (a^2 + b^2 + c^2 + ab(w+w^2) + bc(w+w^2) + ca(w+w^2))
= (a^2+b^2 + c^2 - ab - bc - ca)

so (x+y)^2 - 3xy
= (2a-b -c)^2 - 3(a^2+b^2 + c^2 - ab - bc - ca)
= (4a^2 + b^2 + c^2 - 4ab + 2bc - 4ac)- 3(a^2+b^2 + c^2 - ab - bc - ca)
= a^2 -2b^2 - 2c ^2 - ab - ac + 5 bc
= a^2 - a(b+c) - (2b^2 + 2c^2 - 5bc)
= a^2 -(a (2b-c) + (2c-b)) - (2b-c)(b-2c))
= a^2 -(a (2b-c) + (2c-b)) + (2b-c)(2c-b))
= (a - 2b + c )(a-2c + b)= (2b-a -c)(2 c-a -b )... 5


from 3 4 and 5 we get the result

2011/047) the value of the expression

2(1+w)(1+w^2) + 3(2w+1)(2w^1+1) + ... (n+1)(nw+1)(nw^2+1) when w is cube root of unit is

1) n^2 (n + 1)^2/4
2) n^2 (n + 1)^2/4 -n
3) n^2 (n + 1)^2/4 + n
4_ no of these

ans 3) because

kth term = (k+1)(kw+1)(kw^2+1) = (k+1) (k^2w^3+ k(w+w^2) + 1)
= (k+1)(k^2 -k +1) as w^3 = 1 and w + w^2- 1
= (k^3 + 1)

so sum = (sum of cubes upto n) + n
= (n(n+1)/2)^2 + n
= n^2(n+1)^2 /4 + n

2011/046) let n be any positive integer >1. then show that n^4+4^n is a composite number

there are 2 cases

1) n is even.

n^4+4^n > 2 and even so composite

2) n is odd

n^4 + 4^n
= n^4 + 2n^2*2^n + 4^n - 2n^2*2^n
= (n^2 + 2^n)^2 - 2n^2*2^n
= (n^2 + 2^n)^2 - n^2*2^(n+1)
= (n^2 + 2^n - n * 2^((n+1)/2)) * (n^2 + 2^n + n * 2^((n+1)/2)). Therefore,

if n is odd it has 2 factors so composite if 1st term is not 1 as second term > 1

for n > 1 1st term is > 1 and hence the given expression is composite

2011/045) What is the remainder when x^81 + x^49 + x^25 + x^9 + x is divided by x^3 - 1

when we divide by x^3-1 it shall be a maximal quadratic polynomial

ax^2 + bx + c

x^81 - 1 is divisible by x^3 -1 so x^81 divided by x^3-1 leaves 1

x^48 -1 divided by x^3 -1 leaves 0

so x(x^48-1) divided by x^3-1 leaves 0
x^49 divided by x^3-1 leaves x

x^25 divided by x^3-1 leaves x( similarly)
x^9 divided by x^3-1 leaves 1
so adding all the above we get remainder 1 + x + x + 1+ x = 3x + 2

2011/044) show that $\sin \frac{\pi}{14} \sin \frac{3\pi}{14} \sin\frac{5\pi}{14} = \frac{1}{8}$

LHS
$= \sin \frac{\pi}{14} \cos (\frac{\pi}{2} -\frac{3\pi}{14}) \cos(\frac{\pi}{2} -\frac{5\pi}{14})$

$= \sin \frac{\pi}{14} \cos (\frac{4\pi}{14})  \cos (\frac{2 \pi}{14})$

$= \sin \frac{\pi}{14} \cos (\frac{2\pi}{14})  \cos (\frac{4 \pi}{14})$
$= \dfrac{\cos \frac{\pi}{14} \sin \frac{\pi}{14} \cos (\frac{2\pi}{14})  \cos (\frac{4 \pi}{14})}{\cos \frac{\pi}{14}} $
$=\dfrac{1}{2}\frac{\sin \frac{2\pi}{14} \cos (\frac{2\pi}{14})  \cos (\frac{4 \pi}{14})}{\cos \frac{\pi}{14}} $
$=\dfrac{1}{4}\dfrac{\sin \frac{4\pi}{14}  \cos (\frac{4 \pi}{14})}{\cos \frac{\pi}{14}} $
$=\dfrac{1}{8}\dfrac{\sin \frac{8\pi}{14}}{\cos \frac{\pi}{14}} $
$=\dfrac{1}{8}\dfrac{\sin(\pi -  \frac{8\pi}{14})}{\cos \frac{\pi}{14}} $
$=\dfrac{1}{8}\dfrac{\sin(\frac{6\pi}{14}}{\cos \frac{\pi}{14}} $
$=\dfrac{1}{8}\dfrac{\cos(\frac{\pi}{2} -  \frac{6\pi}{14})}{\cos \frac{\pi}{14}} $
$=\dfrac{1}{8}\dfrac{\cos(\frac{\pi}{2} -  \frac{6\pi}{14})}{\cos \frac{\pi}{14}} $
 $=\dfrac{1}{8}\frac{\cos \frac{\pi}{14}}{\cos \frac{\pi}{14}} $
$=\frac{1}{8}$

edited the above based on the comment to keep the flow

2011/043) find limit $\lim_{n\to\infty} \sqrt[n]{3^n+4^n}$

We can we take 3 or 4 out but taking 3 it diverges

we get $3 \sqrt[n]{1^n+(\frac{4}{3})^n}$ and reach no where

taking 4 we get
$4 \sqrt[n]{1^n+(\frac{3}{4})^n}$

now $1 < 1+(\frac{3}{4})^n < 2$

so  $\sqrt[n]{1} < \sqrt[n]{1+(\frac{3}{4})^n} < \sqrt[n]{2} $ and as $\sqrt[n]{2}$ as n goes to infinite goes to 1

so result = 4 * 1 or 4.

2011/042) Given $x^2+y^2 = 1$ and $r^2 + s^2 =1$ maximise $rx + sy$

solution:
this can be done by AM GM inequality also but here I show a different way

as $x^2+y^2 = 1 $so let $x = \sin\, t , y= \cos\, t$
and $r^2+s^2 = 1$ so let $r = \sin\, a , s = \cos\, a$

$xr+ sy = \sin\, t \sin\, a+ \cos\, t \cos\, a = cos (t-a)$

maximum when $t = a$ that is 1 ( it is possible when x = r = 1 and y = s = 0)
so maximum 1 is possible
so 1 is the result

2011/041) factoring by diminishing power $6bc - 9c² - 12cd - 8be + 12ce +16de$

one of the factoring that I saw in the net was factor

$6bc - 9c² - 12cd - 8be + 12ce +16de$

...the "clue" you need to recognize is that the coefficients of the first three terms have a common factor...and the coefficients of the last three terms have a common factor...and, the ratios are constant...so, group accordingly...

$(6bc - 9c^2 - 12cd) - (8be - 12ce - 16de) = 3c(2b - 3c - 4d) - 4e(2b - 3c - 4d) =$
$(2b - 3c - 4d)(3c - 4e)$

the above is correct and it was luck that grouping was there but what is luck is not there
then see that highest power of c is 2 and keep them in descending order

$ 6bc - 9c² - 12cd - 8be + 12ce +16de$
$= - 9c² + 6bc - 12cd + 12ce - 8be + 16de$
now you can factor as quadratic in c eliminating the luck factor
$= - 9c² - + 6c( b - 2d + 2e) -8e(b- 2d)$
$= - 9c² - + 6c(( b - 2d) + 2e)) -8e(b- 2d)$
letting b - 2d = a we get
$= - 9c^2 + 6c(a + 2e) - 8ea$
$= - 9c^2 + 6ca + 12ec - 8ea$
$= - 3c(3c - 2a) + 4e(3c-2a)$
$= (3c-2a)(4e - 3c)$
$= (3c-2b+4d)(4e-3c)$
which is same as 1st one

I do not mean to say that 2nd one is preferable to 1st but 2nd one can be used when 1st one does not work

2011/040) Boolean algebra simplify A'(A+B) + (B+AA) (A+B')

A'(A+B)
= A'A + A'B
= A'B as A'A = 0 for any A
(B+AA) (A+B')
= ( B + A) (A + B') as AA = A
= ( B A + AA + BB' + AB'
= BA + A + AB' as AA = A and BB' = 0
= A + BA as A + AB' = A
so A'(A+B) + (B+AA) (A+B')
= A'B + A + BA
= A + B(A'+A)
= A + B as A' + A = 1

2011/039) prove that (-1)* (-1) = 1

a lot of proof of the same is there including the proof below

we know
(-1) + 1 = 0

multiply both sides by -1 to get
(-1)*(-1) + 1*(-1) = 0
add 1* (1) on both sides

to get (-1)*(-1) + 1*(-1) + 1* (1) = 1
or (-1)*(-1) + (1*(-1) + 1* (1))= 1
or(-1)*(-1) + 0= 1

or (-1)*(-1) = 1

proved

However all the proofs including the above come from back ward calculation from LDMA
The LDMA shall fail unless we define it the above way which is a big sacrifice and hence it need to be defined this way to maintain the consistency of LDMA.

( similarly 0! is defined to be 1 so that law of combinatorial holds

2011/038) Find lim n-> inf (n+1)^1/2 – n ^(1/2)

This is of the form inf- inf

We can solve it in 2 ways

1) binomial expansion
(n+1)^(1/2) – n ^(1/2) = n^(1/2) + ½(1/n(^(1/2)) + … - n^(/12)
= ½(1/n(^(1/2)) = 0 as n -> inf

2) by rationalising the numerator
(n+1)^(1/2) – n ^(1/2) = ((n+1)^(1/2) – n ^(1/2) ((n+1)^(1/2) + n ^(1/2) / ((n+1)^(1/2) + n ^(1/2)
= 1/(n+1)^(1/2) + n ^(1/2))
= 1/(inf + inf) = 0

The above method shows rationalising the numerator instead of denominator which is generally used can be used as a tool