Sunday, August 31, 2008

Exponent of Prime p in n! - July-Dec Revision

The exponent of prime number of 3 in 100! is 48.
It means 100! is divisible by 348

How do you find it?
Let p be a prime number and n be a positive integer. Then find the last integer in the sequence 1,2,…,n which is divisible by p.
Express this integer as [n/p]p.
[n/p] denotes the greatest integer less than or equal to n/p

In case of 3 (p) and 100 (n); [n/p] is 33 and n/p is 33 and 1/3.

Let Ep(n) denote the exponent of the prime p in the positive integer n. Then,

Ep(n!) = Ep(1.2.3…(n-1).n)

This will be equal to Ep(p.2p.3p…[n/p]p)
= [n/p]+ Ep(1.2.3...[n/p])

This process continues and we get the answer

Ep(n!) = [n/p] + [n/p²]+…+[n/ps]
Where s is the largest positive integer such that ps≤n≤ps+1

Hence applying the formula to find exponent of prime 3 in 100!

E3(100!) = [100/3] + [100/3²] + [100/3³] + [100/34]
= 33+11+3+1 = 48

Note: remember the meaning of notation [100/3] or [n/p]

Chapter: Permutations

Saturday, August 30, 2008

Binomial theorem - July Dec Revision

If x and a are real numbers, then for all n Є N

= nC0xna0 + nC1xn-1a1 +nC2xn-2a2 + ...+nCrxn-rar+ ...+nCn-1x1an-1+nCnx0an

= (r = 0 to n)Σ nCrxn-rar

Some conclusions

1. Total number of terms in the expansion = n+1

2. The sum of indices of x and a in each term is n.

3. The coefficients of terms equidistant from the beginning and the end are equal.

4. (x-a)n
= (r = 0 to n)Σ ((-1) r*nCrxn-rar

The terms in the expansion of(x-a)n are alternatively positive and negative, the last term is positive or negative according as n is even or odd.

5. (1+x) n = (r = 0 to n)Σ nCrxr

you get it by putting x =1 and a = x in the expression for (x+a)n.

6. (x+1) n = (r = 0 to n)Σ nCrxn-r

7. (1-x) n = (r = 0 to n)Σ(-1)r* nCrxr

8. (x+a) n +(x-a) n = 2[nC0xna0 +
nC2xn-2a2 +
nC4xn-4a4+ ...]

9. General term in a binomial expansion

(r+1) term in (x+a) n
= nCrxn-rar

10. Another form of binomial expansion

= (r = 0 to n) and r+s = n Σ (n!/r!s!) xras

11. Coefficient of (r+1)th term in the binomial expression of (1+x)n is nCr

12. Algorithm to find the greatest term

(i) Write term r+1 = T(r+1) and term r = T® from the given expression.
(ii) Find T(r+1)/T®
(iii) Put T(r+1)/T®>1
(iv) Solve the inequality for r to get an inequality of the form rm

If m is an integer, the mth and (m+1)th terms are equal in magnitude and these two are the greatest terms.

If m is not an integer, then obtain integral part of m, say, k. In this case (k+1) term is the greatest term

13 Properties binomial coefficients

(i) the sum of binomial coefficients in the expansion of (1+x)n is 2n

(ii) the sum of odd binomial coefficients in the expansion of (1+x)n is equal to the sum of the coefficients of even terms and each is equal to 2n-1.

14. Middle terms in binomial expression

If n is even the ((n/2) +1) th term is middle term.
If n is odd then ((n+1)/2) th and ((n+3)/2)th terms are two middle terms

Multinomial theorem

(x1+x2)n = (r1 = 0 to n) and r1+r2 = nΣ (n!/r1!r2!) x1r1x2r2

Past JEE Problem

The coefficient of x4 in [(x/2)- (3/x²)]10 is

a. 405/256
b. 504/259
c. 450/263
d. none of these

(JEE 1983)

Answer: a

Use the formula

(r+1) term in (x+a) n
= nCrxn-rar

to find r which will give x4 term in expansion.

For some more problems of JEE

Wednesday, August 27, 2008

Hyperbola – Definitions - July Dec Revision

A hyperbola is the locus of a point in a plane which moves in the plane in such a way that the ratio of its distance from a fixed point (called focus) in the same plane to its distance from a fixed line (called directrix) is always constant which is always greater than unity.

The constant ratio is generally denoted by ‘e’ and is known as the eccentricity of the hyperbola.

Every hyperbola has a second focus and second directrix.

The difference of the focal distances of any point on a hyperbola is constant and is equal to the length of transverse axis of the hyperbola.

On account of this property, a second definition of the hyperbola is:

A hyperbola is the locus of a point which moves in such a way that the difference of its distance from two fixed points (foci) is always constant.

Conjugate hyperbola: The hyperbola whose transverse and conjugate axes are respectively the conjugate and transverse axes of a given hyperbola is called the conjugate hyperbola of the given hyperbola.

Director circle: is the locus of points from which perpendicular tangents are drawn to the given hyperbola.

Asymptotes: An asymptote to a curve is a straight line, at a finite distance from the origin, to which the tangent to a curve tends as the point of contact goes to infinity. In other words, asymptote to a curve touches the curve at infinity.

Rectangular hyperbola: A hyperbola whose asymptotes are right angles to each other is called a rectangular hyperbola.

Tuesday, August 26, 2008

Ellipse - Definitions - July Dec Revision

1. An ellipse is the locus of a point which moves in a plane such that its distance from a fixed point (the focus) is always 'e' times its distance from a fixed line (the directrix), where e<1 is the eccentricity.

le the focus be at S(ae,0) and the directrix be x = a/e.

Then the equations of the ellipse is

x²/a² + y² /[a²(1-e²)] = 1

As e<1, 1-e²>0, we can write a² = b²/(1-e²) or

b² = a²(1-e²)

x²/a² + y² /b² = 1

The piont (0,0) is the centre of the ellipse.

The major axis will be along x axis and its length is 2a

The minor axis will be along y axis and its length is 2b.

The point (a,0) and (-a,0) are the two vertices of the ellipse.

A line through the focus of the ellipse perpendicular to the its major axis is called its latus recturm. Its equation is x = ae and its length is 2b²/a.

If we take (-ae,0) as the focus and the line x = -a/e as the directrix, we get the same equation of the ellipse. Therefore ellipse has two foci, (ae,0) and (-ae,0) and it has two directrices, x = a/e and x = -a/e respectively. It has also two latera recta whose equations are, x = ae and x = -ae.