Paper 4: Fundamentals of Business Mathematics & Statistic

2.34 I FUNDAMENTALS OF BUSINESS MATHEMATICS AND STATISTICS

Algebra

(3) a−k=a^1 k for any non zero real number a and any integar k.

For any positive real number a and any positive integar m,the mth root of a is defined and denoted by

a^1 m or ^1 = 

 

ma a am m and is a real number .But when m is an even positive integar and a is a negative real

number .then a^1 m is not a real number.

e.g,

51 1

(^1) ,( 3) 4
2

 − −

   etc. are not real numbers.

In general a a am n=n m=( )mn^1

Now (am/n)n = am/n x am/n x am/n ........ n factors

=am m mn n n+ + .......nterms

= amn/n = am

Taking nth root of both sides, we get

= amn=nam

Note 1. If in an equation base on both sides is the same,powers can be equated, i e. if x x then a ba= b =

SOLVED EXAMPLES

Example 43 : Show that the expression

2 2 2

2.

2 .3 .5 .6

6 .10 15

+ − + +

+

m m n m n n

m n m is indepedent of m and n.

Solution : L.H.S.=

2 2 2

2.

2 .3 .5 .6

6 .10 15

+ − + +

+

m m n m n n

m n m

=2 .2 .3 .3 .2 .5 .5 .3 .5 .2 .3m+^2 −m m m n n−^2 − − +( 2) ( 2)−n m n+ + +^2 −m m n n−

− − − + −+ − −

+

(^) = = =
(^)
( 2) ( 2)
2
(^1) 2 .3. (^1) 2 .5 , (^1) 3 .5
6 10 15
m m n n m m
ô m n m
∴ L.H.S. = (^) = 2 m m n n m m n m n n m n m+ −^2 − − +^2 .3− +^2 − − +.5− − +^2 + + −^2
=2 3 5 1o× o ox = which is independant of m and n.
Example 44: SImplify
1 23 3 54 34 4
13

2 .8 .6 .3

9 16

− − −

−

 

 

 

Solution:

The given expression

− − −

−

(^) ×
=^
(^)
13 323 54 34 4
2 1 4^13

2 .(2 ) .(2 3) .3.

(3 ) .(2 )