Chapter 6.3 - Linear Inequalities in Two Variables

In the previous section, we completed a discussion on solving linear inequalities in one variable. We saw some solved examples also. In this section, we will see linear inequalities in two variables.

• In the previous sections, we saw that graphical method is very effective to show all the acceptable values for linear inequalities in one variable. Graphical method is effective for two variables also. Some basics can be written in 5 steps:
1. In fig.6.7(a) below, the vertical red line divides the Cartesian plane into two parts.
• Each part is known as a half plane.
   ♦ Half plane on the left is known as the left half plane I.  
   ♦ Half plane on the right is known as the right half plane II.

Fig.6.7

2. In fig.6.7(b) above, the red line is a sloping line. It slopes upwards as we move from left to right along the x axis. This sloping line divides the Cartesian plane into two parts. Here also, each part is known as a half plane.
   ♦ Half plane on the lower side is known as the lower half plane I. 
   ♦ Half plane on the upper side is known as the upper half plane II. 
3. In fig.6.7(c) above, the red line is a sloping line. It slopes downwards as we move from left to right along the x axis. This sloping line divides the Cartesian plane into two parts. Here also, each part is known as a half plane.
   ♦ Half plane on the lower side is known as the lower half plane I.
   ♦ Half plane on the upper side is known as the upper half plane II.
4. Consider the equation ax + by = c
• There will be a large number of ordered pairs (x,y) which will satisfy this equation.
• We know that ‘ordered pairs’ are points on the Cartesian plane.
   ♦ All points which satisfy the above equation will lie on a line.
   ♦ That line is the graph of the above equation.
5. So when we consider the infinite number of points in the Cartesian plane, there are three possibilities:
(i) A large number of those points will satisfy the equation ax + by = c
(ii) Another large number of points will satisfy the inequality ax + by < c
(iii) The remaining points will satisfy the inequality ax + by > c

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◼ So now we can think about solving the inequalities:
• We want to be able to mark those points which satisfy the inequality ax + by < c.
    ♦ Those points will form the solution set of this inequality.  
• We want to be able to mark those points also which satisfy the inequality ax + by > c.
    ♦ Those points will form the solution set of this inequality.
• The following 18 steps will help us to mark the required points.
1. Consider the equation ax + by = c
• We know that, the general form of a line is: y = mx +c
• We can transform ax + by = c into the general form:
$\begin{array}{ll}
{}&ax+by &{}={}& {c} &{} \\
\Rightarrow &ax+by-ax&{}={}& c-ax &{} \\
\Rightarrow &by&{}={}& -ax+c &{} \\
\Rightarrow &\frac{by}{b}&{}={}& \frac{-ax+c}{b} &{} \\
\Rightarrow &y&{}={}& {\frac{-a}{b}}x+\frac{c}{b} &{} \\
\Rightarrow &y&{}={}& {\frac{-a}{b}}x+c &{} \\
{} &{}&{}& \color {green}{\frac{c}{b}~\text{is a constant. So it can be denoted by 'c'}}&{} \\
\Rightarrow &y&{}={}& -mx+c &{} \\
{} &{}&{}& \color {green}{\frac{a}{b}~\text{is a constant. So it can be denoted by 'm'}}&{} \\
\end{array}$
2. Consider the equation ax - by = c
• We know that, the general form of a line is: y = mx +c
• We can transform ax - by = c into the general form:
$\begin{array}{ll}
{}&ax-by &{}={}& {c} &{} \\
\Rightarrow &ax-by-ax&{}={}& c-ax &{} \\
\Rightarrow &-by&{}={}& -ax+c &{} \\
\Rightarrow &\frac{-by}{-b}&{}={}& \frac{ax+c}{-b} &{} \\
\Rightarrow &y&{}={}& {\frac{-a}{-b}}x + \frac{c}{-b}&{} \\
\Rightarrow &y&{}={}& {\frac{a}{b}}x - \frac{c}{b}&{} \\
\Rightarrow &y&{}={}& {\frac{a}{b}}x-c &{} \\
{} &{}&{}& \color {green}{\frac{c}{b}~\text{is a constant. So it can be denoted by 'c'}}&{} \\
\Rightarrow &y&{}={}& mx-c &{} \\
{} &{}&{}& \color {green}{\frac{a}{b}~\text{is a constant. So it can be denoted by 'm'}}&{} \\
\end{array}$
3. We can write:
   ♦ ax + by = c in general form is: y = -mx - c
   ♦ ax - by = c in general form is: y = mx + c
• We see that, the sign of b will determine the sign of m.
• From our earlier classes, we know that:
   ♦ If m is +ve, the line will slope upwards.
   ♦ If m is -ve, the line will slope downwards. (Details here)
• Thus we get:
   ♦ If b is +ve, the line will slope downwards.
   ♦ If b is -ve, the line will slope upwards.
• What about the signs of a and c?
The answer can be written in 2 steps:
(i) The sign of 'a' can be always made +ve. This is by multiplying the whole equation by -1 if necessary.
(ii) The sign of 'c' can be +ve or -ve. It will not affect the slope of the line.
• Now we have a good knowledge about the shape of the line. We can proceed to investigate inequalities.
• The steps from (4) to (9) will help us to understand the inequality when b is +ve. That is, b > 0
• The steps from (10) to (16) will help us to understand the inequality when b is -ve. That is, b < 0
4. Consider the equation ax + by = c
• Since 'b' is +ve, the line will slope downwards. This is shown in fig.6.8 below:

Fig.6.8

• Let P(𝛼,β) be a point on the line. Then both 𝛼 and β will satisfy the equation ax+by = c
We can write: a𝛼 + bβ = c
5. Now consider a point Q in such a way that:
   ♦ Q lies in the vertical line through P
   ♦ Q is in the half plane II
• Since Q and P are in the same vertical line, both of them will have the same x coordinate .
   ♦ Let the y coordinate be 𝛾
   ♦ Then the coordinates of Q will be (𝛼,𝛾)
6. Since Q is in the half plane II, it is at a higher level than P.
   ♦ So the y coordinate of Q
   ♦ will be greater than
   ♦ the y coordinate of P
• We can write: 𝛾 > β 
   ♦ Multiplying both sides by b, we get: b𝛾 > bβ (Rule 2) 
   ♦ Adding a𝛼 on both sides we get: a𝛼+b𝛾 > a𝛼+bβ (Rule 1)
7. But from step (4), we have: a𝛼+bβ = c
• So the result in (6) becomes: a𝛼+b𝛾 > c
• That means:
The point Q(𝛼,𝛾) will satisfy the inequality ax + by > c
• Q(𝛼,𝛾) is a general point that we took in half plane II
• So we can write:
All points that lie in the half plane II will satisfy the inequality ax + by > c, where b >0
8. Now we have to prove the converse:
If a point Q(𝛼,𝛾) satisfies the inequality ax+by > c, where b>0, then Q(𝛼,𝛾) will be on the half plane II.
This can be done in 4 steps:
(i) Given that: Q(𝛼,𝛾) satisfies the inequality ax+by >c
Then we can write: a𝛼 + b𝛾 > c
(ii) But from step (4), we have: a𝛼+bβ = c
• So the result in (i) becomes: a𝛼+b𝛾 > a𝛼+bβ
• Subtracting a𝛼 from both sides, we get: 
b𝛾 > bβ (Rule 1)
(iii) Dividing both sides by b, we get: 𝛾 > β (Rule 2)
• The sign will not reverse because b is greater than zero.
(iv) Since P and Q are on the same vertical line, 𝛾 > β implies that Q lies in the half plane II.
9. So when b > 0, we have the details about all three regions:
(i) The region which lies exactly on the line ax+by=c,
   ♦ will contain all those points
   ♦ which satisfy the equation ax+by=c
(ii) The region of half plane II,
   ♦ will contain all those points
   ♦ which satisfy the inequality ax+by > c
(iii) Obviously the remaining region of half plane I,
   ♦ will contain all those points
   ♦ which satisfy the inequality ax+by < c

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10. Now we can investigate the case when b < 0
• Consider the equation ax + by = c
• Since 'b' is -ve, the line will slope upwards. This is shown in fig.6.9 below:

Fig.6.9

• Let P(𝛼,β) be a point on the line. Then both 𝛼 and β will satisfy the equation ax+by = c
We can write: a𝛼 + bβ = c
11. Now consider a point Q in such a way that:
   ♦ Q lies in the vertical line through P
   ♦ Q is in the half plane II
• Since Q and P are in the same vertical line, both of them will have the same x coordinate .
   ♦ Let the y coordinate be 𝛾
   ♦ Then the coordinates of Q will be (𝛼,𝛾)
12. Since Q is in the half plane II, it is at a higher level than P.
   ♦ So the y coordinate of Q
   ♦ will be greater than
   ♦ the y coordinate of P
• We can write: 𝛾 > β 
   ♦ Multiplying both sides by b, we get: b𝛾 < bβ (Rule 3) 
• The sign will reverse because b is less than zero.
   ♦ Adding a𝛼 on both sides we get: a𝛼+b𝛾 < a𝛼+bβ (Rule 1)
13. But from step (4), we have: a𝛼+bβ = c
• So the result in (12) becomes: a𝛼+b𝛾 < c
• That means:
The point Q(𝛼,𝛾) will satisfy the inequality ax + by < c
• Q(𝛼,𝛾) is a general point that we took in half plane II
• So we can write:
All points that lie in the half plane II will satisfy the inequality ax + by < c, where b < 0
14. Now we have to prove the converse:
If a point Q(𝛼,𝛾) satisfies the inequality ax+by < c, where b<0, then Q(𝛼,𝛾) will be on the half plane II.
• This can be done in 4 steps:
(i) Given that: Q(𝛼,𝛾) satisfies the inequality ax+by < c
Then we can write: a𝛼 + b𝛾 < c
(ii) But from step (4), we have: a𝛼+bβ = c
• So the result in (i) becomes: a𝛼+b𝛾 < a𝛼+bβ
• Subtracting a𝛼 from both sides, we get: 
b𝛾 < bβ (Rule 1)
(iii) Dividing both sides by b, we get: 𝛾 > β (Rule 3)
• The sign will reverse because b is less than zero.
(iv) Since P and Q are on the same vertical line, 𝛾 > β implies that Q lies in the half plane II.
15. So when b < 0, we have the details about all three regions:
(i) The region which lies exactly on the line ax+by=c,
   ♦ will contain all those points
   ♦ which satisfy the equation ax+by=c
(ii) The region of half plane II,
   ♦ will contain all those points
   ♦ which satisfy the inequality ax+by < c
(iii) Obviously the remaining region of half plane I,
   ♦ will contain all those points
   ♦ which satisfy the inequality ax+by > c
16. Based on the results in steps (9) and (15), we can prepare a table:

Table 6.1

• Let us see a sample application of this table. We will consider the same inequality that we saw at the beginning of this chapter.
Example 4: 20x + 8y ≤ 120 
It can be written in 4 steps:
(i) Suppose we want to solve the inequality 20x+8y < 120
• Let us first compare it with the standard form: ax +by < c
   ♦ We see that, b is +ve. And the sign before c is '<'
   ♦ So from the table, we get: Lower plane I
(ii) In fig.6.10 below, the red line is the graph of 20x + 8y = 120

Fig.6.10

• This line divides the Cartesian plane into Lower half plane I and upper half plane II
• All points in the lower half plane I will satisfy the inequality 20x+8y < 120
• We can check this using any convenient point. O(0,0) is convenient for calculations. So we will use that point.
• If we input (0,0) in the inequality, we will get:
   ♦ (20 × 0) + (8 × 0) < 120
   ♦ ⇒ 0 + 0 < 120
   ♦ ⇒ 0 < 120, which is true.
(iii) O(0,0) lies in the lower half plane I. The check confirms that, all points in the lower half plane I will satisfy the inequality 20x+8y < 120
• So we hatch the entire lower half plane I using green lines.
• The area hatched using green lines is the graph of the inequality 20x+8y < 120
• Note that:
   ♦ Graph of an "inequality in one variable" is a line.
   ♦ Graph of an "inequality in two variables" is not a line, but an area.
(iv) On a sheet of paper or on the computer screen, we can hatch only a limited area.
◼ But the actual hatched area extends up to infinity.
• The red line goes upwards up to (-∞,∞)
   ♦ So the left boundary of the hatched area is -∞
   ♦ Also the top boundary of the hatched area is ∞
• The red line goes downwards up to (∞,-∞)
   ♦ So the right boundary of the hatched area is ∞
   ♦ Also the bottom boundary of the hatched area is -∞
(v) Now let us consider the fact that the sign of our present inequality is '≤' not '<'. For simplicity we ignored '≤' in step (i)
• Since the sign to be considered is '≤', the points on the red line representing 20x + 8y = 120 are also acceptable.
• So we draw the line of the equation using a bold red line.
• If the sign is '<', we would draw the line of the equation using a dashed red line.
◼ The rule is:
   ♦ Bold line for '≤' and '≥'
   ♦ Dashed line for '<' and '>'
17. So now we have a basic idea about how to draw the graph of an inequality.
• In the above step (16), we saw that,
   ♦ all points in the hatched area
   ♦ and all point in the red line
   ♦ are solutions of the inequality 20x+8y ≤ 120
For example, (-√2,1) is a point in the hatched area.
• If we input (-√2,1) in the inequality, we will get:
   ♦ (20 × -√2) + (8 × 1) < 120
   ♦ ⇒ (20 × -1.414) + (8 × 1) < 120
   ♦ ⇒ -28.284 + 8 < -20.284, which is true
• For our particular problem, we need to make an adjustment. It can be written in 5 steps:
(i) For our particular problem, x and y are to be whole numbers. Because they represent the numbers of books and pens that are purchased.
(ii) So we have to:
   ♦ pick out those points from the hatched area
   ♦ in such a way that,
   ♦ both x and y coordinates of those points are whole numbers.
• Such points are shown as magenta dots in fig.6.11 below:

Fig.6.11

(iii) All points in the graph of 20x+8y = 120 are not formed by x and y whole numbers. So we cannot draw it as a bold line. We must draw it as a dashed line.
(iv) All magenta dots in fig.6.11 will satisfy the inequality 20x+8y ≤ 120
For example, consider (3,6)
• If we input (0,0) in the inequality, we will get:
   ♦ (20 × 3) + (8 × 6) ≤ 120
   ♦ ⇒ 60 + 24 ≤ 120
   ♦ ⇒ 84 ≤ 120, which is true.
(v) Only five points on the dashed red line satisfy the inequality.
They are: (6,0), (4,5), (2,10) and (0,15)
18. Based on the above step (17), we can write:
• If x and y are real numbers, graph of the inequality can be drawn by hatching the entire area of the appropriate half plane.
• If x and y are integers or whole numbers, graph of the inequality can be drawn by putting dots at the appropriate points.

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• In the next section, we will see a few more examples.

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Chapter 6.2 - Solved Examples on Linear Inequalities in One Variable

In the previous section, we saw the rules for solving linear inequalities in one variable. We saw some solved examples also. In this section, we will see a few more solved examples.

Solved example 6.5
Solve 7x+3 < 5x+9
Solution:
• Given inequality is: 7x+3 < 5x+9
• This can be simplified as follows:
$\begin{array}{ll}
{}&7x+3 &{}<{}& {5x+9} &{} \\
\Rightarrow &7x+3-3&{}<{}& 5x+9-3 &\color {green}{\text{(Rule 1)}} \\
\Rightarrow &7x&{}<{}& 5x+6 &{} \\
\Rightarrow &7x-5x&{}<{}& 5x+6-5x &\color {green}{\text{(Rule 1)}} \\
\Rightarrow &2x&{}<{}& 6 &{} \\
\Rightarrow &\frac{2x}{2}&{}<{}& \frac{6}{2} &\color {green}{\text{(Rule 2)}} \\
\Rightarrow &x&{}<{}& 3 &{} \\
\end{array}$
• So all real numbers less than 3 are the solutions of this inequality.
• The solution set is: (-∞,3)
• The graph is shown in fig.6.4 below:

Fig.6.4

Solved example 6.6
Solve $\frac{3x-4}{2}\ge\frac{x+1}{4} -1$
Solution:
• Given inequality is: $\frac{3x-4}{2}\ge\frac{x+1}{4} -1$
• This can be simplified as follows:
$\begin{array}{ll}
{}&\frac{3x-4}{2}&{}\ge{}& \frac{x+1}{4} -1 &{} \\
\Rightarrow &4 \left(\frac{3x-4}{2}\right)&{}\ge{}& 4\left(\frac{x+1}{4} -1\right) &\color {green}{\text{(Rule 2)}} \\
\Rightarrow &2 \left(3x-4 \right)&{}\ge{}& x+1-4 &{} \\
\Rightarrow &6x-8&{}\ge{}& x-3 &{} \\
\Rightarrow &6x-8+8&{}\ge{}& x-3+8 &\color {green}{\text{(Rule 1)}} \\
\Rightarrow &6x&{}\ge{}& x+5 &{} \\
\Rightarrow &6x-x&{}\ge{}& x+5-x &\color {green}{\text{(Rule 1)}} \\
\Rightarrow &5x&{}\ge{}& 5 &{} \\
\Rightarrow &x&{}\ge{}& 1 &{} \\
\end{array}$
• So all real numbers greater than or equal to 1 are the solutions of this inequality.
• The solution set is: [1,∞)
• The graph is shown in fig.6.5 below:

Fig.6.5

Solved example 6.7
The marks obtained by a student of class XI in first and second terminal examination are 62 and 48 respectively. Find the minimum marks he should get in the annual examination to have an average of at least 60 marks.
Solution:
1. Let x be the marks in the annual examination. Then we can write:
$\frac{62+48+x}{3} \ge 60$
• Given inequality is: $\frac{3x-4}{2}\ge\frac{x+1}{4} -1$
• This can be simplified as follows:
$\begin{array}{ll}
{}&\frac{62+48+x}{3}&{}\ge{}&60 &{} \\
\Rightarrow &3 \left(\frac{62+48+x}{3}\right)&{}\ge{}& 3 × 60 &\color {green}{\text{(Rule 2)}} \\
\Rightarrow &62+48+x&{}\ge{}& 180 &{} \\
\Rightarrow &110+x&{}\ge{}& 180 &{} \\
\Rightarrow &110+x-110&{}\ge{}& 180-110 &\color {green}{\text{(Rule 1)}} \\
\Rightarrow &x&{}\ge{}& 70 &{} \\
\end{array}$
• So the student must obtain at least 70 marks.
• If the maximum marks is 100, the solution set will be: [70,100]
• The graph is shown in fig.6.6 below:

Fig.6.6

Solved example 6.8
Find all pairs of consecutive odd natural numbers, both of which are larger than 10, such that their sum is less than 40.
Solution:
1. The consecutive odd numbers starting from 1 are: 1, 3, 5, 7, . . .
2. The consecutive odd numbers greater than 10 are: 11, 13, 15, 17, . . .
• We have to pick out pairs from this list.
• Some of the possible pairs are: (11,13), (13,15) etc.,
• The sum of the two members of any pair that we pick, must be less than 40
3. If x is the first member of a pair, the second member will be (x+2)
• So we can write: x+(x+2) < 40
4. Given that both members of the pairs are to be greater than 10.
• So the smaller one which is x, must be larger than 10. This will ensure that, the larger one which is (x+2) will also be greater than 10.
• We can write: x > 10
5. From step (3) we get: 2x +2 < 40
• This can be simplified as follows:
$\begin{array}{ll}
{}&2x+2 &{}<{}& 40 &{} \\
\Rightarrow &2x+2-2&{}<{}& 40-2 &\color {green}{\text{(Rule 1)}} \\
\Rightarrow &2x&{}<{}& 38 &{} \\
\Rightarrow &\frac{2x}{2}&{}<{}& \frac{38}{2} &\color {green}{\text{(Rule 2)}} \\
\Rightarrow &x&{}<{}& 19 &{} \\
\end{array}$
6. So if the sum is to be less than 40, x must be less than 19
• Also from step (4), we have: x must be greater than 10
• The consecutive odd natural numbers satisfying both the above conditions are:
11, 13, 15 and 17
• So the possible pairs are:
(11, 11+2), (13, 13+2), (15, 15+2), (17, 17+2)
• That is: (11,13), (13,15), (15,17) and (17,19)

Solved example 6.9
Find all pairs of consecutive odd natural numbers, both of which are smaller than 18, such that their sum is more than 20.
Solution:
1. The consecutive odd natural numbers starting from 1 and smaller than 18 are:
1, 3, 5, 7, . . . , 17
2. Let x and x+2 form the pair.
• Given that both of them must be smaller than 18.
• So the larger one which is (x+2), must be smaller than 18. This will ensure that, the smaller one x will also be smaller than 18
• Thus we can write: x+2 < 18
• This can be simplified as follows:
$\begin{array}{ll}
{}&x+2 &{}<{}& 18 &{} \\
\Rightarrow &x+2-2&{}<{}& 18-2 &\color {green}{\text{(Rule 1)}} \\
\Rightarrow &x&{}<{}& 16 &{} \\
\end{array}$
• So first member of all pairs must be less than 16
3. Given that, the sum must be larger than 20.
So we can write: x+(x+2) > 20
• This can be simplified as follows:
$\begin{array}{ll}
{}&2x+2 &{}>{}& 20 &{} \\
\Rightarrow &2x+2-2&{}<>{}& 20-2 &\color {green}{\text{(Rule 1)}} \\
\Rightarrow &2x&{}>{}& 18 &{} \\
\Rightarrow &\frac{2x}{2}&{}>{}& \frac{18}{2} &\color {green}{\text{(Rule 2)}} \\
\Rightarrow&x&{}>{}& 9 &{} \\
\end{array}$
4. So if the sum is to be greater than 40, x must be greater than 9
• Also from step (2), we have: x must be less than 16
• The consecutive odd natural numbers satisfying both the above conditions are:
11, 13, and 15
• So the possible pairs are:
(11, 11+2), (13, 13+2) and (15, 15+2)
• That is: (11,13), (13,15) and (15,17)

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The link below gives some more solved examples:

Exercise 6.1

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• In the next section, we will see linear inequalities in two variables.

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Chapter 6.1 - Linear Inequalities in One Variable

In the previous section, we saw the disadvantages of using trial and error method for solving inequalities. In this section, we will see a systematic method.

For developing a systematic method, we must first learn some basic properties of inequalities. We can use those basic properties as rules for solving inequalities. This can be explained in 7 steps:

1. Let us recall the two rules that we used while solving linear equations:
First rule:
• Any number can be added to any side of the equation. The same number should be added to the other side also.
• Any number can be subtracted from any side of the equation. The same number should be subtracted from the other side also.
Second rule:
• Any side of the equation can be multiplied by a non-zero number. The other side also should be multiplied by the same number.  
• Any side of the equation can be divided by a non-zero number. The other side also should be divided by the same number.
2. In the same way, three rules can be developed for inequalities.
• If we add the same number to both sides of the inequality, there will be no change for the sign.
    ♦ For example, consider the inequality 2 < 8.
    ♦ Let us add 7 on both sides. We get 9 < 15.
    ♦ The ‘<’ sign has not changed.   
• If we subtract the same number from both sides of the inequality, there will be no change for the sign.
    ♦ For example, consider the inequality 12 < 21.
    ♦ Let us subtract 4 from both sides. We get 8 < 17.
    ♦ The ‘<’ sign has not changed.
3. So we can write the first rule for solving inequalities.
Rule 1 for solving inequalities:
• Any number can be added to any side of the inequality. The same number should be added to the other side also. Then the sign will not change.
• Any number can be subtracted from any side of the inequality. The same number should be subtracted from the other side also. Then the sign will not change.
4. If we multiply both sides of an inequality by the same +ve number, there will be no change for the sign.
    ♦ For example, consider the inequality 2 < 8.
    ♦ Let us multiply both sides by 3. We get 6 < 24.
    ♦ The ‘<’ sign has not changed.
• If we divide both sides of an inequality by the same +ve number, there will be no change for the sign.
    ♦ For example, consider the inequality 9 < 12.
    ♦ Let us divide both sides by 4. We get 2.25 < 3.
    ♦ The ‘<’ sign has not changed.
5. So we can write the second rule for solving inequalities.
Rule 2 for solving inequalities:
• Any side of the inequality can be multiplied by a +ve number. The other side also should be multiplied by the same number. The sign will not change.  
• Any side of the inequality can be divided by a +ve number. The other side also should be divided by the same number. The sign will not change.
6. If we multiply both sides of an inequality by the same -ve number, the sign will be reversed.
    ♦ For example, consider the inequality 2 < 8.
    ♦ Let us multiply both sides by -3. We get -6 > -24.
    ♦ The ‘<’ sign has become ‘>’.
• If we divide both sides of an inequality by the same -ve number, the sign will be reversed.
    ♦ For example, consider the inequality 9 < 12.
    ♦ Let us divide both sides by -4. We get -2.25 > -3.
    ♦ The ‘<’ sign has become ‘>'.
7. So we can write the third rule for solving inequalities.
Rule 3 for solving inequalities:
• Any side of the inequality can be multiplied by a -ve number. The other side also should be multiplied by the same number. The sign will be reversed.  
• Any side of the inequality can be divided by a -ve number. The other side also should be divided by the same number. The sign will be reversed.
• Reversal of sign means:
    ♦ < will become >      
    ♦ > will become <
    ♦ ≤ will become ≥     
    ♦ ≥ will become ≤

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Now we will see some solved examples

Solved example 6.1
Solve 30x < 200 when (i) x is a natural number  (ii) x is an integer
Solution:
• Given inequality is: 30x < 200
• Applying Rule 2, we get:
$\frac{30x}{30}~<~\frac{200}{30}$
$\Rightarrow x~<~\frac{20}{3}$
$\Rightarrow~x~<~6.667$
• So whichever value we select for x, must be less than 6.667
Part (i):
• The set of natural number is {1, 2, 3, 4, . . .}
• We must select the appropriate values from this set. The solution set will contain those appropriate values.
• So the solution set is {1, 2, 3, 4, 5, 6}
Part (ii):
• The set of integers is {. . . , -4, -3, -2, -1, 0, 1, 2, 3, . . .}
• We must select the appropriate values from this set. The solution set will contain those appropriate values.
• So the solution set is {. . . , -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6}

---

Why do we specifically say natural numbers, integers etc.,?
The answer can be written in 2 steps:
1. In some problems, x may represent number of pens, number of books, number of cars etc., In such cases, x must be a natural number (or whole number).
2. In some problems, x may represent the floor level of a building.
    ♦ The floor immediately below the ground level is indicated by -1
    ♦ The floor below that is indicated by -2 and so on . . .
• In such cases, x must be an integer.

---

Solved example 6.2
Solve 5x-3 < 3x+1 when (i) x is an integer  (ii) x is a real number
Solution:
• Given inequality is: 5x-3 < 3x+1
• This can be simplified as follows:
$\begin{array}{ll}
{}&5x-3 &{}<{}& {3x+1} &{} \\
\Rightarrow &5x-3+3&{}<{}& 3x+1+3 &\color {green}{\text{(Rule 1)}} \\
\Rightarrow &5x&{}<{}& 3x+4 &{} \\
\Rightarrow &5x-3x&{}<{}& 3x+4-3x &\color {green}{\text{(Rule 1)}} \\
\Rightarrow &2x&{}<{}& 4 &{} \\
\Rightarrow &\frac{2x}{2}&{}<{}& \frac{4}{2} &\color {green}{\text{(Rule 2)}} \\
\Rightarrow &x&{}<{}& 2 &{} \\
\end{array}$

Part (i):
• The set of integers is {. . . , -4, -3, -2, -1, 0, 1, 2, 3, . . .}
• We must select the appropriate values from this set. The solution set will contain those appropriate values.
• So the solution set is {. . . , -4, -3, -2, -1, 0, 1}
Part (ii):
• We know that the set of real numbers contain all types of numbers.
    ♦ They include natural numbers, whole numbers, integers
    ♦ They also include rational numbers and irrational numbers.
    ♦ So there will be numbers like -√3, -√2, √3, √2, -π, π etc.,
• Since they do not occur at regular intervals, we pick out the "relevant portion of the number line" and write it as an interval.
• In our present case, the interval will be: (-∞,2)
• That means, the relevant portion on the number line in our case, begins from -∞ at the extreme left end. It extends upto +2.
• But the exact +2 should not be included. Numbers like 1.999, 1.9999 etc., are allowed. This is indicated by the ')' on the right side of 2.
• This interval can be shown graphically as in fig.6.1 below:

Fig.6.1

• The yellow line represents the number line. The red line is the graph.
• The arrow at the left end of the red line indicates that, the left boundary of the interval is at -∞.
• The circle at the right end of the red line indicates that, the right boundary of the interval is at 2.
   ♦ Since 2 is not included in the interval, it is an ordinary circle.
   ♦ If 2 is also included, we give a filled circle.
• All points on the red line is a solution of the inequality. We can write:
When x is a real number, the solution set is: (-∞,2)

---

Why do we specifically say integers, real numbers etc.,?
The answer can be written in 2 steps:
1. In solved example 6.1 above, we saw the situations where integers are specified. So we need not discuss about it again.
2. In some problems, x may represent lengths or distances. Such quantities are not always available as integers.
• We may encounter lengths like 3.5 cm, 121.667 meter etc.,
• Some times distances towards the left are considered -ve and those towards the right are considered +ve
   ♦ Then we may encounter lengths like -3.5 cm, -121.667 meter etc.,
• Also it is possible to encounter lengths like -2√3 meter, 5√2 meter, -3π cm etc.,
• In some problems, x may represent temperature.
• In some problems, x may represent volume.
• In all such cases, x must be a real number.

---

We have seen the significance of specifying whether x is natural number, integer or real number. For the rest of our discussion in this chapter, the variables x, y etc., will be considered as representing real numbers.

---

Solved example 6.3
Solve 4x+3 < 6x+7
Solution:
• Given inequality is: 4x+3 < 6x+7
• This can be simplified as follows:
$\begin{array}{ll}
{}&4x+3 &{}<{}& {6x+7} &{} \\
\Rightarrow &4x+3-3&{}<{}& 6x+7-3 &\color {green}{\text{(Rule 1)}} \\
\Rightarrow &4x&{}<{}& 6x+4 &{} \\
\Rightarrow &4x-6x&{}<{}& 6x+4-6x &\color {green}{\text{(Rule 1)}} \\
\Rightarrow &-2x&{}<{}& 4 &{} \\
\Rightarrow &\frac{-2x}{-2}&{}<{}& \frac{4}{-2} &\color {green}{\text{(Rule 3)}} \\
\Rightarrow &x&{}>{}& -2 &{} \\
\end{array}$
• So all real numbers greater than -2 are the solutions of this inequality.
• The solution set is: (-2,∞)
• The graph is shown in fig.6.2 below:

Fig.6.2

Solved example 6.4
Solve $\frac{5-2x}{3}\le\frac{x}{6} -5$
Solution:
• Given inequality is: $\frac{5-2x}{3}\le\frac{x}{6} -5$
• This can be simplified as follows:
$\begin{array}{ll}
{}&\frac{5-2x}{3}&{}\le{}& \frac{x}{6} -5 &{} \\
\Rightarrow &6 \left(\frac{5-2x}{3}\right)&{}\le{}& 6\left(\frac{x}{6} -5\right) &\color {green}{\text{(Rule 2)}} \\
\Rightarrow &2 \left(5-2x \right)&{}\le{}& x-30 &{} \\
\Rightarrow &10-4x&{}\le{}& x-30 &{} \\
\Rightarrow &10-4x+4x&{}\le{}& x-30+4x &\color {green}{\text{(Rule 1)}} \\
\Rightarrow &10&{}\le{}& 5x-30 &{} \\
\Rightarrow &10+30&{}\le{}& 5x-30+30 &\color {green}{\text{(Rule 1)}} \\
\Rightarrow &40&{}\le{}& 5x &{} \\
\Rightarrow &\frac{40}{5}&{}\le{}& \frac{5x}{5} &\color {green}{\text{(Rule 2)}} \\
\Rightarrow &8&{}\le{}& x &{} \\
\Rightarrow &x&{}\ge{}& 8 &{} \\
\end{array}$
• So all real numbers greater than or equal to 8 are the solutions of this inequality.
• The solution set is: [8,∞)
• The graph is shown in fig.6.3 below:

Fig.6.3

---

• In the next section, we will see a few more solved examples.

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Chapter 5 - Complex Numbers and Quadratic Equations

In the previous section, we completed a discussion on mathematical induction. In this chapter, we will see complex numbers and quadratic equations.

Let us recall the three types of equations that we have seen in our earlier classes.
(i) Linear equations in one variable
Example:
2x+5 = 17
• We know how to solve such equations. (Details here)
(ii) Linear equations in two variables
Example:
4x+3y = 43
3x - 2y = 11
• We know how to solve such equations. (Details here)  
(iii) Quadratic equations in one variable
Example: x² + 2x – 224 = 0
• We know how to solve such equations. (Details here)

---

• Let us consider the quadratic equations again.
We know that, the solutions of a quadratic equation can be found out by using the equation: $\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
• Consider the portion $b^2 - 4ac$.
If this portion is -ve, we will not be able to calculate $\sqrt{b^2 - 4ac}$.
Because, root of -ve numbers do not exist.
(Mark any point on the number line. Even if that number is -ve, it’s square will be a +ve number. That is why we say that, root of a -ve number is not a real number)
• So in this chapter we try an alternate method to solve quadratic equations, when $b^2 - 4ac$ is -ve.

Significance of i

Some basics about $i$ can be written in 6 steps:
1. Consider the quadratic equation: $x^2 + 1 = 0$   
• Rearranging it, we get: $x^2 = -1$
• So $x = \pm \sqrt{-1}$
2. We know that $\sqrt{-1}$ is not a real number.
• Let us denote $\sqrt{-1}$ by the symbol $i$
• Then we get: $x = \pm i$
• That is: x = $i$ or $-i$
(i) Let us substitute $i$ in the place of x in the given equation. We get: $i^2 +1 = 0$
   ♦ That is., $(\sqrt{-1})^2 + 1 = 0 ~\Rightarrow~ -1 + 1 = 0$, which is true.
   ♦ So we can write: $i$ is a solution of the equation $x^2 + 1 = 0$  
(ii) Let us substitute $-i$ in the place of x in the given equation. We get: $(-i)^2 +1 = 0$
   ♦ That is., $(-\sqrt{-1})^2 + 1 = 0 ~\Rightarrow~ [(-1 × -1) × -1] + 1 = 0$
   ♦ That is.,$[(1) × -1]+1 = 0~\Rightarrow~ -1+1 =0$, which is true.
   ♦ So we can write: $-i$ is also a solution of the equation $x^2 + 1 = 0$  
(iii) Thus we get:
   ♦ Solutions of the equation $x^2 + 1 = 0$ are $\sqrt{-1}$ and $-\sqrt{-1}$
   ♦ In other words:
   ♦ Solutions of the equation $x^2 + 1 = 0$ are $i$ and $-i$

• While solving the equation $x^2 + 1 = 0$, we are actually calculating the square roots of -1. This is clear from the fact that, the equation is rearranged into the form: $x^2 = -1$ • From the detailed steps that we wrote, we come to the conclusion that: $\sqrt{-1}$ can be $i$  or $-i$ • However, when we write $\sqrt{-1}$, we would mean $i$ only.    ♦ If we want $-i$, we must specifically write $-\sqrt{-1}$

3. We denoted $\sqrt{-1}$ by the symbol $i$. What about $\sqrt{-2}$ ?
(i) Consider the quadratic equation: $x^2 + 2 = 0$   
• Rearranging it, we get: $x^2 = -2$
• So $x = \pm \sqrt{-2}$
(ii) We can write $\pm \sqrt{-2}$ as:
$\pm \sqrt{-1 \times 2}~=~\pm \sqrt{-1} \times \sqrt{2}~=~ \pm i \times \sqrt{2}$
• So we get: $\sqrt{-2}~=~\pm \sqrt{2}\,i$  
(iii) Let us substitute $\sqrt{2}\,i$ in the place of x in the given equation. We get: $(\sqrt{2}\,i)^2 +2 = 0$
   ♦ That is., $[(\sqrt{2})^2~ × ~i^2] + 2 = 0$
   ♦ That is., $[(\sqrt{2})^2~ × ~(\sqrt{-1})^2] + 2 = 0 ~\Rightarrow~ [2 × -1] + 2 = 0$, which is true.
   ♦ So we can write: $\sqrt{2}\;i$ is a solution of the equation $x^2 + 2 = 0$  
(iv) Let us substitute $-\sqrt{2}\,i$ in the place of x in the given equation. We get:
$(-\sqrt{2}\,i)^2 +2 = 0$
   ♦ That is., $[(-1 × -1) × ~ (\sqrt{2})^2~ × ~i^2] + 2 = 0$
   ♦ That is., $[(-1 × -1) × ~ (\sqrt{2})^2 × ~(\sqrt{-1})^2] + 2 = 0$
   ♦ That is., $[(1) × 2 × -1] + 2 = 0~\Rightarrow~[-2]+2=0$, which is true.
   ♦ So we can write: $-\sqrt{2}\;i$ is also a solution of the equation $x^2 + 2 = 0$
(v) Thus we get:
   ♦ Solutions of the equation $x^2 + 2 = 0$ are $\sqrt{-2}$ and $-\sqrt{-2}$
   ♦ In other words:
   ♦ Solutions of the equation $x^2 + 2 = 0$ are $\sqrt{2}\;i$ and $-\sqrt{2}\;i$

• While solving the equation $x^2 + 2 = 0$, we are actually calculating the square roots of -2. This is clear from the fact that, the equation is rearranged into the form: $x^2 = -2$ • From the detailed steps that we wrote, we come to the conclusion that: $\sqrt{-2}$ can be $\sqrt{2}\,i$  or $-\sqrt{2}\,i$ • However, when we write $\sqrt{-2}$, we would mean $\sqrt{2}\,i$ only.    ♦ If we want $-\sqrt{2}\,i$, we must specifically write $-\sqrt{-2}$

4. We denoted $\sqrt{-1}$ by the symbol $i$. What about $\sqrt{-3}$ ?
(i) Consider the quadratic equation: $x^2 + 3 = 0$   
• Rearranging it, we get: $x^2 = -3$
• So $x = \pm \sqrt{-3}$
(ii) We can write $\sqrt{-3}$ as:
$\pm \sqrt{-1 \times 3}~=~\pm \sqrt{-1} \times \sqrt{3}~=~ \pm i \times \sqrt{3}$
• So we get: $\sqrt{-3}~=~\pm \sqrt{3} \,i$  
(iii) Let us substitute $\sqrt{3}\,i$ in the place of x in the given equation. We get: $(\sqrt{3}\;i)^2 +3 = 0$
   ♦ That is., $[(\sqrt{3})^2~ × ~i^2] + 3 = 0$
   ♦ That is., $[(\sqrt{3})^2~ × ~(\sqrt{-1})^2] + 3 = 0 ~\Rightarrow~ [3 × -1] + 3 = 0$, which is true.
   ♦ So we can write: $\sqrt{3}\,i$ is a solution of the equation $x^2 + 3 = 0$  
(iv) Let us substitute $-\sqrt{3}\,i$ in the place of x in the given equation. We get:
$(-\sqrt{3}\,i)^2 +3 = 0$
   ♦ That is., $[(-1 × -1) × ~ (\sqrt{3})^2~ × ~i^2] + 3 = 0$
   ♦ That is., $[(-1 × -1) × (\sqrt{3})^2 × (\sqrt{-1})^2] + 3 = 0$
   ♦ That is., $[(1) × 3 × -1] + 3 = 0~\Rightarrow~[-3]+3=0$, which is true.
   ♦ So we can write: $-\sqrt{3}\,i$ is also a solution of the equation $x^2 + 3 = 0$
(v) Thus we get:
   ♦ Solutions of the equation $x^2 + 3 = 0$ are $\sqrt{-3}$ and $-\sqrt{-3}$
   ♦ In other words:
   ♦ Solutions of the equation $x^2 + 3 = 0$ are $\sqrt{3}\,i$ and $-\sqrt{3}\,i$

• While solving the equation $x^2 + 3 = 0$, we are actually calculating the square roots of -3. This is clear from the fact that, the equation is rearranged into the form: $x^2 = -3$ • From the detailed steps that we wrote, we come to the conclusion that: $\sqrt{-3}$ can be $\sqrt{3}\,i$  or $-\sqrt{3}\,i$ • However, when we write $\sqrt{-3}$, we would mean $\sqrt{3}\,i$ only.    ♦ If we want $-\sqrt{3}\,i$, we must specifically write $-\sqrt{-3}$

5. We denoted $\sqrt{-1}$ by the symbol $i$. What about $\sqrt{-4}$ ?
(i) Consider the quadratic equation: $x^2 + 4 = 0$   
• Rearranging it, we get: $x^2 = -4$
• So $x = \pm \sqrt{-4}$
(ii) We can write $\pm \sqrt{-4}$ as:
$\pm \sqrt{-1 \times 4}~=~\pm \sqrt{-1} \times \sqrt{4}~=~ \pm i \times \sqrt{4}$
• So we get: $\sqrt{-4}~=~\pm \sqrt{4}\,i~=~\pm 2i$  
(iii) Let us substitute $2i$ in the place of x in the given equation. We get: $(2i)^2 +4 = 0$
   ♦ That is., $[(2)^2~ × ~i^2] + 4 = 0$
   ♦ That is., $[(2)^2~ × ~(\sqrt{-1})^2] + 4 = 0 ~\Rightarrow~ [4 × -1] + 4 = 0$, which is true.
   ♦ So we can write: $2i$ is a solution of the equation $x^2 + 4 = 0$  
(iv) Let us substitute $-2i$ in the place of x in the given equation. We get:
$(-2i)^2 +4 = 0$
   ♦ That is., $[(-1 × -1) × ~ (2)^2~ × ~i^2] + 4 = 0$
   ♦ That is., $[(-1 × -1) × (2)^2~ × ~(\sqrt{-1})^2] + 4 = 0$
   ♦ That is., $[(1) × 4 × -1] + 4 = 0~\Rightarrow~[-4]+4=0$, which is true.
   ♦ So we can write: $-2i$ is also a solution of the equation $x^2 + 4 = 0$
(v) Thus we get:
   ♦ Solutions of the equation $x^2 + 4 = 0$ are $\sqrt{-4}$ and $-\sqrt{-4}$
   ♦ In other words:
   ♦ Solutions of the equation $x^2 + 4 = 0$ are $2i$ and $-2i$

• While solving the equation $x^2 + 4 = 0$, we are actually calculating the square roots of -4. This is clear from the fact that, the equation is rearranged into the form: $x^2 = -4$ • From the detailed steps that we wrote, we come to the conclusion that: $\sqrt{-4}$ can be $2i$  or $-2i$ • However, when we write $\sqrt{-4}$, we would mean $2i$ only.    ♦ If we want $-2i$, we must specifically write $-\sqrt{-4}$

6. In this way, we can write the square roots of -5, -6, -7, . . . so on.
• In fact, we can write the square roots of any -ve real number.
• Generally, if '$a$' is a positive real number, $\sqrt{-a}=\sqrt{a} × \sqrt{-1}=\sqrt{a}\,i$

---

• So now we know the significance of $i$.
• Using $i$, we can do problems involving the square roots of -ve real numbers. We will see those problems in later sections.
• In the next section, we will see complex numbers.

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Chapter 3.9 - Domain and Range of Cosine and Secant Functions

In the previous section, we saw the domain and range of sine and cosecant functions. In this section, we will see the domain and range of cosine and secant functions.

Let us find the domain and range of f(x) = cos x. It can be written in 2 steps:
1. First we will find the domain:
• We have seen that:
In the case of f(x) = cos x, any real number can be used as input x.
• So we can write: Domain of f(x) = cos x, is the set R.
(Recall that, R is the set of real numbers)
2. Let us find the range of f(x) = cos x
• The range in this case can be better understood if we analyze the graph of f(x) = cos x. It is shown in fig.3.29 below:

Fig.3.29

• The graph of f(x) = sin x, which we saw in the previous section, is given again below. This is for a comparison.

Fig.3.26

• An analysis of the cosine graph can be written in 3 steps:
(i) We already know the reason for the markings $\frac{\pi}{2},\; \pi,\; \frac{3\pi}{2},\; 2\pi,\; \frac{5\pi}{2}$ so on . . . on the x axis.
(ii) Now let us consider the output y values.
• We see that, whatever be the value of input x,
   ♦ Output y value (that is., cosine of x) never becomes greater than 1
   ♦ Output y value never becomes lesser than -1
(iii) We can mark any point on the red curve in the graph. The y coordinate of that point will be either 1 or -1 or a value between 1 and -1.
◼ So the range can be written as [-1, 1]
• That means:
   ♦ All real values between -1 and 1 are included in the range.
   ♦ Left side '[' indicates that -1 is also included in the range.
   ♦ Right side ']' indicates that +1 is also included in the range.

---

• We just saw that, cosine value can never rise above -1. Neither can it fall below -1.
• The reason can be written in 8 steps:
1. In the animation in fig.3.30 below, the thick white horizontal line is the base of the triangle in our familiar unit circle.

Fig.3.30

• We know that, this base is related to the cosine value of the angle.
2. The red ray starts to rotate from the +ve side of the x axis.
• Since the green circle is a unit circle, the length of the horizontal white line at this initial point will be 1
• As the rotation proceeds, the length of the horizontal line gradually decreases.
• It becomes zero when the ray completes a rotation of ${\frac{\pi}{2}}^c$. That is, when the ray coincides with the y axis.
• So we can write:
(i) The cosine value start to decrease from 1 (when the ray coincides with the +ve side of x axis)
(ii) The cosine value attains a value of 0 (when the ray coincides with the +ve side of y axis)
(iii) This decrease of cosine value, from 1 to zero, is indicated by the falling portion between x = 0 and x = $\frac{\pi}{2}$ in the graph in fig.3.29 above.
3. Next, the red ray proceeds to rotate from x = $\frac{\pi}{2}$ to x = π.
• We see that, the length of the horizontal white line increases from zero to 1.
• Though it is an 'increase in length', it happens on the negative side of the x axis. On this negative side, every x coordinate is negative, That means, on the negative side of the x axis, every cosine value is negative.
• As the length increases, the negative value increases. So in effect, it is a decrease.
• This decrease is indicated by the falling portion between x = $\frac{\pi}{2}$ and x = π in the graph in fig.3.29 above.
4. Next, the red ray proceeds to rotate from x = π to x = $\frac{3\pi}{2}$.
• We see that, the length of the horizontal white line decreases from 1 to zero.
• Though it is a 'decrease in length', it happens on the negative side of the x axis. On this negative side, every x values is negative, That means, on the negative side of the x axis, every cosine value is negative.
• As the length decreases, the negative value decreases. So in effect, it is an increase.
• This increase is indicated by the rising portion between x = π and x = $\frac{3\pi}{2}$ in the graph in fig.3.29 above.
5. Next, the red ray proceeds to rotate from x = $\frac{3\pi}{2}$ to x = 2π.
• We see that, the length of the horizontal white line increases from zero to 1.
• This increase is indicated by the rising portion between x = $\frac{3\pi}{2}$ and x = 2π in the graph in fig.3.29 above.
6. As the ray continues to rotate, this pattern repeats again and again. Thus we get the wave form on the positive side of the x axis in the graph.
7. Similar steps can be written for rotation in the anticlockwise direction also. Based on those steps, we will be able to explain the wave form on the negative side of the x axis in the graph.
8. Note that:
• In the first ${\frac{\pi}{2}}^c$ rotation in the anticlockwise direction, there is a fall in cosine value.
• In the first ${\frac{\pi}{2}}^c$ rotation in the clockwise direction also, there is a fall in cosine value.
• So there is a smooth transition between the two waves on either sides of the x axis.

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Let us find the domain and range of f(x) = sec x. It can be written in 4 steps:
1. First we will find the domain:
• We have seen that:
    ♦ In the case of f(x) = sec x,
    ♦ $\frac{(2n+1)\pi}{2}$ where n is any integer, should not be used as input x
• So we can write:
Domain of f(x) = sec x is $R-\{x:x=\frac{(2n+1)\pi}{2},\; n\, \in \,Z\}$
• That means, we must subtract the set $\{x:x=\frac{(2n+1)\pi}{2},\; n\, \in \,Z\}$ from R. The resulting set after subtraction, is the domain of f(x) = sec x
• $\{x:x=\frac{(2n+1)\pi}{2},\; n\, \in \,Z\}$ is the set containing all $\frac{(2n+1)\pi}{2}$, where n is any integer.
2. Let us find the range of f(x) = sec x
• The range in this case can be better understood if we analyze the graph of f(x) = cos x and f(x) = sec x together. It is shown in fig.3.31 below:

Fig.3.31

An analysis of this graph can be written in 6 steps:
(i) Consider the segment from 0 to $\frac{\pi}{2}$ on the x axis.
• In this segment,
   ♦ the cosine curve falls
   ♦ but the secant curve rises.
• That means, when the angle increases from 0 to $\frac{\pi}{2}$,
   ♦ the cosine value decreases
   ♦ but the secant value increases.
• This is obvious because, secant is the reciprocal of cosine.
• In the segment from 0 to $\frac{\pi}{2}$, the maximum value of cosine occurs at zero
   ♦ So the minimum value of secant should also be at zero. 
   ♦ Indeed we see that, the rise of the secant curve occurs after zero.
• The value of cosine at zero is 1
   ♦ The value of secant at zero should be the reciprocal of 1, which is 1.
   ♦ Indeed we see that, the value of secant at zero is 1.
• The value of cosine at $\frac{\pi}{2}$ is 0
   ♦ The value of secant at $\frac{\pi}{2}$ should be the reciprocal of 0, which is not defined
   ♦ Indeed we see that, the value of secant curve never touches the vertical line through $\frac{\pi}{2}$
• As the angle approaches $\frac{\pi}{2}$, the cosine value becomes very small and consequently the secant which is the reciprocal, becomes very large.

◼ As seen before, we cannot put x = $\frac{\pi}{2}$ in $f(x) = \sec x = \frac{1}{\cos x}$. We can write: • As x approaches $\frac{\pi}{2}$, cos x becomes smaller and smaller, getting closer and closer to zero. (For example, values like 0.00001, 0.000001 are very close to zero) • As cos x becomes smaller and smaller, the reciprocal sec x becomes larger and larger. • This is indicated by the rising portion of the secant curve between 0 and $\frac{\pi}{2}$. • As angle becomes closer and closer to $\frac{\pi}{2}$, this rising portion gets closer and closer to the vertical line through $\frac{\pi}{2}$. (why does this happen? The reader may write the answer in his/her own notebooks) • But it never touches that vertical line. • If it touch, it would mean that, x = $\frac{\pi}{2}$ is a point in the secant curve. We know that, it cannot happen.

(ii) Consider the segment from $\frac{\pi}{2}$ to π on the x axis.
• In this segment,
   ♦ the cosine curve falls
   ♦ but the secant curve rises.
• That means, when the angle increases from $\frac{\pi}{2}$ to π,
   ♦ the cosine value decreases (increases negatively)
   ♦ but the secant value increases (decreases negatively).
• This is obvious because, secant is the reciprocal of cosine
Note that here, all the cosine values are -ve. Consequently, all the secant values will also be -ve.
• In the segment from $\frac{\pi}{2}$ to π, the minimum value of cosine occurs at π.
  ♦ So the maximum value of secant should also be at π.
  ♦ Indeed we see that, at π, the secant is at it's peak point in this segment.
  ♦ The value is reciprocal of -1, which is -1
(iii) Consider the segment from π to $\frac{3\pi}{2}$ on the x axis.
• In this segment,
   ♦ the cosine curve rises
   ♦ but the secant curve falls.
• That means, when the angle increases from π to $\frac{3\pi}{2}$,
   ♦ the cosine value increases
   ♦ but the secant value decreases.
• This is obvious because, secant is the reciprocal of cosine.
Note that here, all the cosine values are -ve. Consequently, all the secant values will also be -ve.
• In the segment from π to $\frac{3\pi}{2}$, the maximum value of cosine occurs at $\frac{3\pi}{2}$.
  ♦ So the minimum value of secant should also be at $\frac{3\pi}{2}$.
  ♦ Indeed we see that, as the angle approaches $\frac{3\pi}{2}$, the secant curve goes further and further down.

◼ As seen before, we cannot put x = $\frac{3\pi}{2}$ in $f(x) = \sec x = \frac{1}{\cos x}$. We can write: • As x approaches $\frac{3\pi}{2}$, cos x becomes smaller and smaller (negatively), getting closer and closer to zero. (For example, values like -0.00001, -0.000001 are very close to zero) • As cos x becomes smaller and smaller, the reciprocal sec x becomes larger and larger (negatively). • This is indicated by the falling portion of the secant curve between π and $\frac{3\pi}{2}$. • As angle becomes closer and closer to $\frac{3\pi}{2}$, this falling portion gets closer and closer to the vertical line through $\frac{3\pi}{2}$. (why does this happen? The reader may write the answer in his/her own notebooks) • But it never touches that vertical line. • If it touch, it would mean that, x = $\frac{3\pi}{2}$ is a point in the secant curve. We know that, it cannot happen.

(iv) Consider the segment from $\frac{3\pi}{2}$ to 2π on the x axis.
• In this segment,
   ♦ the cosine curve rises
   ♦ but the secant curve falls.
• That means, when the angle increases from $\frac{3\pi}{2}$ to 2π,
   ♦ the cosine value increases
   ♦ but the secant value decreases.
• This is obvious because, secant is the reciprocal of cosine.
Note that here, all the cosine values are +ve. Consequently, all the secant values will also be +ve.

(v) Now we have a basic idea about how the 'U' shapes and 'inverted U' shapes are formed in positive side of the x axis.
As the angle increases beyond 2π, this pattern repeats again and again.
(vi) Similar steps can be written for negative angles also. Those steps will explain the 'U' shapes and 'inverted U' shapes in the negative side of the x axis.
3. Based on the analysis of the graph, we can write:
• The output of f(x) = sec x can be any +ve real number starting from +1 and higher.
• The output of f(x) = sec x can be any -ve real number starting from -1 and lower.
• The values lying between -1 and +1 cannot be output values.
4. So the range of f(x) = sec is R - (-1, +1)
• That means, we have to subtract the set (-1, +1) from the the set of real numbers R
• The resulting set after subtraction is the range of f(x) = sec x
• (-1, +1) indicates that:
   ♦ all values between -1 and +1 are included in the set.
   ♦ '(' on the left side indicates that -1 is not included in the set.
   ♦ ')' on the right side indicates that +1 is not included in the set.
• So it is clear that, the two values -1 and +1 should not be subtracted from R.

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In the next section, we will see domain and range of tangent and cotangent functions.

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Chapter 3.8 - Domain and Range of Sine and Cosecant Functions

In the previous section, we saw that any real number can be used as input x for f(x) = sin x and f(x) = cos x. So we get the feeling that, any real number can be used as input x for tan x, sec x etc., also. But it is not true. There are some exceptions. It can be written in 4 steps:

1. Consider f(x) = tan x.
• This can be written as $f(x)=\frac{\sin x}{\cos x}$
    ♦ So it is obvious that, cos x should not become zero.
• If we use an input x which makes cos x equal to zero, we will not get an output for the function f(x) = tan x.
• We know the exact x values for which cos x becomes zero. They are: $(2n+1)\frac{\pi}{2}$, where n is any integer.
◼ So we can write:
f(x) = tan x = $\frac{\sin x}{\cos x},\;x \neq (2n+1)\frac{\pi}{2}$ where n is any integer.
2. Consider f(x) = csc x.
• This can be written as $f(x)=\frac{1}{\sin x}$
    ♦ So it is obvious that, sin x should not become zero.
• If we use an input x which makes sin x equal to zero, we will not get an output for the function f(x) = csc x.
• We know the exact x values for which sin x becomes zero. They are: nπ, where n is any integer.
◼ So we can write:
f(x) = csc x = $\frac{1}{\sin x},\;x \neq n \pi$ where n is any integer.
3. Consider f(x) = sec x.
• This can be written as $f(x)=\frac{1}{\cos x}$
    ♦ So it is obvious that, cos x should not become zero.
• If we use an input x which makes cos x equal to zero, we will not get an output for the function f(x) = sec x.
• We know the exact x values for which cos x becomes zero. They are: $(2n+1)\frac{\pi}{2}$, where n is any integer.
◼ So we can write:
f(x) = sec x = $\frac{1}{\cos x},\;x \neq (2n+1)\frac{\pi}{2}$ where n is any integer.
4. Consider f(x) = cot x.
• This can be written as $f(x)=\frac{\cos x}{\sin x}$
    ♦ So it is obvious that, sin x should not become zero.
• If we use an input x which makes sin x equal to zero, we will not get an output for the function f(x) = cot x.
• We know the exact x values for which sin x becomes zero. They are: nπ, where n is any integer.
◼ So we can write:
f(x) = cot x = $\frac{\cos x}{\sin x},\;x \neq n \pi$ where n is any integer.

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Sign of trigonometric functions

This can be written in 4 steps:
1. Consider the unit circle that we saw in the previous sections. In that circle, we have three important items:
(i) The tip of the ray.
   ♦ We denoted it as P.
(ii) Base of the triangle OMP.
   ♦ We denoted it as a
(iii) Altitude of the triangle OMP.
   ♦ We denoted it as b
2. We know that:
Coordinates of P are written using a and b.
3. We also know that:
  ♦ a is related to cosine
  ♦ b is related to sine
4. Based on steps (1), (2) and (3), we can write about the four quadrants:
First quadrant
(i) If P is in the I quadrant, the coordinates of P will be (a, b)
(ii) We know that:
   ♦ x coordinate is the cosine.
   ♦ y coordinate is the sine.
(iii) So in the I quadrant, cosine will be +ve, sine will be +ve.
• Consequently,
sec will be +ve, csc will be +ve.
• Also,
tan will be +ve, cot will be +ve
(iv) Thus we get the column named I in the table 3.1 below:

Table 3.1

• Note that, the ray can rotate in the anticlockwise direction (to give a positive angle) and reach the I quadrant.
• The ray can also rotate in the clockwise direction (to give a negative angle) and reach the I quadrant.
• What ever be the direction of rotation, the coordinates of P will not change. So the signs written in the column named I of table 3.1 is valid for both +ve and -ve angles.

Second quadrant
(i) If P is in the II quadrant, the coordinates of P will be (-a, b)
(ii) We know that:
   ♦ x coordinate is the cosine.
   ♦ y coordinate is the sine.
(iii) So in the II quadrant, cosine will be -ve, sine will be +ve.
• Consequently,
sec will be -ve, csc will be +ve.
• Also,
tan will be -ve, cot will be -ve
(iv) Thus we get the column named II in the table 3.1 above.
• Note that, the ray can rotate in the anticlockwise direction (to give a positive angle) and reach the II quadrant.
• The ray can also rotate in the clockwise direction (to give a negative angle) and reach the II quadrant.
• What ever be the direction of rotation, the coordinates of P will not change. So the signs written in the column named II of table 3.1 is valid for both +ve and -ve angles. 

Third quadrant
(i) If P is in the III quadrant, the coordinates of P will be (-a, -b)
(ii) We know that:
   ♦ x coordinate is the cosine.
   ♦ y coordinate is the sine.
(iii) So in the III quadrant, cosine will be -ve, sine will be -ve.
• Consequently,
sec will be -ve, csc will be -ve.
• Also,
tan will be +ve, cot will be +ve
(iv) Thus we get the column named III in the table 3.1 above.
• Note that, the ray can rotate in the anticlockwise direction (to give a positive angle) and reach the III quadrant.
• The ray can also rotate in the clockwise direction (to give a negative angle) and reach the III quadrant.
• What ever be the direction of rotation, the coordinates of P will not change. So the signs written in the column named III of table 3.1 are valid for both +ve and -ve angles. 

Fourth quadrant
(i) If P is in the IV quadrant, the coordinates of P will be (a, -b)
(ii) We know that:
   ♦ x coordinate is the cosine.
   ♦ y coordinate is the sine.
(iii) So in the IV quadrant, cosine will be +ve, sine will be -ve.
• Consequently,
sec will be +ve, csc will be -ve.
• Also,
tan will be -ve, cot will be -ve
(iv) Thus we get the column named IV in the table 3.1 above.
• Note that, the ray can rotate in the anticlockwise direction (to give a positive angle) and reach the IV quadrant.
• The ray can also rotate in the clockwise direction (to give a negative angle) and reach the IV quadrant.
• What ever be the direction of rotation, the coordinates of P will not change. So the signs written in the column named IV of table 3.1 are valid for both +ve and -ve angles.

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Domain and range of trigonometric functions

This can be explained in 2 steps:
1. We know that:
• Domain of a function is the set containing all the input x values.
• Range of a function is the set containing all the output y [y is same as f(x)] values.
2. We need to have a basic idea about the domain and range of the various trigonometric functions like f(x) = sin x, f(x) = cos x, f(x) = sec x etc.,

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Let us find the domain and range of f(x) = sin x. It can be written in 2 steps:
1. First we will find the domain:
• We have seen that:
In the case of f(x) =sin x, any real number can be used as input x.
• So we can write: Domain of f(x) = sin x, is the set R.
(Recall that, R is the set of real numbers)
2. Let us find the range of f(x) = sin x
The range in this case can be better understood if we analyze the graph of f(x) = sin x. It is shown in fig.3.26 below:

Fig.3.26

• An analysis of this graph can be written in 4 steps:
(i) We see that, on the x axis, the markings are: $\frac{\pi}{2},\; \pi,\; \frac{3\pi}{2},\; 2\pi,\; \frac{5\pi}{2}$ so on . . .
• Normally, we expect markings as: 1, 2, 3, . . . OR  5, 10, 15, . . .
• The markings in the above graph are no different from 1, 2, 3, . . . OR  5, 10, 15, .
• If we use the real number line as the x axis, this type of marking can be achieved by putting marks at $\frac{3.14}{2},\; 3.14,\; \frac{3\times 3.14}{2},\; 2\times 3.14,\; \frac{5\times 3.14}{2}$ so on . . .
• Here each unit on the x axis is ${\frac{\pi}{2}}^c$
• These markings can go up to +∞. That means, all positive real numbers can be used as input x.
• They are positive angles obtained when the ray rotates in the anticlockwise direction.
(ii) Similar is the case with markings on the negative side of the x axis.
• They are negative angles, which are obtained when the ray rotates in the clockwise direction.
• The -ve markings can go up to -∞. That means, all negative real numbers can be used as input x.
(iii) Now let us consider the output y values.
• We see that, whatever be the value of input x,
   ♦ Output y value (that is., sine of x) never becomes greater than 1
   ♦ Output y value never becomes lesser than -1
(iv) We can mark any point on the red curve in the graph. The y coordinate of that point will be either 1 or -1 or a value between 1 and -1.
◼ So the range can be written as [-1, 1]
• That means:
   ♦ All real values between -1 and 1 are included in the range.
   ♦ Left side '[' indicates that -1 is also included in the range.
   ♦ Right side ']' indicates that +1 is also included in the range.

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• We just saw that sine value can never rise above +1. Neither can it fall below -1.
• The reason can be written in 8 steps:
1. In the animation in fig.3.27 below, the thick white vertical line is the altitude of the triangle in our familiar unit circle.

Fig.3.27

• We know that, this altitude is related to the sine value of the angle.
2. The red ray starts to rotate from the +ve side of the x axis.
• As the rotation proceeds, the length of the white vertical line gradually increases.
• It becomes maximum when the ray completes a rotation of ${\frac{\pi}{2}}^c$. That is, when the ray coincides with the y axis.
• Since the green circle is a unit circle, the length of the vertical white line at this point will be 1
• So we can write:
(i) The sine value start to increase from zero (when the ray coincides with the +ve side of x axis)
(ii) The sine value attains a maximum value of 1 (when the ray coincides with the +ve side of y axis)
(iii) This increase of sine value, from zero to 1, is indicated by the rising portion between x = 0 and x = $\frac{\pi}{2}$ in the graph in fig.3.26 above.
3. Next, the red ray proceeds to rotate from x = $\frac{\pi}{2}$ to x = π.
• We see that, the length of the vertical white line decreases from 1 to zero.
• This decrease is indicated by the falling portion between x = $\frac{\pi}{2}$ and x = π in the graph in fig.3.26 above.
4. Next, the red ray proceeds to rotate from x = π to x = $\frac{3\pi}{2}$.
• We see that, the length of the vertical white line increases from zero to 1.
• Though it is an 'increase in length', it happens below the x axis. Below the x axis, every y coordinate is negative. That means, below the y axis, every sine value is negative.
• As the length increases, the negative value increases. So in effect, it is a decrease.
• This decrease is indicated by the falling portion between x = π and x = $\frac{3\pi}{2}$ in the graph in fig.3.26 above.
5. Next, the red ray proceeds to rotate from x = $\frac{3\pi}{2}$ to x = 2π.
• We see that, the length of the vertical white line decreases from 1 to zero.
• Though it is a 'decrease in length', it happens below the x axis. Below the x axis, every y coordinate is negative, That means, below the y axis, every sine value is negative.
• As the length decreases, the negative value decreases. So in effect, it is an increase.
• This increase is indicated by the rising portion between x = $\frac{3\pi}{2}$ and x = 2π in the graph in fig.3.26 above.
6. As the ray continues to rotate, this pattern repeats again and again. Thus we get the wave form on the positive side of the x axis in the graph.
7. Similar steps can be written for rotation in the anticlockwise direction also. Based on those steps, we will be able to explain the wave form on the negative side of the x axis in the graph.
8. Note that:
• In the first ${\frac{\pi}{2}}^c$ rotation in the anticlockwise direction, there is a rise in sine value.
• In the first ${\frac{\pi}{2}}^c$ rotation in the clockwise direction, there is a fall in sine value.
• So there is a smooth transition between the two waves on either sides of the x axis.

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Let us find the domain and range of f(x) = csc x. It can be written in 4 steps:
1. First we will find the domain:
• We have seen that:
    ♦ In the case of f(x) = csc x,
    ♦ nπ where n is any integer, should not be used as input x
• So we can write:
Domain of f(x) = csc x is R -{x : x = nπ, n ∈ Z}
• That means, we must subtract {x : x = nπ, n ∈ Z} from R. The resulting set after subtraction, is the domain of f(x) = csc x
• {x : x = nπ, n ∈ Z} is the set containing all nπ, where n is any integer.
2. Let us find the range of f(x) = csc x
We know that, csc is the reciprocal of sine. So the range of f(x) = csc x can be better understood if we analyze the graph of f(x) = sin x and f(x) = csc x together. It is shown in fig.3.28 below:

Fig.3.28

An analysis of this graph can be written in 6 steps:
(i) Consider the segment from 0 to $\frac{\pi}{2}$ on the x axis.
• In this segment,
   ♦ the sine curve rises
   ♦ but the csc curve falls.
• That means, when the angle increases from 0 to $\frac{\pi}{2}$,
   ♦ the sine value increases
   ♦ but the csc value decreases.
• This is obvious because, csc is the reciprocal of sine
• In the segment from 0 to $\frac{\pi}{2}$, the maximum value of sine occurs at $\frac{\pi}{2}$
   ♦ So the minimum value of csc should also be at $\frac{\pi}{2}$   
   ♦ Indeed we see that, the fall of the csc curve occurs upto $\frac{\pi}{2}$.
   ♦ Thereafter, it rises.
• The value of sine at $\frac{\pi}{2}$ is 1
   ♦ The value of csc at $\frac{\pi}{2}$ should be the reciprocal of 1, which is 1
   ♦ Indeed we see that, the value of csc at $\frac{\pi}{2}$ is 1
(ii) Consider the segment from $\frac{\pi}{2}$ to π on the x axis.
• In this segment,
   ♦ the sine curve falls
   ♦ but the csc curve rises.
• That means, when the angle increases from $\frac{\pi}{2}$ to π,
   ♦ the sine value decreases
   ♦ but the csc value increases.
• This is obvious because, csc is the reciprocal of sine
• In the segment from $\frac{\pi}{2}$ to π, the minimum value of sine occurs at π.
  ♦ So the maximum value of csc should also be at π.
  ♦ Indeed we see that, as π approaches, the csc curve rises further and further up.

◼ As seen before, we cannot put x = π in $f(x) = \csc x = \frac{1}{\sin x}$. We can write: • As x approaches π, sin x becomes smaller and smaller, getting closer and closer to zero. (For example, values like 0.00001, 0.000001 are very close to zero) • As sin x becomes smaller and smaller, the reciprocal csc x becomes larger and larger. • This is indicated by the rising portion of the csc curve between $\frac{\pi}{2}$ and π. • As angle becomes closer and closer to π, this rising portion gets closer and closer to the vertical line through π. (why does this happen? The reader may write the answer in his/her own notebooks) • But it never touches that vertical line. • If it touch, it would mean that, x = π is a point in the csc curve. We know that, it cannot happen.

(iii) Consider the segment from π to $\frac{3\pi}{2}$ on the x axis.
• In this segment,
   ♦ the sine curve falls
   ♦ but the csc curve rises.
• That means, when the angle increases from π to $\frac{3\pi}{2}$,
   ♦ the sine value decreases
   ♦ but the csc value increases.
• This is obvious because, csc is the reciprocal of sine.
Note that here, all the sine values are -ve. Consequently, all the csc values will also be -ve.
• In the segment from π to $\frac{3\pi}{2}$, the minimum value of sine occurs at $\frac{3\pi}{2}$.
  ♦ So the maximum value of csc should also be at $\frac{3\pi}{2}$.
   ♦ Indeed we see that, the rise of the csc curve occurs upto $\frac{3\pi}{2}$.
   ♦ Thereafter, it falls.
• The value of sine at $\frac{3\pi}{2}$ is -1
   ♦ The value of csc at $\frac{3\pi}{2}$ should be the reciprocal of -1, which is -1.
   ♦ Indeed we see that, the value of csc at $\frac{3\pi}{2}$ is -1.
• Consider the portion just after π. The sine values here will be very small negative values. Consequently, the csc values here will be very large negative values. This fact can be clearly seen in the graphs.
(iv) Consider the segment from $\frac{3\pi}{2}$ to 2π on the x axis.
• In this segment,
   ♦ the sine curve rises
   ♦ but the csc curve falls.
• That means, when the angle increases from $\frac{3\pi}{2}$ to 2π,
   ♦ the sine value increases
   ♦ but the csc value decreases.
• This is obvious because, csc is the reciprocal of sine.
Note that here, all the sine values are -ve. Consequently, all the csc values will also be -ve.

◼ As seen before, we cannot put x = 2π in $f(x) = \csc x = \frac{1}{\sin x}$. We can write: • As x approaches 2π, sin x becomes smaller and smaller (negatively), getting closer and closer to zero. (For example, values like -0.00001, -0.000001 are very close to zero) • As sin x becomes smaller and smaller (negatively), the reciprocal csc x becomes larger and larger (negatively). • This is indicated by the falling portion of the csc curve between $\frac{3\pi}{2}$ and 2π. • As angle becomes closer and closer to 2π, this falling portion gets closer and closer to the vertical line through 2π. (why does this happen? The reader may write the answer in his/her own notebooks) • But it never touches that vertical line. • If it touch, it would mean that, x = 2π is a point in the csc curve. We know that, it cannot happen.

(v) Now we have a basic idea about how the 'U' shapes and 'inverted U' shapes are formed in positive side of the x axis.
• As the angle increases beyond 2π, this pattern repeats again and again.
(vi) Similar steps can be written for negative angles also. Those steps will explain the 'U' shapes and 'inverted U' shapes in the negative side of the x axis.
3. Based on the analysis of the graph, we can write:
• The output of f(x) = csc x can be any +ve real number starting from +1 and higher.
• The output of f(x) = csc x can be any -ve real number starting from -1 and lower.
• The values lying between -1 and +1 cannot be outpit values.
4. So the range of f(x) = csc is R - (-1, +1)
• That means, we have to subtract the set (-1, +1) from the the set of real numbers R
• The resulting set after subtraction is the range of f(x) = csc x
• (-1, +1) indicates that:
   ♦ all values between -1 and +1 are inculded in the set (-1, +1).
   ♦ '(' on the left side indicates that -1 is not included in the set (-1, +1).
   ♦ ')' on the right side indicates that +1 is not included in the set (-1, +1).
• So it is clear that, the two values -1 and +1 should not be subtracted from R.

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In the next section, we will see domain and range of cosine and secant functions.

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