Q) Find the coefficient of x^4 in (1 + 2x + 3x^2)^5 .

A) [(1+2x)+3x^2)]^5

'r'th Term = 5Cr (1+2x)^r . (3x^2)^{5-r}

To form net effective 4th powers of x:

from the righthand term, we can only use x^0 x^2 and x^4

so from the lefthand term (1+2x)^r ... we use x^4, x^2 and x^0 respectively.

5C5.(3x^2)^0 . (1+2x)^5 -----------> 5C5. 5C4.(2x)^4

and

5C4.(3x^2).(1+2x)^4 --------> 5C4.(3x^2).[4C2.(2x)^2]

and

5C3.(3x^2)^2 .(1+2x)^3 ---------> 5C3.(3x^2)^2 [3C0. (2x)^0]

So, coefficient of x^4 :

5C5.5C4.2^4 + 5C4.3.4C2. 2^2 + 5C3.3^2. 1

= 5. 16 + 5.3.6.4 + 10.9

= 80+ 360 + 90

= 530

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## Sunday, November 2, 2008

Q) Find the sum of :

1 + 1/2 + 1/3 + 1/4 .... up to infinite terms.

A)

1/1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + ...

= ( 1/1 ) + ( 1/2 ) + ( 1/3 + 1/4 ) + ( 1/5 + 1/6 + 1/7 + 1/8 ) + ...

(where each set of parenthesis stops at a reciprocal of a power of two.) Replacing the terms in each set of parenthesis by the smallest term (in those parenthesis) we get a smaller sum:

1/1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + ...

> ( 1/1 ) + ( 1/2 ) + ( 1/4 + 1/4 ) + ( 1/8 + 1/8 + 1/8 + 1/8 ) + ...

= 1/1 + 1/2 + 1/2 + 1/2 + 1/2 + ...

which clearly gets arbitrarily large! So it is a diverging series and will not have a definite value since it is not converging. The sum will go on approaching infinity.

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1 + 1/2 + 1/3 + 1/4 .... up to infinite terms.

A)

1/1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + ...

= ( 1/1 ) + ( 1/2 ) + ( 1/3 + 1/4 ) + ( 1/5 + 1/6 + 1/7 + 1/8 ) + ...

(where each set of parenthesis stops at a reciprocal of a power of two.) Replacing the terms in each set of parenthesis by the smallest term (in those parenthesis) we get a smaller sum:

1/1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + ...

> ( 1/1 ) + ( 1/2 ) + ( 1/4 + 1/4 ) + ( 1/8 + 1/8 + 1/8 + 1/8 ) + ...

= 1/1 + 1/2 + 1/2 + 1/2 + 1/2 + ...

which clearly gets arbitrarily large! So it is a diverging series and will not have a definite value since it is not converging. The sum will go on approaching infinity.

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Q) Find the number of real pairs (x,y) satisfying

A) y = 2xy

y (1-2x) = 0..... So, either y=0 or x = 1/2

For y = 0,

x = x^3

x^3 - x = 0 = (x - x^1/3) (x^2 + x^{2/3} + x^{4/3} )

x = -1,1, 0 for y = 0 ............ 3 real roots.

AND for x = 1/2

1/2 - 1/8 = y^4

0.375 = y^4

y^4 - 0.375 = 0 = (y^2 + rt(0.375) ) (y^2 - rt(0.375) )

which has 2 real and 2 imaginary roots.

So, total 3 + 2 = 5 real solutions.

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**and****.**A) y = 2xy

y (1-2x) = 0..... So, either y=0 or x = 1/2

For y = 0,

x = x^3

x^3 - x = 0 = (x - x^1/3) (x^2 + x^{2/3} + x^{4/3} )

x = -1,1, 0 for y = 0 ............ 3 real roots.

AND for x = 1/2

1/2 - 1/8 = y^4

0.375 = y^4

y^4 - 0.375 = 0 = (y^2 + rt(0.375) ) (y^2 - rt(0.375) )

which has 2 real and 2 imaginary roots.

So, total 3 + 2 = 5 real solutions.

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Q) Find the integral of :

A)

=

=

Now take out the 1 and you get:

e^x + e^x [ f(x) + f'(x) ] ..... where f(x) = -4/(x+4)

And we know that integral of e^x [ f(x) + f'(x) ] = e^x. f(x) + C

So the integral is :

e^x + e^x.[-4 / (x+4) ] + C

OR

e^x . [x/ (x+4)] + C

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Q) Find the number of 5-digit numbers of the form xyzyx in which x is less than y.

A) Number of ways of selecting x and y: 9C2 . 2!

Out of these, 1/2 will be such that x than y, and 1/2 such that x greater than y

So for x less than y: 9C2 selections

And z can be any number from 0 to 9, so 10 ways.

So total number of such numbers is 9C2 . 10

= 360 numbers

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A) Number of ways of selecting x and y: 9C2 . 2!

Out of these, 1/2 will be such that x than y, and 1/2 such that x greater than y

So for x less than y: 9C2 selections

And z can be any number from 0 to 9, so 10 ways.

So total number of such numbers is 9C2 . 10

= 360 numbers

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## Saturday, November 1, 2008

Q) If the arithmetic mean of two numbers 'a' and 'b' is 'n' times their harmonic mean, find a/b in terms of 'n'.

A) AM = n. HM

(a+b) / 2 = n. [2ab/(a+b)]

a^2 + b^2 + 2ab = 4*n*ab

(a+b)^2 = 4*n*ab

(a/b + 1)^2 = 4*n*a/b

Let a/b = x

(x + 1)^2 = 4nx

x^2 + 1 + 2x - 4nx = 0

x^2 + x(2-4n) + 1 = 0

x = 4n-2 +- sqrt(16n^2 + 4 - 16n -4) / 2

x = 4n-2 +- 4.sqrt(n^2 - n) / 2

x = 2n - 1 +- 2.sqrt (n^2 - n)

OR

a/b = 2n - 1 +- 2.sqrt (n^2 - n)

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A) AM = n. HM

(a+b) / 2 = n. [2ab/(a+b)]

a^2 + b^2 + 2ab = 4*n*ab

(a+b)^2 = 4*n*ab

(a/b + 1)^2 = 4*n*a/b

Let a/b = x

(x + 1)^2 = 4nx

x^2 + 1 + 2x - 4nx = 0

x^2 + x(2-4n) + 1 = 0

x = 4n-2 +- sqrt(16n^2 + 4 - 16n -4) / 2

x = 4n-2 +- 4.sqrt(n^2 - n) / 2

x = 2n - 1 +- 2.sqrt (n^2 - n)

OR

a/b = 2n - 1 +- 2.sqrt (n^2 - n)

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Q) A particle is thrown upwards from the ground. It experiences constant air resistance which produces a retardation of 2 m/s^2. Find the ratio of time of ascent to time of descent.

A)

v = u + at

0 = u + -(g + 2)t

t = u / (g + 2)

0 - u^2 = 2as

- u^2 = 2 *-(g + 2)s

s = u^2 / 2(g + 2)

s = 1/2 aT^2

u^2 / 2(g+2) = (1/2) * (g-2)T^2

u / sqrt(g^2 - 4) = T

So t/T = sqrt (g-2) / sqrt (g+2) = sqrt (8/12) = sqrt (2/3)

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A)

v = u + at

0 = u + -(g + 2)t

t = u / (g + 2)

0 - u^2 = 2as

- u^2 = 2 *-(g + 2)s

s = u^2 / 2(g + 2)

s = 1/2 aT^2

u^2 / 2(g+2) = (1/2) * (g-2)T^2

u / sqrt(g^2 - 4) = T

So t/T = sqrt (g-2) / sqrt (g+2) = sqrt (8/12) = sqrt (2/3)

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