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# Algebra

FUNCTIONS-1

Function is basically a relation between two sets, set of inputs and a set of output where an element of the set containing the inputs corresponds to not more than one output. Let us say we have a set of four persons namely called A whose elements are P,Q, R and T and set a set B containing names of five companies namely X1,X2,X3,X4 and X5. Now a relation between A to B is a function if a member of A is linked to not more than one member of B.

Here the elements X4 and X5 are not mapped but none of the elements P,Q,R and S are mapped to more than one element in the set B.

This means every algebraic expression may not be a function. For one value of x one must get only one value of y. It can so happen that for more than one value of x, we can get one value of y. This means for an algebraic expression, to be a function, a line drawn parallel to the Y axis must cut the curve at only one point. Consider the expression x^{3}, its curve is as shown below and lines drawn parallel to the Y axis will cut the curve at only one point.

Thus, y = x^{3} is a function.

Consider the expression: y^{2 }= 4ax. Here the lines drawn parallel to the Y axis cut the curve at more than one point

Hence y^{2 }= 4ax is not a function.

Function can be viewed as a machine which takes input, processes it and gives output. The set of input for which the function is defined, is known as Domain while the set of values of output which we get is known as Range.

For e.g. let’s say we have a function y = x^{2} +1 , then the function takes the input as the values of x, squares it and adds 1 to it and gives the output.

As one can see, the input can be any real number so the Domain of the given function is R (Set of all real numbers) while the range is positive number starting from 1.

__Calculation of Domain: __

For the calculation of domain, we take the values of x for which the function is defined. It can also be calculated by finding the values of x or the intervals of x where the function is not defined and then subtracting that interval from the set of real numbers. Let’s see the cases where the function becomes not defined

a. If the expression under the square root is negative. For example consider

c. logx is defined for only positive values of x, for example consider log(3x-4) , if 3x -4 < 0 or x < 4/3, then the function becomes not defined.

d.sin^{-1}x or cos-1x is defined only when -1 ≤ x ≤ 1, consider the function

sin^{-1}(2x+1), if 2x+1 > 1 i.e. x > 0 or 2x+1 < -1 or x <-1 then the function becomes not defined.

**Example 1:** Find the domain of the following functions:

a. log_{10}(x^{3} –x)

b. 1/[x-3]

c. sin^{-1}{log_{3}(x/3)}

Solution :

a.log_{10}(x^{3} –x)

For the function to be defined x^{3} –x > 0 or x(x^{2}-1) >0 or x(x-1)(x+1) > 0

b. 1/[x-3] where [ ] represents greatest integer function.

Now the denominator cannot be equal to zero. So, [x-3] ≠ 0

Or let’s find out the interval in which it is not defined. So for the function to be undefined, [x-3] = 0 or 0 ≤ x-3 < 1 or 3≤ x <4

**Thus, domain will be R –[3,4) **

c. f(x) = sin^{-1}{log_{3}(x/3)}

For the function to be defined -1 ≤ log_{3}(x/3) ≤ 1

Or 3^{-1} ≤ (x/3) ≤ 3^{1 }Or 1/3 ≤ ( x/3) ≤ 3 or 1 ≤ x ≤ 9

**So, domain of f(x) is 1 ≤ x ≤ 9**

**Example 2:** Find the range of the following functions:

- F(x) = 2sinx + 3cosx
- F(x) = 2x/(1+x)
- F(x) = (2x+1)/(x
^{2}+1)

**Solution:**

b. f(x) = 2x/(1+x)

Let f(x) = y = 2x/(1+x) or y +xy = 2x

Or y = x(2-y)

x = y/(2-y)

As x is real, so y≠ 2

So, range is all real values except 2.

c . f(x) = (x^{2} +x +2)/(x^{2}+x+1)

y = (x^{2} +x +2)/(x^{2}+x+1)

x^{2}y +xy +y = x^{2} + x+2

x^{2}(y-1) + x(y-1) + y-2 = 0

As x is real, so discriminant must be positive

(y-1)^{2} -4(y-1)(y-2) > 0

(y-1)(y-1-4y +8) > 0

(y-1)(-3y +7) ≥ 0 or (y-1)(y-7/3) < 0

1 < y < 7/3

Thus, range of the function becomes (1,7/3)

__Types of functions:__

**One –One Function:**

A function f: A →B is called a one- one function if each element in A corresponds to a different element in B. So, every object in A would have a distinct image in the set B. Such a function is also known as injective function. It is not necessary that all the elements in the co-domain of set B are mapped

Eg. f(x) = logx is a one-one function

If f(x) is a continuous function, either increasing or decreasing, then it is a one-one function.

Draw a horizontal lines, if they intersect the curve at only one point respectively, then the function is one –one.

__Many-one function__

The function f : A→ B is called many-one function, if two or more than two different elements in A have the same image in B.

- Onto Function:

In a function f: A→B, if the range is equal to its co-domain, then it is known as onto function. In other words, every element in the set B must be mapped.

Here in this case each element of B is mapped, so it is onto. At the same time it is also many one function. Onto functions are also known as surjective functions.

Eg : f: R→R is given by f(x) = x^{3}, then here the range is equal to all real numbers which is also the co domain of the function.

Every polynomial f: R→R is onto if the degree of the polynomial is odd.

**Into Function:**

Into function If f: A→ B is such that there exists atleast one element in codomain which is not the image of any element in domain, then f (x )is into. i.e. Range is a subset of Co-domain

Eg. f: R→R, such that f(x) = x^{2}, then the range is all positive numbers including zero but the co-domain is set of all real numbers, so range is a subset of co-domain here and hence it is an into function.

f: R →R , f(x) = x^{2 }

*Note: If the function is both injective as well as surjective, then it is known as bijective function. *

Example 3 : The function f : [0,∞) → [0,∞) is defined by

f(x) = 2x/(1+2x) is

- one-one and onto
- one-one but not onto
- not one-one but onto
- Neither one-one nor onto

Solution: First we need to check whether the function is on-one or many-one

f(x_{1}) = 2x_{1}/(1+2x_{1}) and f(x_{2}) = 2x_{2}/(1+2x_{2})

For F(x_{1}) = F(x_{2}) , if x_{1} = x_{2}, then the function is one- one otherwise it is many one.

x_{1} + 2x_{1}x_{2} = x_{2} + 2x_{1}x_{2}

Or x_{1} = x_{2}

Thus, it is a one-one function

Now for onto functions, range = co-domain

y = 2x/(1+2x)

y+2xy = 2x or y = 2x(1-y)

X = y/2(1-y)

As x is real so, y cannot be one

Hence the range is R –{1}

But the codomain includes the number 1, thus, co-domain is not equal to range.

Thus, the function is not onto

So, the function is one-one into.

Example 4:The mapping f: N→ N given by f(n) = 1+n^{2} , n ? N, where N is the set of natural numbers, is

- one-one and onto c. one-one but not onto
- onto but not one-one d. neither one-one nor onto

**Solution:** Here the domain is all natural numbers while the range is also all natural numbers.

Clearly, it is not defined for negative values, so it is an increasing function and hence every element in the domain gives a unique value in the output. So, it a one-one function

The lowest value of f(n) is obtained as 2 when n = 1

So, Range will be all natural numbers starting from 2, which means the co-domain is not equal to the range. Thus, it is not an onto function

So, f(n) is a one-one but not onto function.

**Example 5:** If A = {a,b,c,d,e} while B is a set { p,q,r,s}, then how many onto function can be formed if f: A→ B ?

**Solution:** For a function to be onto, every element in the set B must be mapped to elements in the set A.

As there are five elements in the set A that need to be mapped, this can happen when two elements in A are mapped to one element in B and the rest of the elements in A are mapped to a different element in the set B. One such case could as shown in the figure:

So, two elements in set A can be selected in ^{5}C_{2} ways while the one element to which they will be mapped can be selected in ^{4}C_{1} ways. Now the rest of the three elements in A must be mapped to a different element in B, which can be done in 3! Ways.

Total ways = ^{5}C_{2} * ^{4}C_{1} * 3! = 240

Thus, there are 240 such onto functions that can be formed.