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

2011/037) If α is a root of equation x² + x + 1 = 0 then show that α is a root of equation x^(3p) + x^(3q+1) + x^(3m+ 2) = 0

Proof:
α is a root of equation x² + x + 1 = 0

so α^2 + α + 1 = 0 and hence α^3 =1

now there are 2 approaches

1)

α ^(3p) + α ^(3q+1) + α ^(3m+ 2)
= (α ^3)^p + (α ^3)^3q* α +( α ^3)3m* α ^2
= 1 + α+ α^2
= 0
Hence α is a root

2)

α ^(3p) + α ^(3q+1) + α ^(3m+ 2)
= α ^(3p) + α ^(3p+1) + α ^(3p+ 2) + α ^(3q+1) - α ^(3p+ 1) + α ^(3m+ 2) - α ^(3p+ 2)
= α ^(3p)( 1+ α + α ^ 2) + α (α ^(3q) - α ^(3p)) + + α^2 (α ^(3m) - α ^(3p))
= 0 + 0+ 0 = 0 as each part is zero
Hence α is a root


as a corollary we see that x² + x + 1 devides x^(3p) + x^(3q+1) + x^(3m+ 2)