Q13/041)Find integral solution of (1-i)^n = 2^n

We have 1-i = (sqrt(2) cis (-pi/4)

So  (1-i)^n = 2^(n/2) cis (-npi/4) = 2^n

The modulo of LHS = 2^(n/2) and rhs = 2^n and both are same if

2^(n/2) = 2^n or n = 0

Then = 2^(n/2) cis (-npi/4) = 1 = RHS

So  n = 0  is the only ans

Q13/040)The fourth power of the common difference of an AP with integer entries is added to the product of any four consecutive terms of it. Prove that the resulting sum is the square of an integer.

proof

Without loss of generality we can take the 1^st term of the 4 consecutive terms to be a and let the difference be t

a and t are integers

We have the 4 terms = a, a+t, a+2t, a+ 3t and 4^th power of difference is t^4

Now a(a+t)(a+2t)(a+3t) + t^4

= (a(a+3t))((a+t)(a+2t)) + t^4

= (a^2+3ta)(a^2+3ta + 2t^2) + t^4

Letting a^2 + 3ta = p we get

= p(p+2 t^2) + t^ 4= (p^2+ 2pt^2+t^4) = (p+t^2)^2 = (a^2+3at+t^2)^2

Which is square of integer as a^2+3at+t^2 is integer

Q13/039) Show that 2^(1/2) 3^(1/2) and 5^(1/2) cannot be 3 terms of an AP

We know 2^(1/2) < 3^(1/2) < 5^(1/2)

Without loss of generality we can assume that 2^(1/2) is the 1^st term and < 3^(1/2) is the pth and 5^(1/2) is the qth term

If r is difference

So 2^(1/2) + (p-1) r = 3^(1/2)

2^(1/2) + (q-1) r = 5^(1/2)

Subtract to get (q-p) r = 5^(1/2) – 3^(1/2)

Or (q-p) r + 3^(1/2) = 5^(1/2)

Square both sides (q-p)^2 r + 3+ 2 sqrt(3)(q-p) = 5

Or sqrt(3) = (2-(q-p)^2)/2((q-p))

Right hand side is rational and LHS is irrational hence a contradiction

Hence proved

Q13/038) Evaluate the following limit without using calculus:

(2−√(2+x))/((2^(1/3)−(4−x)^(1/3))

Solution: 

I would rationalize the denominator to get

(2−√(2+x))(2^(2/3) + 2^(1/3)  (4-x)^(1/3) + (4-x)^(1/3)) /( 2 – (4-x))

Now 2- (4-x) = - ( 4- (2+x) = - (2−√(2+x))( (2+√(2+x))

So we get - (2^(2/3) + 2^(1/3)  (4-x)^(1/3) + (4-x)^(1/3))/ ((2+√(2+x))

Putting x =2 we get - (2^(2/3) + 2^(2/3) + 2^(2/3))/( 2 + 2) = - 3 * 2^(2/3) / 4

Q13/037) The remainder of f(x)/(x^2+x+1) and f(x)/[(x+1)^2] are x+5 and x-1 respectively.? What is the remainder of f(x)/[(x^2+x+1)(x+1)]

We have

f(x) = P(x)(x^2+x + 1)+ (x+5)  ..(1)
_
and f(x) = Q(x)(x+1)^2 + (x-1) = (Q(x)(x+1) +1)(x +1) - 2 ..(2)

now f(x) divided by (x^2+x+1)(x+1) the remainder shall be a quadratic polynomial say

A(x^2 + x + 1) + Bx + C

from (1) B= 1 and C = 5

so remainder = A(x^2+x + 1) + x + 5
from (2) we should have A + 4 = - 2 or A = - 6

so remainder = - 6 x^2 - 5 x -1

Q13/036) if a,b,c,d are in harmonic progression then prove ab+bc+cd=3ad.

a b c d are in HP

so 1/a , 1/b. 1/c , 1/d are in AP

let common difference be m

m = 1/b - 1/a = (a-b)/ab

or ab = (a-b)/m

similarly
bc = ( b-c)/m

and
cd = (c-d)/m

adding we get ab + bc + cd = (a-d )/m ... 1

further 1/d - 1/a = 3m or (a-d) = 3mda ...2

for (1) and (2) we get ab + bc + cd = (a-d )/m = 3mad / m = 3ad

proved

Q13/035)Prove that 8^91 > 7^92?

Proof:

8 = 7 + 1
= 7(1+ 1/7)

8^91 = 7^91( 1+1/7)^91

now (1+ 1/7)^ 91 > 1 + 1/7*91 ignoring rest parts after binomial expansion as (1+a)^n > 1+na

> 1 + 13 > 14

so 8^91 ? 7^91* 14 > 7^91 *7 or 7^92 ( actually it is greater than 2 * 7^92)

Q13/034) Multiplying Polynomials

Given two linear polynomials (a+bx) and (c+dx), their product is given by the quadratic polynomial
bd x² + (bc+ad) x + ac


Let this polynomial be represented by Ax² + Bx + C.
Thus A = bd, B=bc+ad, C = ac.

You are given the values of a,b,c,d and your task is to calculate the values of A,B,C. The constraint is that you can carry out only three multiplications.

solution

A = bd

B=bc+ad

C = ac.

 There are  4 multiplications

However if we convert B = (a+b)(c+d) – bd – ac = (a+b)(c+d) – A – C we have 3 multiplications

 

Q13/033) Find a 3 digit number which equals sum 17 times the hundredth digit, 34 times tens digits and 51 times the units digit.

Let the digits be xyz

So the number = 100x + 10y + z = 17 x + 34 y + 51 z

Or 83 x = 24 y + 50 z  (1)

Working in mode 24 we have 11 x = 2z

So x is even

  X =2 => z = 11 not possible

X= 4 => z = 10 not possible

X = 6 => Z = 9 ( and from (1) 24 y = 83 x- 50 z = 498 – 540 = 48 or y = 2)

X = 8=> z = 8 ( and from(1) 24 y = 33 * 8 or y > 10)

So the number is 629

Q13/032) Find lim n-> inf (n/(n+2))^n

(n/(n+2))^n

= (1/(1+2/n)^n

= ((1/(1+2/n)^n/2)^2

As (1+1/x) ^x as x -> infinite is  e so /(1+2/n)^n/2 = e so given limit = 1/e^2