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