17.13 - More Miscellaneous Examples

In the previous section, we saw some miscellaneous examples. In this section, we will see a few more miscellaneous examples.

Solved example 17.47
Let A = {1,2,3}. Then show that the number of relations containing (1,2) and (2,3) which are reflexive and transitive but not symmetric is three.
Solution:
1. The fig.17.12 below shows all the 9 possible pairs.

Fig.17.12

2. We need to pick suitable pairs from the above nine and form various relations.
• The relations thus formed should:
    ♦ be reflexive
    ♦ be transitive
    ♦ contain (1,2) and (2,3)
    ♦ not be symmetric.

3. The pairs in the diagonal yellow cells are to be included because, the relation should be reflexive.
4. The relation should be transitive. Also (1,2) and (2,3) must be present. So (1,3) also must be present. Thus the three red cells will be present.
5. The relation should not be symmetric. So all green cells must be avoided. If we include the green cells, then:
(1,2) will get the symmetric pair (2,1)
(1,3) will get the symmetric pair (3,1)
(2,3) will get the symmetric pair (3,2)
6. So the smallest relation possible is obtained by including yellow and red.
We get: R₁ = {(1,1),(2,2),(3,3),(1,2),(1,3),(2,3)}
7. In (5), we wrote that, all greens must be avoided.
In fact, it is true that, we must not include the three greens together. But we can pick (2,1) and (3,2) one by one.
8. Picking (2,1) and adding it to R₁, we get:
R₂ = {(1,1),(2,2),(3,3),(1,2),(1,3),(2,3),(2,1)}
9. Picking (3,2) and adding it to R₁, we get:
R₃ = {(1,1),(2,2),(3,3),(1,2),(1,3),(2,3),(3,2)}
10. We cannot add (3,1) to R₁. This is because, it will group with (1,2). So we will be forced to add (3,2) to maintain transitivity. This will give:
R₄ = {(1,1),(2,2),(3,3),(1,2),(1,3),(2,3),(3,1),(3,2)}
11. Now, (2,3) will group with (3,1) and force us to include (2,1)
This will give:
R₅ = {(1,1),(2,2),(3,3),(1,2),(1,3),(2,3),(3,1),(3,2),(2,1)}
This contain all greens.
So R₄ and R₅ are not possible.
12. In (10), we wrote that, the green (3,1) must never be added.
Can we add the remaining two greens together?
We get: R6q = {(1,1),(2,2),(3,3),(1,2),(1,3),(2,3),(2,1),(3,2)}
Here, (3,2) will group with (2,1) and force us to add (3,1).
So R6q is also not possible.
13. Thus the only possible relations are: R₁, R₂ and R₃.
We can write:
The number of relations which satisfy the given conditions is three.

Solved example 17.48
Show that the number of equivalence relation in the set {1,2,3} containing (1,2) and (2,1) is two.
Solution:
1. The fig.17.13 below shows all the 9 possible pairs.

Fig.17.13

2. We need to pick suitable pairs from the above nine and form various relations.
• The relations thus formed should:
    ♦ be equivalence relations
    ♦ contain (1,2) and (2,1).
3. The pairs in the diagonal yellow cells are to be included because, the relation should be reflexive.
4. The pairs (1,2) and (2,1) in the red cells are to be included because, it is a given condition.
5. So the smallest relation is R₁ = {(1,1), (2,2),(3,3),(1,2),(2,1)}
This is an equivalence relation.
6. Now there are four pairs remaining. If we add any one of those remaining pairs, we will be forced to add all of them.
For example, if we add (1,3), we get the group: (2,1), (1,3). Then we have to add (2,3) to make it transitive.
As a consequence, we are forced to add (3,2) to make it symmetric.
Also, when (1,3) is added, we have to add (3,1)
7. When the remaining four pairs are added, we get the universal relation.
It is an equivalence relation.
8. So there are two relations which satisfy the given conditions.
R₁ and the universal relation.

Solved example 17.49
Show that the number of binary operations on {1,2} having 1 as identity and having 2 as the inverse of 2 is exactly one.
Solution:
1. A binary operation is a function.
• In our present case, the domain is {1,2} × {1,2}.
 ♦ That means, the domain is {(1,1),(1,2),(2,1),(2,2)}
 ♦ That means, the only possible input pairs are (1,1),(1,2),(2,1) and (2,2)
• The range is {1,2}
 ♦ That means, the output will be either 1 or 2
2. Let us see what happens when each of the pairs mentioned in (1) are given as input.
(i) First we input (1,1)
• So the operation is 1∗1
• Given that, 1 is the identity (e).
• Recall that, if e is the identity, then:
a∗e = a = e∗a for all a ∈{1,2}
• So we can write:
1∗1 = 1 = 1∗1 (Here we apply e to the element 1)
• That means:
For the input (1,1), the output is 1
(ii) Next we input (1,2)
• So the operation is 1∗2
Here also, since ‘1’ is the identity, we can write:
2∗1 = 1 = 1∗2 (Here we apply e to the element 2)
• That means:
For the input (1,2), the output is 1
(iii) Next we input (2,1)
• So the operation is 2∗1
We already saw that, for this operation, the result is 1
• That means:
For the input (2,1), the output is 1
(iv) Finally, we input (2,2)
So the operation is 2∗2
• Here we cannot use the property of identity because, identity is ‘1’. This operation does not involve ‘1’.
• So we use the property of inverse.
• Given that, 2 is the inverse of 2.
• Recall that, if b is the inverse of a, then:
a∗b = e = b∗a for all a ∈{1,2}
• So we can write:
2∗2 = 1 = 2∗2 (Here we apply the inverse ‘2’  to the element 2)
• That means:
For the input (2,2), the output is 1
3. Now consider the given operation:
The operation has ‘1’ as identity and ‘2’ as the inverse of 2.
• We will denote this operation as '∗'. We used this operation and found out the outputs for all possible inputs.
4. Our next task is to prove that, ∗ is unique. It can be done in steps.
(i) In step (2), we saw the inputs and the corresponding outputs. We can write them as a set:
∗ = {[(1,1),1], [(1,2),1], [(2,1),1], [(2,2),1]}
(ii) Suppose that, ∗' is another operation having 1 as identity and having 2 as the inverse of 2.
• Here also, we will get the same set. That means:
∗' = {[(1,1),1], [(1,2),1], [(2,1),1], [(2,2),1]}
• The same set is obtained because, all possible inputs are already present in ∗.
(iii) Since both sets are same, we can write: ∗ = ∗'.
Therefore, ∗is unique.

Solved example 17.50
Consider the identity function I_N : N → N defined as I_N(x) = x ∀ x ∈ N.
Show that although I_N is onto but I_N + I_N : N → N defined as
(I_N + I_N ) (x) = I_N (x) + I_N (x) = x + x = 2x is not onto.
Solution:
1. Consider the function
I_N : N → N defined as I_N(x) = x ∀ x ∈ N.
• Let y be any element in the codomain N
Then we can write: I_N(x) = x = y
• So it is clear that:
   ♦ If we want any ‘y’ (in the codomain N) to be an image,
   ♦ we need to pick an ‘x’ (from the domain N) in such a way that,
   ♦ that ‘x’ is equal to 'y'.
• In this way, all y in the codomain can become an image. So this function is an onto function.

2. Consider the function
I_N + I_N : N → N defined as
(I_N + I_N ) (x) = I_N (x) + I_N (x) = x + x = 2x.
• Let y be any element in the codomain N
Then we can write: (I_N + I_N )(x) = 2x = y
• So it is clear that:
   ♦ If we want any ‘y’ (in the codomain N) to be an image,
   ♦ we need to pick an ‘x’ (from the domain N) in such a way that,
   ♦ that ‘x’ is equal to 'y/2'.
• Suppose that, y = 3. Then y/2 = 1.5.
• This 1.5 is not present in the domain N. So all elements in the codomain N cannot become images. That means, this function is not an onto function.

Solved example 17.51
Consider a function $f:~\left[0,\frac{\pi}{2} \right] \to R$ given by f(x) = sin x and another function $g:~\left[0,\frac{\pi}{2} \right] \to R$ given by g(x) = cos x. Show that f and g are one-one. But (f+g) is not one-one.
Solution:
1. Both f and g have the same domain.
• We can use any real number from 0 to $\frac{\pi}{2}$ (both included) as the input.
2. First consider f.
• In a one-one function, if f(x₁) is to be equal to f(x₂), then x₁ must be equal to x₂. This condition can be used to prove that, a given function is a one-one function.
• In our present case, suppose that, f(x₁) is equal to f(x₂). Then we can write:
$\begin{array}{ll}{}    &{f(x_1)}    & {~=~}    &{f(x_2)}    &{} \\
{\Rightarrow}    &{\sin x_1}    & {~=~}    &{\sin x_2}    &{} \\
{\Rightarrow}    &{x_1}    & {~=~}    &{x_2}    &{} \\
\end{array}$

• So f is a one-one function.

◼ Let us see an example:
• We know that $\frac{\pi}{6}$ lies between 0 and $\frac{\pi}{2}$.
• If we use $\frac{\pi}{6}$ as the input, we will get:
output = $f \left(\frac{\pi}{6} \right) ~=~\sin \left(\frac{\pi}{6} \right)~=~\frac{1}{2}$
• Only the input $\frac{\pi}{6}$ will give $\frac{1}{2}$ as the output. We will never find another input value (from 0 to $\frac{\pi}{2}$) which will give $\frac{1}{2}$ as the output.

3. Next, consider g.
• In a one-one function, if g(x₁) is to be equal to g(x₂), then x₁ must be equal to x₂. This condition can be used to prove that, a given function is a one-one function.
• In our present case, suppose that, g(x₁) is equal to g(x₂). Then we can write:
$\begin{array}{ll}{}    &{g(x_1)}    & {~=~}    &{g(x_2)}    &{} \\
{\Rightarrow}    &{\cos x_1}    & {~=~}    &{\cos x_2}    &{} \\
{\Rightarrow}    &{x_1}    & {~=~}    &{x_2}    &{} \\
\end{array}$

• So g is a one-one function.

◼ Let us see an example:
• We know that $\frac{\pi}{6}$ lies between 0 and $\frac{\pi}{2}$.
• If we use $\frac{\pi}{6}$ as the input, we will get:
output = $g \left(\frac{\pi}{6} \right) ~=~\cos \left(\frac{\pi}{6} \right)~=~\frac{\sqrt{3}}{2}$
• Only the input $\frac{\pi}{6}$ will give $\frac{\sqrt{3}}{2}$ as the output. We will never find another input value (from 0 to $\frac{\pi}{2}$) which will give $\frac{\sqrt{3}}{2}$ as the output.

4. Finally, we consider (f+g).
• We have: (f+g)(x) = sin x + cos x
• We want to prove that, (f+g) is not a one-one function.
• Let us use 0 as the input. We get:
(f+g)(0) = sin 0 + cos 0 = (0+1) = 1  
• Let us use $\frac{\pi}{2}$ as the input. We get:
$(f+g) \left(\frac{\pi}{2} \right)~=~\sin \left(\frac{\pi}{2} \right)~+~ \cos \left(\frac{\pi}{2} \right)~=~ (1+0)~=~1$
• So ‘1’ is the image of both 0 and $\frac{\pi}{2}$
Therefore, (f+g) is not a one-one function.

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The link below gives a few more solved examples

Miscellaneous Exercise 17 

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In the next chapter, we will see inverse trigonometric functions.

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17.12 - Miscellaneous Examples

In the previous section, we saw identity element and inverse element. In this section, we will see some miscellaneous examples.

Solved example 17.41
If R₁ and R₂ are equivalence relations in a set A, show that R₁∩R₂ is also an equivalence relation.
Solution:
Let us assume that, the set A = {a₁, a₂, a₃, a₄, . . . , a_n}
Part (i): Proving that, the intersection is reflexive.
1. R₁ is an equivalence relation in set A.
• So all reflexive pairs like (a₁,a₁), (a₂,a₂), (a₃,a₃), . . .(a_n,a_n) will be present in the set R₁.
2. R₂ is also an equivalence relation in set A.
• So all reflexive pairs like (a₁,a₁), (a₂,a₂), (a₃,a₃), . . .(a_n,a_n) will be present in the set R₂ also.
3. The same reflexive pairs are present in both R₁ and R₂.
• So those reflexive pairs will be present in R₁∩R₂ also.
Therefore, R₁∩R₂ is reflexive.

Part (ii): Proving  that, the intersection is symmetric.
1. Suppose that, a random pair (a₃,a₇) is present in the intersection.
   ♦ Then this (a₃,a₇) will be present in the set R₁.
   ♦ This (a₃,a₇) will be present in the set R₂ also.
2. (a₃,a₇) is present in R₁.
• But R₁ is an equivalence relation. So the symmetric pair (a₇,a₃) will be present in R₁.
3. (a₃,a₇) is present in R₂.
• But R₂ is an equivalence relation. So the symmetric pair (a₇,a₃) will be present in R₂.
4. From (2) and (3), we see that:
(a₇,a₃) is present in both R₁ and R₂.
5. From (1) and (4) we see that:
The symmetric pairs (a₃,a₇) and (a₇,a₃) are present in the intersection.
6. In this way, all symmetric pairs will be present in the intersection.
So the intersection is a symmetric relation.

Part (iii): Proving that, the intersection is transitive.
1. Suppose that, two random pairs (a₃,a₇) and (a₇,a₂) are present in the intersection.
   ♦ Then these two pairs will be present in R₁.
   ♦ These two pairs will be present in R₂ also.
2. (a₃,a₇) and (a₇,a₂) are present in R₁.
• But R₁ is a symmetric relation. So the transitive pair (a₃,a₂) will be present in R₁.   
3. (a₃,a₇) and (a₇,a₂) are present in R₂.
• But R₂ is a symmetric relation. So the transitive pair (a₃,a₂) will be present in R₂.
4. From (2) and (3), we see that:
• (a₃,a₂) will be present in both R₁ and R₂.
• So it will be present in the intersection also.
5. From (1) and (4),we see that:
• The intersection contains (a₃,a₇), (a₇,a₂) and (a₃,a₂)
Therefore, the intersection is transitive.     
◼ From parts (i), (ii) and (iii), we see that, the intersection is reflexive, symmetric and transitive. So the intersection is an equivalence relation.

Solved example 17.42
Let R be the relation on set A of ordered pairs of positive integers defined by (x,y)R(u,v) if and only if xv = yu. Show that R is an equivalence relation.
Solution:
1. Set A is a set of ordered pairs.
• Each of those ordered pairs will contain +ve integers.
For example: (3,5), (1,2), (7,11) etc.,

2. The given relation is on A.
• That means, we need to consider the set A × A
• Each element in A × A will be a pair of ordered pairs.
For example: [(11,3),(5,4)], [(7,5),(8,2)], [(14,8),(9,3)] etc.,
• In general, we can write:
Each element in A × A will be a pair of ordered pairs in the form [(x,y),(u,v)]

3. The set R will contain those elements from A × A, which satisfy the condition: xv = yu.
• We need to prove that R is an equivalence relation.

4. First we check whether R is reflexive.
(i) If R is reflexive, then a possible random element in R is: [(a₃,a₇),(a₃,a₇)].
(ii) Let us check whether this element satisfies the condition xv = yu.
• Here, x = u = a₃ and y = v = a₇
• We can write:
xv = a₃ a₇  and  yu = a₇ a₃ = a₃ a₇
(iii) We see that xv = yu.
• So all reflexive elements are eligible to be included in R.
Therefore, R is a reflexive relation.   

5. Next we check whether R is symmetric.
(i) Suppose that a random element [(a₃,a₇),(a₂,a₉)] is present in R.
• It's symmetric element is [(a₂,a₉),(a₃,a₇)]. Is this symmetric element present in R? Let us check.
(ii) [(a₃,a₇),(a₂,a₉)] is present in R.
• That means, this element satisfies the condition: xv = yu
• That means, a₃ a₉ = a₇ a₂.
(iii) If the symmetric element is to be present in R, it must also satisfy the condition xv = yu.
• That means, a₂ a₇ must be equal to a₉ a₃.
• From (ii), we see that, they are indeed equal.
(iv) So we can write:
• If [(a₃,a₇),(a₂,a₉)] is present in R, then the symmetric element [(a₂,a₉),(a₃,a₇)] will also be present in R.
• So all symmetric elements are eligible to be included in R.
Therefore, R is a symmetric relation.

6. Finally, we check whether R is transitive.
(i) Suppose that two random elements [(a₃,a₇),(a₂,a₉)] and [(a₂,a₉),(a₈,a₁₁)] are present in R.
• Then the transitive element is [(a₃,a₇),(a₈,a₁₁)]. Is this transitive element present in R? Let us check.
(ii) [(a₃,a₇),(a₂,a₉)] is present in R.
• That means, this element satisfies the condition: xv = yu
• That means, a₃ a₉ = a₇ a₂.
(iii) Similarly, [(a₂,a₉),(a₈,a₁₁)] is present in R.
• That means, this element satisfies the condition: xv = yu
• That means, a₂ a₁₁ = a₉ a₈.
(iv) If the transitive element written in (i) is to be present in R, then it should also satisfy the condition xv = yu.
• That means, a₃ a₁₁ must be equal to a₇ a₈
• That means, $\frac{a_3}{a_7}~\text{must be equal to}~\frac{a_8}{a_{11}}$
(v) From (ii) we get: $\frac{a_3}{a_7}~=~\frac{a_2}{a_9}$
(vi) From (iii) we get: $\frac{a_2}{a_9}~=~\frac{a_8}{a_{11}}$
(vii) Combining the results in (v) and (vi), we get:
$\frac{a_2}{a_9}~=~\frac{a_8}{a_{11}}~=~\frac{a_3}{a_7}$
• So the condition mentioned in (iv) is satisfied.
(vii) That means, all transitive elements will be present in R.
Therefore, R is a transitive relation.

◼ Based on the above 6 steps, we see that, R is reflexive, symmetric and transitive. So R is an equivalence relation.

Solved example 17.43
Let X = {1,2,3,4,5,6,7,8,9}. Let R₁ be a relation in X given by R₁ = {(x,y): x-y is divisible by 3} and R₂ be another relation on X given by R₂ ={(x,y): {x,y}⊂{1,4,7} or {x,y}⊂{2,5,8} or {x,y}⊂{3,6,9}}. Show that R₁ = R₂.
Solution:
1. We can write set R₁ easily. Ordered pairs like (1,4), (4,1), (5,8), are some of the elements of R₁.
• But in this problem, we do not have to write the elements of R₁. We just need to know the nature of the elements. We see that, for all those elements, (x-y) will be divisible by 3.
2. Next we consider R₂
(i) Any ordered pair (x,y) for which, {x,y} is a subset of {1,4,7}, will be an element of R₂.
(ii) Any ordered pair (x,y) for which, {x,y} is a subset of {2,5,8}, will be an element of R₂.
(iii) Any ordered pair (x,y) for which, {x,y} is a subset of {3,6,9}, will be an element of R₂.
3. Consider the sets {1,4,7}.
• We can take any two elements from this set. The difference between those two elements will be divisible by 3.
• Similar is the case with the other two sets {2,5,8} and {3,6,9}
• Union of the three sets will give X.
• So R₁ will be a subset of R₂
4. While writing the ordered pairs of R₂, we see that, difference between the members of each ordered pair is divisible by 3.
• So R₂ will be a subset of R₁.
5. Now we can compare the results.
   ♦ In (3), we see that: R₁ ⊂ R₂.
   ♦ In (4), we see that: R₂ ⊂ R₁.
• So we can write: R₁ = R₂

Solved example 17.44
Let f: X→Y be a function. Define a relation R in X given by R = {(a,b): f(a) = f(b)}. Examine whether R is an equivalence relation or not.
Solution:
1. Let X = {x₁, x₂, x₃, . . . } and Y = {y₁, y₂, y₃, . . .}
2. Then a possible example set for f is {(x₁, y₅), (x₂, y₈), (x₃, y₁₁), . . .}
• This means:
    ♦ When x₁ is the input, the function f gives y₅ as the output.
    ♦ When x₂ is the input, the function f gives y₈ as the output.
    ♦ When x₃ is the input, the function f gives y₁₁ as the output.
so on . . .
3. Now we can write the set R.
• Set R will contain ordered pairs of the form (a,b).
• Consider any one ordered pair (a,b). That ordered pair is eligible to be in R because, f(a) = f(b).
• So both a and b will be from set X. This is because, all inputs of f are taken from X.
4. So R is a relation on X.
• That means, the elements of R are taken from the set X×X.
• So a random ordered pair taken from R will be (x₅,x₉)
• This ordered pair is eligible to be in R because, f(x₅) = f(x₉)
5. Now we check whether R is reflexive.
• A possible random element in X×X is (x₅,x₅)
• This element will satisfy the condition f(a) = f(b).
This is because, f(x₅) = f(x₅)
• So all reflexive pairs will be present in R.
Therefore, R is reflexive.
6. Next we check whether R is symmetric.
• If a random pair (x₅,x₉) is present in R, then the symmetric pair (x₉,x₅) will also be present in R.
This is because:
f(x₅) = f(x₉) ⇒ f(x₉) = f(x₅)
• So all symmetric pairs will be present in R.
Therefore, R is symmetric.
7. Finally we check whether R is transitive.
• Consider two random pairs from R: (x₅,x₉) and (x₉,x₂)
• Will the transitive pair (x₅,x₂) be present in R?
• Since both (x₅,x₉) and (x₉,x₂) are present in R, we can write:
f(x₅) = f(x₉) = f(x₂)
• So it is clear that, the transitive pair (x₅,x₂) will be present in R
Therefore, R is transitive.
◼ Since R is reflexive, symmetric and, transitive, it is an equivalence relation.

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Solved example 17.45
Determine which of the following binary operations on the set N are associative and which are commutative.
$(a)~a * b = 1 ~ \forall ~ a,b \in N~~~~~(b)~a * b = \frac{a+b}{2} ~ \forall ~ a,b \in N$
Solution:
Part (i):
1. Checking whether commutative or not.
• For an operation to be commutative, the condition which should be satisfied is:
(a∗b) = (b∗a)
• For our present case, a and b should be natural numbers.
• It is given that, if we perform the operation '*' between any two natural numbers a and b, then the result will be 1.
• So the same operation between b and a will also give 1.
• That means, (a∗b) = (b∗a)
• So the given ∗ is a commutative binary operation.
2. Checking whether associative or not.
• For an operation to be associative, the condition which should be satisfied is:
(a∗b)∗c = a∗(b∗c)
• For our present case, a, b and c should be natural numbers.
• First we calculate (a∗b)∗c:
(a ∗ b) = 1
So (a∗b)∗c = (1∗c) = 1
• Next we calculate a∗(b∗c):
(b ∗ c) = 1
So a∗(b∗c) = (a∗1) = 1
• We see that: (a∗b)∗c = a∗(b∗c)
So the given ∗ is an associative binary operation.
◼ Based on the above 2 steps, we can write:
The given ∗ is both commutative and associative.

Part (ii):
1. Checking whether commutative or not.
• For an operation to be commutative, the condition which should be satisfied is:
(a∗b) = (b∗a)
• For our present case, a and b should be natural numbers.
• For any two natural numbers, $\frac{a+b}{2}$ will be equal to $\frac{b+a}{2}$ .
• That means, (a∗b) = (b∗a)
• So the given ∗ is a commutative binary operation.
2. Checking whether associative or not.
• For an operation to be associative, the condition which should be satisfied is:
(a∗b)∗c = a∗(b∗c)
• For our present case, a, b and c should be natural numbers.
• First we calculate (a∗b)∗c:
$(a*b)~=~\frac{a+b}{2}$
$\text{So}~(a*b)*c~=~\left(\frac{a+b}{2}\right) * c~=~\frac{\frac{a+b}{2}~+~c}{2}~=~\frac{a+b+2c}{4}$
• Next we calculate a∗(b∗c):
$(b*c)~=~\frac{b+c}{2}$
$\text{So}~a*(b*c)~=~a * \left(\frac{b+c}{2}\right)~=~\frac{a~+~\frac{b+c}{2}}{2}~=~\frac{2a+b+c}{4}$
• We see that: (a∗b)∗c ≠ a∗(b∗c)
So the given ∗ is not an associative binary operation.
◼ Based on the above 2 steps, we can write:
The given ∗ is commutative but not associative.

Solved example 17.46
Find the number of all one-one functions from set A = {1,2,3} to itself.
Solution:
1. Take any one element from the domain. It can be mapped to the codomain in 3 different ways.
2. Take a second element from the domain. It can be mapped to the codomain in 2 ways. This is because, the way chosen in step 1 should not be repeated. For example, (1,2) and (2,2) cannot be selected. Then it would not be a one-one function.
3. Take the third element from the domain. It can be mapped to the codomain in one way.
4. So the total number of ways = 3 × 2 × 1 = 3! = 6

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

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17.4 - Composition of Functions

In the previous section, we completed a discussion on one-one and onto functions. In this section, we will see Composition of functions.

Composition of functions

Composition of functions can be explained using an example. It can be written in 8 steps:
1. The display board of a textile shop reads: 14% discount on all items.
2. A person wants to buy a shirt. The marked price of the shirt is Rs 400/-. He wants to know whether he can afford it. He has to consider both the discount and the tax. The tax is to be paid at the rate of 8%.
3. For solving this problem, first we consider the discount.
• We know that:
If x is the marked price, then the cost after 14% discount will be 0.86x
• We can write this as a function: f(x) = 0.86x
This function can be used for all items in that shop.
    ♦ Input ‘x’ is the marked price.
    ♦ Output ‘f(x)’ is the cost after discount.
• In our present case, we can put x₁ = 400
So the cost after discount = f(x₁) = 0.86 × 400 = 344
4. Next we consider the tax.
• We know that:
If x is the cost of an item, the payment to be made considering 8% tax will be 1.08x
• We can write this as a new function: g(x) = 1.08x
This function can be used for all items in that shop.
    ♦ Input ‘x’ is the cost after discount.
    ♦ Output ‘g(x)’ is the payment to be made.
    ♦ In our present case, we can put x₂ = 344
So the payment = g(x₂) = 1.08 × 344 = 371.52
5. Note that, we did two separate calculations:
(i) In the first calculation, we found out the cost after discount.
(ii) In the second calculation, we found out the payment to be made after considering the tax.
6. The two calculations can be done together. It can be explained in 4 steps:
(i) Output of the first function is the input of the second function.
• Output of the first function is 0.86x. Here x is the marked price.
(ii) So input of the second function is 0.86x
• The second function is: g(x) = 1.08x
So we get: g(0.86x) = 1.08(0.86x) = 0.9288x
• Check:
0.9288 × 400 = 371.52
This is the same result that we obtained in (4)
(iii) Now we can write the general form:
• From the above steps, it is clear that:
    ♦ Output of f
    ♦ is the
    ♦ Input of g  
• Using symbols, this can be written in two ways:
    ♦ (g∘f)(x)
    ♦ g(f(x))
We can use any one of the above two ways. Both are read as "g of f of x"
(iv) Note that, the small circle between g and f should not be a filled circle. Then it would look like a dot. If it is a dot, then it means multiplication.
7. (g∘f)(x) is actually a function. It can be plotted just like any other function. The plot in our present case is the magenta line in fig.17.6 below:

Fig.17.6

An explanation of the graph can be written in 5 steps:
(i) Draw the three graphs:
    ♦ f(x) = 0.86x
    ♦ g(x) = 1.08x
    ♦ (g∘f)(x) = 0.9288x
(ii) We want the payment amount when the marked price is 400. So draw a vertical cyan dashed line through 400 on the x-axis.
• This vertical line meets f(x) at a point. Draw a horizontal cyan dashed line through this meeting point.
• This horizontal cyan dashed line meets the y-axis at 344. This is the price after discount.
(iii) We want the payment amount when the cost is 344. So draw a vertical brown dashed line through 344 on the x-axis.
• This vertical line meets g(x) at a point. Draw a horizontal brown dashed line through this point.
• This horizontal brown dashed line meets the y-axis at 371.52. This is the payment amount.
(iv) The horizontal cyan dashed line and vertical brown dashed line, together with the axes, form a square. Why?
(v) In practice, we need not plot f(x) and g(x). All we need is the plot of (g∘f)(x).
• This is because:
The vertical dashed line through 400 meets the magenta line at the white dot. The horizontal dashed line through this white dot meets the y-axis at the same 371.52.
8. (g∘f)(x) can be explained using set theory also. It can be written in 5 steps:
(i) We write set M which contains the marked prices of the various items in the shop. This is shown in fig.17.7(a) below:

Fig.17.7

 
(ii) Each element of M becomes input of the function f. The outputs are the costs after discount. These outputs become elements of set D. Let us see an example:
• 735 is an input for function f. We get: f(735) = 0.86 × 735 = 632.1
• So there is an arrow connecting 735 (in set M) and 632.1 (in set D).
• We can define the function f using sets M and D:
f: M→D defined as f(x) = 0.86x
(iii) Each element of D becomes input of the function g. The outputs are the payment amounts. These outputs become elements of set P. Let us see an example:
• 632.1 is an input for function g. We get: g(632.1) = 1.08 × 632.1 = 682.668
• So there is an arrow connecting 632.1 (in set D) and 682.668 (in set P).
• We can define the function g using sets D and P:
g: D→P defined as g(x) = 1.08x
(iv) Instead of applying two separate functions, we can apply the single function (g∘f). This is shown in fig.17.7(b) above. Let us see an example:
• (g∘f)(735) = 0.9288 × 735 = 682.668
We get the same result as above.
• We see an arrow directly connecting 725 (in set M) and 682.668 (in set P).
• We can define the function (g∘f) using sets M and P:
(g∘f): M→P defined as (g∘f)(x) = 0.9288x
• Note that, M is the domain of the first function and P is the codomain of the second function. So we can write:
The composite function connects the domain of the first function to the codomain of the second function.
(v) So it is clear that, by using composite functions, we can obtain the answers in a single step.    

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Solved example 17.15
Let f : {2, 3, 4, 5} → {3, 4, 5, 9} and g : {3, 4, 5, 9} → {7, 11, 15} be functions defined as:
f(2) = 3,
f(3) = 4,
f(4) = f(5) = 5
and
g(3) = g(4) = 7,
g(5) = g(9) = 11.
Find g∘f.
Solution:
1. We can name the given sets as shown below:
A = {2, 3, 4, 5}
B = {3, 4, 5, 9}
C = {7, 11, 15}
• So the functions can be indicated as:
f : A → B
g : B → C
2. The Venn diagram is shown in fig.17.8(a) below:

Fig.17.8

Let us see an example from the diagram.
• Given that, f(3) = 4,
So there is an arrow connecting 3 (in set A) and 4 (in set B).
• Given that, g(4) = 7,
So there is an arrow connecting 4 (in set B) and 7 (in set C).
3. By following the arrows in fig.a, we get the connection between elements in A and C. This is shown in fig.b above.
• So we can write:
(g∘f)(2) = 7
(g∘f)(3) = 7
(g∘f)(4) = 11
(g∘f)(5) = 11

Alternate method:
(g∘f) means, "output of f" is used as "input of g".
1. Given that, f(2) = 3
So output of f for the input 2 is 3
• This 3 is the input for g.
• Given that, g(3) = 7
So when the input for (g∘f) is 2, the output will be 7
We can write: (g∘f)(2) = 7
2. Given that, f(3) = 4
• Given that, g(4) = 7
So (g∘f)(3) = 7
3. Given that, f(4) = 5
• Given that, g(5) = 11
So (g∘f)(4) = 11
4. Given that, f(5) = 5
• Given that, g(5) = 11
So (g∘f)(5) = 11

Solved example 17.16
Find g∘f and f∘g, if f : R → R and g : R → R are given by f(x) = cos x and g(x) = 3x² . Show that g∘f ≠ f∘g.
Solution:
Part (i): Finding g∘f
1. g∘f means, "output of f" is used as "input of g".
2. Output of f is cos x.
So we can write:
g∘f = g(cos x) = 3(cos x)² = 3 cos²x

Part (ii): Finding f∘g
1. f∘g means, "output of g" is used as "input of f".
2. Output of g is 3x².
So we can write:
f∘g = f(3x²) = cos (3x²)

Part (iii): proving that, g∘f ≠ f∘g
1. We have: g∘f = 3 cos²x
Put x = 0. We get:
g∘f = 3 cos²0  = 3 × 1² = 3
2. We have f∘g = cos (3x²)
Put x = 0. We get:
f∘g = cos (3 × 0) = cos 0 = 1
3. So we can write: g∘f ≠ f∘g

Solved example 17.17
Show that,
• if $f: R- \left \{\frac{7}{5} \right\}~\to~R- \left \{\frac{3}{5} \right\}$ is defined by $f(x)~=~\frac{3x+4}{5x-7}$
• and $g: R- \left \{\frac{3}{5} \right\}~\to~R- \left \{\frac{7}{5} \right\}$ is defined by $f(x)~=~\frac{7x+4}{5x-3}$,
• then f∘g = I_A and g∘f = I_B,
• where
    ♦ $A = R- \left \{\frac{3}{5} \right\}$
    ♦ $B = R- \left \{\frac{7}{5} \right\}$
    ♦ $I_A (x) = x, \forall x \in A$
    ♦ $I_B (x) = x, \forall x \in B$
Solution:
Details about identity function can be seen here.

Part (i): To prove that (f∘g)(x) = I_A(x).
1. First we find (f∘g)
• (f∘g) means, "output of g" is used as "input of f".
Output of g is $\frac{7x+4}{5x-3}$.
So we get:
$$\begin{array}{ll}{}    &{(f \circ g)(x)}    & {~=~}    &{f \left(\frac{7x+4}{5x-3} \right)}    &{} \\
{}    &{}    & {~=~}    &{\frac{3\left(\frac{7x+4}{5x-3} \right)+4}{5\left(\frac{7x+4}{5x-3} \right)-3}}    &{} \\
{}    &{}    & {~=~}    &{\frac{21x+12 + 20x – 12}{35x + 20 -35x + 21}}    &{} \\
{}    &{}    & {~=~}    &{\frac{41x}{41}}    &{} \\
{}    &{}    & {~=~}    &{x}    &{} \\
\end{array}               
$$
2. We see that, (f∘g)(x) is an identity function. This is because, what ever input x we give, the output will be same as the input.
• But we need to specify the the set from which the input values can be taken. In other words, we need to specify the domain.
• We wrote that: $(f \circ g)(x)~=~f \left(\frac{7x+4}{5x-3} \right)$
• When x is $\frac{3}{5}$, the denominator will become zero. So x can be any real number other that $\frac{3}{5}$.
• That means, the domain is $R- \left \{\frac{3}{5} \right\}$
• But we are given that $A = R- \left \{\frac{3}{5} \right\}$.
So the domain is set A.
• We can write:
(f∘g)(x) is an identity function whose domain is A.
• Using symbols, we write this as:
(f∘g)(x) = I_A(x).

Part (ii): To prove that (g∘f)(x) = I_B(x)
1. First we find (g∘f)
• (g∘f) means, "output of f" is used as "input of g".
Output of f is $\frac{3x+4}{5x-7}$.
So we get:
$$\begin{array}{ll}{}    &{(g \circ f)(x)}    & {~=~}    &{g \left(\frac{3x+4}{5x-7} \right)}    &{} \\
{}    &{}    & {~=~}    &{\frac{7\left(\frac{3x+4}{5x-7} \right)+4}{5\left(\frac{3x+4}{5x-7} \right)-3}}    &{} \\
{}    &{}    & {~=~}    &{\frac{21x+28 + 20x – 28}{15x + 20 -15x + 21}}    &{} \\
{}    &{}    & {~=~}    &{\frac{41x}{41}}    &{} \\
{}    &{}    & {~=~}    &{x}    &{} \\
\end{array}               
$$
2. We see that, (g∘f)(x) is an identity function. This is because, what ever input x we give, the output will be same as the input.
• But we need to specify the the set from which the input values can be taken. In other words, we need to specify the domain.
• We wrote that: $(g \circ f)(x)~=~g \left(\frac{3x+4}{5x-7} \right)$
• When x is $\frac{7}{5}$, the denominator will become zero. So x can be any real number other that $\frac{7}{5}$.
• That means, the domain is $R- \left \{\frac{7}{5} \right\}$
• But we are given that $B = R- \left \{\frac{7}{5} \right\}$.
So the domain is set B.
• We can write:
(g∘f)(x) is an identity function whose domain is B.
• Using symbols, we write this as:
(g∘f)(x) = I_B(x)

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In the next section, we will see a few more solved examples. We will also see some important properties of composite functions.

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Chapter 17 - Relations And Functions

In the previous section, we completed appendix B. With that, we have completed all topics in class 11. In this chapter, we will see Relations and Functions which is the first chapter in class 12.

• In class 11, we saw some basic details about relations and functions. See chapter 2.
• Let us recall five important points:
(i) A and B are two sets.
(ii) A × B is a new set formed from A and B. This new set will contain all possible ordered pairs between the elements of A and B.
(iii) The ordered pairs are written in the form (a,b).
   ♦ a is from set A
   ♦ b is from set B
(iv) We pick out some of those ordered pairs and form a new set R.
(v) Take any ordered pair from R. There will be a definite relation between the a and the b of that ordered pair.
• For example,
   ♦ a is the brother of b.
   ♦ a is 2 less than b.
• We write this as a R b.

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In this section, we will learn more details about relations and functions.

Types of relations

First we will see empty relation. It can be explained in 4 steps:
1. We have seen the relation from set A to itself. It is written as R in A.
2. Relation in A will be a subset of A × A
• But Φ (empty set) is a subset of any set. So there exists a relation R in A such that R is a empty set.
3. Let us see an example. It can be written in 4 steps:
(i) Let A = {1,2,3,4}
(ii) R in A in set builder form is:
R = {(a,b): a-b = 10}
(iii) a-b = 10 is same as b = a-10
• Take the first element 1. Ten subtracted from 1 is -9. This -9 is not present in A. So when a = 1, we cannot write b. In other words, no ordered pair in R can have a = 1.
• Take the second element 2. Ten subtracted from 2 is -8. This -8 is not present in A. So when a = 2, we cannot write b. In other words, no ordered pair in R can have a = 2.
• Similarly, we will find that:
   ♦ No ordered pair in R can have a = 3
   ♦ No ordered pair in R can have a = 4
(iv) None of the elements in A can become 'a' of the ordered pairs in R. So there will be no element in R. In other words, R is an empty set.
4. If R = Φ, then that relation is called an empty relation.

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Now we will see universal relation. It can be explained in 4 steps:
1. We have seen the relation from set A to itself. It is written as R in A.
2. Relation in A will be a subset of A × A
• But a set itself is a subset of any set. So there exists a relation R in A such that R is A × A.
3. Let us see an example. It can be written in 4 steps:
(i) Let A = {1,2,3,4}
(ii) R' in A in set builder form is:
R' = {(a,b): |a-b| ≥ 0}
(iii) A has four elements. So A × A will have 16 elements.
• Take the first element (1,1). Here a = 1 and b = 1
|a-b| = |1-1| = |0| = 0
So (1,1) is eligible to be included in the set R'.
• Take the second element (1,2). Here a = 1 and b = 2
|a-b| = |1-2| = |-1| = 1 ≥ 0
So (1,2) is eligible to be included in the set R'.
• Take the third element (1,3). Here a = 1 and b = 3
|a-b| = |1-3| = |-2| = 2 ≥ 0
So (1,3) is eligible to be included in the set R'.   
• Take any one of the 16 elements, say (4,1). Here a = 4 and b = 1
|a-b| = |4-1| = |3| = 3 ≥ 0
So (4,1) is eligible to be included in the set R'
• In this way we can check all the 16 elements. We will see that, all of them are eligible to be included in the set R'.
(iv) We can write: R' = A × A
4. If R' = A × A, then that relation is called a universal relation.
• Here every element in A has the relation R' with every other element of A.

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Both the empty relation and the universal relation are some times called trivial relations.

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Solved example 17.1
A is the set of all students of a boys school. Show that the relation R in A given by R={(a,b) : a is sister of b} is the empty relation and R’ = {(a,b) : the difference between heights of a and b is less than 3 meters} is the universal relation.
Solution:
Part (i):
1. We can try to write the elements in R. Those elements will be in the form (a,b)
‘a’ will be from set A. ‘b’ will also be from set A.
2. Set A contains only boys.
So ‘a’ can never be the sister of ‘b’.
3. As a consequence, we will not be able to write a single element of the form (a,b). That means, R is an empty set.
• So the relation R is an empty relation.
Part (ii):
1. Heights greater than 2 m are very rare. We can safely assume that, the tallest student has a height of 2.5 m.
2. Heights smaller than 1.5 are very rare. We can safely assume that, the shortest student has a height of 1 m.
3. So the maximum possible difference between the heights is (2.5 – 1) = 1.5 m.
• None of the “differences” will be greater than 1.5 m.
• Consequently, none of the “differences” will be greater than 3 m.
• In other words, all differences will be less than 3 m.
4. We can take any (a,b) from A × A.
• The difference in that pair will be less than 3 m.
• So all elements in A × A are eligible to be included in R`
• Thus we get R` = A × A.
• That means, R` is a universal relation.

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Next we will see reflexive relation. It can be explained in 3 steps:
1. Let A = {1,2,3,4}
• We know that there will be 16 elements in A × A.
• (1,1), (2,2), (3,3) and (4,4) will be among those 16 elements.
2. Suppose that, there is a relation R in A in such a way that, (1,1), (2,2), (3,3) and (4,4) are elements of R.
• Then that relation is called a reflexive relation.
3. We can write the definition in two steps:
(i) R is a relation in A
(ii) For every a ∈ A, if (a,a) ∈ R, then R is a reflexive relation.

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Next we will see symmetric relation. It can be explained in 3 steps:
1. Let A = {1,2,3,4}
• We know that there will be 16 elements in A × A.
• (1,2), (2,1), (1,3), (3,1) etc., will be among those 16 elements.
2. Suppose that, there is a relation R in A which satisfies the following conditions:
• If (1,2) is an element of R, then (2,1) is also an element of R.
• If (1,3) is an element of R, then (3,1) is also an element of R.

- - - - 

• If (2,4) is an element of R, then (4,2) is also an element of R.

so on . . .

• If this is true for all such pairs, then that relation is called a symmetric relation.
3. We can write the definition in two steps:
(i) R is a relation in A
(ii) For all a₁, a₂ ∈ A, if (a₁,a₂) ∈ R ⇒ (a₂,a₁) ∈ R, then R is a symmetric relation.

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Next we will see transitive relation. It can be explained in 3 steps:
1. Let A = {1,2,3,4}
• We know that there will be 16 elements in A × A.
• (1,2), (2,3), (1,3), (3,4) etc., will be among those 16 elements.
2. Suppose that, there is a relation R in A which satisfies the following conditions:
• If (1,2) and (2,3) are elements of R, then (1,3) is also an element of R.
• If (1,3) and (3,4) are elements of R, then (1,4) is also an element of R.

- - - - 

• If (2,4) and (4,3) are elements of R, then (2,3) is also an element of R.

so on . . .

• If this is true for all such pairs, then that relation is called a transitive relation.
3. We can write the definition in two steps:
(i) R is a relation in A
(ii) For all a₁, a₂, a₃∈ A,
if (a₁,a₂) ∈ R and (a₂,a₃) ∈ R ⇒ (a₁,a₃) ∈ R, then R is a transitive relation.

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Now we can write the definition of an equivalence relation. It can be written in 2 steps:
1. R is a relation in A.
2. This R is an equivalence relation if all three conditions below are satisfied.
(i) R is a reflexive relation.
(ii) R is a symmetric relation.
(iii) R is a transitive relation.

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Solved example 17.2
Let T be the set of all triangles in a plane. R is a relation in T.
R = {(T₁,T₂) : T₁ is congruent to T₂}. Show that R is an equivalence relation.
Solution:
• Congruent triangles are those triangles which have the same sides and same corresponding angles.
• Given that, T is the set of all triangles in a plane. There will be infinite number of triangles in that set. Let us number them as: 1, 2, 3, 4, . . .
All those triangles will be present in T.
• R is a relation in T. So we must consider T × T
• (1,1), (1,2), (1,3), (1,4), . . . , (2,1), (2,2), (2,3), . . .  are all elements of T × T.
There will be infinite elements in T × T
1. First we check whether R is reflexive. It can be written in 3 steps:
(i) (1,1) is eligible to be included in R. This is because, the first triangle is congruent to itself. In fact, any triangle is congruent to itself.
(ii) In this way, (2,2), (3,3), (4,4), . . . are eligible to be included in R.
(iii) Therefore, R is a reflexive function.
2. Now we check whether R is symmetric. It can be written in 3 steps:
(i) Suppose that (3,9) is eligible to be included in R.
• Then it means that, the third triangle is congruent to the ninth triangle.
(ii) Now, (9,3) is an element of T × T.
• The element (9,3) is eligible to be included in R. This is because:
If third triangle is congruent to the ninth triangle, then ninth triangle will be congruent to the third.
(iii) So in general, if (T₁, T₂) is an element of R, then (T₂,T₁) will also be an element of R
• Therefore, R is a symmetric relation.
3. Now we check whether R is transitive. It can be written in 4 steps:
(i) Suppose that (4,7) is eligible to be included in R.
• Then it means that, the fourth triangle is congruent to the seventh triangle.
(ii) Also suppose that (7,10) is eligible to be included in R.
• Then it means that, the seventh triangle is congruent to the tenth triangle.
(iii) It is clear that, fourth, seventh and tenth triangles are congruent.
Then fourth triangle is congruent to the tenth triangle.
So (4,10) is eligible to be included in R.
(iv) In general, if both (T₁,T₂) and (T₂,T₃) are elements of R, then (T₁,T₃) will also be an element of R.
Therefore R is a transitive relation.
4. We see that:
R is reflexive, symmetric and transitive. So R is an equivalence relation.   

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

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Chapter 13.1 - When Limit At A Point Does Not Exist

In the previous section, we saw two examples which demonstrated the basics of limits. In this section, we will see a few more examples.

Example 3:
This can be written in 3 steps:
1. Fig.13.3 below shows the graph of the function y = f(x) = x²

Fig.13.3

2. Let us find the limit of f(x) as x approaches 0 from the left. For that, we prepare table 13.5 below:

Table 13.5

$\begin{array}{cc}
{}&{\color{green}x}
& {-2}& {\color{green}{-1}}& {-0.5}& {\color{green}{-0.1}}& {-0.01}& {\color{green}{-0.001}}& {-0.0001}
&{} \\

{}&{\color{green}{f(x)}}
& {4}& {\color{green}{1}}& {0.25}& {\color{green}{0.01}}& {0.0001}& {\color{green}{10^{-6}}}& {10^{-8}}
&{} \\

\end{array}$

• It is clear that, as x become closer and closer to 0 from the left, f(x) becomes closer and closer to 0
• We can write:
$$\lim_{x\rightarrow 0^{-}} f(x) = 0$$      

3. Let us find the limit of f(x) as x approaches 0 from the right. For that, we prepare table 13.6 below:

Table 13.6 

$\begin{array}{cc}
{}&{\color{green}x}
& {2}& {\color{green}{1}}& {0.5}& {\color{green}{0.1}}& {0.01}& {\color{green}{0.001}}& {0.0001}
&{} \\

{}&{\color{green}{f(x)}}
& {4}& {\color{green}{1}}& {0.25}& {\color{green}{0.01}}& {0.0001}& {\color{green}{10^{-6}}}& {10^{-8}}
&{} \\

\end{array}$

• It is clear that, as x become closer and closer to 0 from the right, f(x) becomes closer and closer to 0
• We can write:
$$\lim_{x\rightarrow 0^{+}} f(x) = 0$$

---

• In this example, we see that:
   ♦ Left side limit of f(x) at 0 is 0
   ♦ Right side limit of f(x) at 0 is 0
   ♦ Value of f(x) at 0 is also 0
         ✰ That means, limit at 0 can be obtained by putting x = 0.
         ✰ f(x) = x²
         ✰ f(0) = 0² = 0  

---

Example 4

This can be written in 3 steps:
1. Fig.13.4 below shows the graph of the function: f(x) = 3

Fig.13.4

2. Let us find the limit of f(x) as x approaches 0 from the left. For that, we prepare table 13.7 below:

Table 13.7

$\begin{array}{cc}
{}&{\color{green}x}
& {-2}& {\color{green}{-1}}& {-0.5}& {\color{green}{-0.1}}& {-0.01}& {\color{green}{-0.001}}& {-0.0001}
&{} \\

{}&{\color{green}{f(x)}}
& {3}& {\color{green}{3}}& {3}& {\color{green}{3}}& {3}& {\color{green}{3}}& {3}
&{} \\

\end{array}$ 

• It is clear that:
   ♦ x become closer and closer to 0 from the left
   ♦ f(x) is 3 for all those increasing values of x
• We can write:
$$\lim_{x\rightarrow 0^{-}} f(x) = 3$$
◼ In fact, we can write:
$$\lim_{x\rightarrow a^{-}} f(x) = 3$$
   ♦ Where 'a' is any real number

• The reader may think that:
   ♦ f(x) is not approaching '3'.
   ♦ '3' is already present.
• This thought is correct.
Even then, we can consider 3 as the limit because, as x approaches 0, there is no value other than '3'. Also, the value '3' will not be exceeded.
    

3. Let us find the limit of f(x) as x approaches 0 from the right. For that, we prepare table 13.8 below:

Table 13.8

$\begin{array}{cc}
{}&{\color{green}x}
& {2}& {\color{green}{1}}& {0.5}& {\color{green}{0.1}}& {0.01}& {\color{green}{0.001}}& {0.0001}
&{} \\

{}&{\color{green}{f(x)}}
& {3}& {\color{green}{3}}& {3}& {\color{green}{3}}& {3}& {\color{green}{3}}& {3}
&{} \\

\end{array}$  

• It is clear that:
   ♦ x become closer and closer to 0 from the right
   ♦ f(x) is 3 for all those decreasing values of x
• We can write:
$$\lim_{x\rightarrow 0^{+}} f(x) = 3$$
◼ In fact, we can write:
$$\lim_{x\rightarrow a^{+}} f(x) = 3$$

   ♦ Where 'a' is any real number

---

• In this example, we see that:
   ♦ Left side limit of f(x) at 0 is 3
   ♦ Right side limit of f(x) at 0 is 3
   ♦ Value of f(x) at 0 is also 3
         ✰ That means, limit at 0 can be obtained by putting x = 0.
         ✰ f(x) = 3
         ✰ f(0) = 3  

---

Example 5

This can be written in 3 steps:
1. Fig.13.5 below shows the graph of the function:

$$f(x) =
\begin{cases}
1,  & {x \le 0} \\[2ex]
2, & {x > 0}
\end{cases}$$

Fig.13.5

 

• Note that:
   ♦ the point (0,1) is marked with a filled circle.
   ♦ the point (0,2) is marked with an ordinary circle.
• We saw the significance of such markings in fig.13.2 of the previous section.
2. Let us find the limit of f(x) as x approaches 0 from the left. For that, we prepare table 13.9 below:

Table 13.9

$\begin{array}{cc}
{}&{\color{green}x}
& {-2}& {\color{green}{-1}}& {-0.5}& {\color{green}{-0.1}}& {-0.01}& {\color{green}{-0.001}}& {-0.0001}
&{} \\

{}&{\color{green}{f(x)}}
& {1}& {\color{green}{1}}& {1}& {\color{green}{1}}& {1}& {\color{green}{1}}& {1}
&{} \\

\end{array}$  

• It is clear that:
   ♦ x become closer and closer to 0 from the left
   ♦ f(x) is 1 for all those increasing values of x
• We can write:
$$\lim_{x\rightarrow 0^{-}} f(x) =1$$      

3. Let us find the limit of f(x) as x approaches 0 from the right. For that, we prepare table 13.10 below:

Table 13.10

$\begin{array}{cc}
{}&{\color{green}x}
& {2}& {\color{green}{1}}& {0.5}& {\color{green}{0.1}}& {0.01}& {\color{green}{0.001}}& {0.0001}
&{} \\

{}&{\color{green}{f(x)}}
& {2}& {\color{green}{2}}& {2}& {\color{green}{2}}& {2}& {\color{green}{2}}& {2}
&{} \\

\end{array}$  

• It is clear that:
   ♦ x become closer and closer to 0 from the right
   ♦ f(x) is 2 for all those decreasing values of x
• We can write:
$$\lim_{x\rightarrow 0^{+}} f(x) = 2$$

---

• In this example, we see that:
   ♦ Left side limit of f(x) at 0 is 1
   ♦ Right side limit of f(x) at 0 is 2
   ♦ Left side and right side limits are different.
   ♦ In such cases, we say that: limit does not exist.
• In this example,
   ♦ The limit as x approaches zero does not exist.
   ♦ But other limits exist.
         ✰ For example: the limit as x approaches 5 is 2 
         ✰ Another example: the limit as x approaches -7 is 1 

---

Example 6

This can be written in 3 steps:
1. Fig.13.6 below shows the graph of the function:

$$f(x) =
\begin{cases}
x-2,  & {x < 0} \\[2ex]
0,  & {x = 0} \\[2ex]
x+2, & {x > 0}
\end{cases}$$

Fig.13.6

• Note that:
   ♦ the point (0,2) is marked with an ordinary circle.
   ♦ the point (0,-2) is also marked with an ordinary circle.
• We saw the significance of such markings in fig.13.2 of the previous section.
2. Let us find the limit of f(x) as x approaches 0 from the left. For that, we prepare table 13.11 below:

Table 13.11

$\begin{array}{cc}
{}&{\color{green}x}
& {-2}& {\color{green}{-1}}& {-0.5}& {\color{green}{-0.1}}& {-0.01}& {\color{green}{-0.001}}& {-0.0001}
&{} \\

{}&{\color{green}{f(x)}}
& {-4}& {\color{green}{-3}}& {-2.5}& {\color{green}{-2.1}}& {-2.01}& {\color{green}{-2.001}}& {-2.0001}
&{} \\

\end{array}$  

• It is clear that, as x become closer and closer to 0 from the left, f(x) becomes closer and closer to -2
• We can write:
$$\lim_{x\rightarrow 0^{-}} f(x) = -2$$      

3. Let us find the limit of f(x) as x approaches 0 from the right. For that, we prepare table 13.12 below:

Table 13.12

$\begin{array}{cc}
{}&{\color{green}x}
& {2}& {\color{green}{1}}& {0.5}& {\color{green}{0.1}}& {0.01}& {\color{green}{0.001}}& {0.0001}
&{} \\

{}&{\color{green}{f(x)}}
& {4}& {\color{green}{3}}& {2.5}& {\color{green}{2.1}}& {2.01}& {\color{green}{2.001}}& {2.0001}
&{} \\

\end{array}$

• It is clear that, as x become closer and closer to 0 from the right, f(x) becomes closer and closer to 2
• We can write:
$$\lim_{x\rightarrow 0^{+}} f(x) = 2$$

---

• In this example, we see that:
   ♦ Left side limit of f(x) at 0 is -2
   ♦ Right side limit of f(x) at 0 is 2
   ♦ Left side and right side limits are different.
   ♦ In such cases, we say that: limit does not exist
• In this example,
   ♦ The limit as x approaches zero does not exist.
   ♦ But other limits exist.
         ✰ For example: the limit as x approaches 5 is 7 
         ✰ Another example: the limit as x approaches -8 is -10 

---

Example 7

This can be written in 3 steps:
1. Fig.13.7 below shows the graph of the function:
$$f(x) =
\begin{cases}
x+2,  & {x \ne 1} \\[2ex]
0,  & {x = 1}
\end{cases}$$

Fig.13.7

• Note that:
   ♦ the point (1,0) is marked with a filled circle.
   ♦ the point (1,3) is marked with an ordinary circle.
• We saw the significance of such markings in fig.13.2 of the previous section.
2. Let us find the limit of f(x) as x approaches 0 from the left. For that, we prepare table 13.13 below:

Table 13.13

$\begin{array}{cc}
{}&{\color{green}x}
& {-1}& {\color{green}{0.5}}& {0}& {\color{green}{0.5}}& {0.9}& {\color{green}{0.99}}& {0.999}
&{} \\

{}&{\color{green}{f(x)}}
& {1}& {\color{green}{1.5}}& {2}& {\color{green}{2.5}}& {2.9}& {\color{green}{2.99}}& {2.999}
&{} \\

\end{array}$  

• It is clear that, as x become closer and closer to 1 from the left, f(x) becomes closer and closer to 3
• We can write:
$$\lim_{x\rightarrow 1^{-}} f(x) = 3$$      

3. Let us find the limit of f(x) as x approaches 1 from the right. For that, we prepare table 13.14 below:

Table 13.14

$\begin{array}{cc}
{}&{\color{green}x}
& {3}& {\color{green}{2}}& {1.5}& {\color{green}{1.1}}& {1.01}& {\color{green}{1.001}}& {1.0001}
&{} \\

{}&{\color{green}{f(x)}}
& {5}& {\color{green}{4}}& {3.5}& {\color{green}{3.1}}& {3.01}& {\color{green}{3.001}}& {3.0001}
&{} \\

\end{array}$

• It is clear that, as x become closer and closer to 1 from the right, f(x) becomes closer and closer to 3
• We can write:
$$\lim_{x\rightarrow 1^{+}} f(x) = 3$$

---

• In this example, we see that:
   ♦ Left side limit of f(x) at 1 is 3
   ♦ Right side limit of f(x) at 1 is 3
   ♦ Value of f(x) at 1 is not 3

---

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

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Chapter 3.11 - Trigonometric Identities

In the previous section, we saw the domain and range of tangent and cotangent functions. In this section, we will derive some trigonometric identities.

First we will derive two simple identities. They can be derived in 21 steps:
1. In fig.3.34(a) below, a ray has rotated by x^c in the +ve direction.
• Another ray has rotated by the same x^c in the negative direction.

Fig.3.34

2. Since the angles are same, the points P, Q and M are on the same vertical line.
• The x axis bisects the base PQ of the triangle OPQ
• Thus we get: QM = PM = b
3. So the coordinates of Q are: (a, -b)
• But x coordinate is cosine and y coordinate is sine.
• So we get:
sin (-x) = -b and cos (x) = a
4. For the +ve angle x^c, the coordinates of P are: (a, b)
• But x coordinate is cosine and y coordinate is sine.
• So we get:
sin x = b and cos x = a
5. comparing the results in (3) and (4), we get:
sin (-x) = -sin x and cos (-x) = cos x

---

6. In fig.3.34(b) above, a ray has rotated by x^c in the +ve direction.
    ♦ x is greater than $\frac{\pi}{2}$
• Another ray has rotated by the same x^c in the negative direction.
7. From the fig., it is clear that:
    ♦ ∠MOP = (π - x)
    ♦ ∠MOQ = (π - x)
• Since these two angles are same, the points P, Q and M are on the same vertical line.
• The x axis bisects the base PQ of the triangle OPQ
• Thus we get: QM = PM = b
8. So the coordinates of Q are: (-a, -b)
• But x coordinate is cosine and y coordinate is sine.
• So we get:
sin (-x) = -b and cos (-x) = -a
9. For the +ve angle x^c, the coordinates of P are: (-a, b)
• But x coordinate is cosine and y coordinate is sine.
• So we get:
sin x = b and cos x = -a
10. comparing the results in (8) and (9), we get:
sin (-x) = -sin x and cos (-x) = cos x

---

11. In fig.3.35(a) below, a ray has rotated by x^c in the +ve direction.
    ♦ x is greater than π
• Another ray has rotated by the same x^c in the negative direction.

Fig.3.35

12. From the fig., it is clear that:
    ♦ ∠MOP = (x - π)
    ♦ ∠MOQ = (x - π)
• Since these two angles are same, the points P, Q and M are on the same vertical line.
• The x axis bisects the base PQ of the triangle OPQ
• Thus we get: QM = PM = b
13. So the coordinates of Q are: (-a, b)
• But x coordinate is cosine and y coordinate is sine.
• So we get:
sin (-x) = b and cos (-x) = -a
14. For the +ve angle x^c, the coordinates of P are: (-a, -b)
• But x coordinate is cosine and y coordinate is sine.
• So we get:
sin x = -b and cos x = -a
15. comparing the results in (13) and (14), we get:
sin (-x) = -sin x and cos (-x) = cos x

---

16. In fig.3.35(b) above, a ray has rotated by x^c in the +ve direction.
    ♦ x is greater than $\frac{3\pi}{2}$
• Another ray has rotated by the same x^c in the negative direction.
17. From the fig., it is clear that:
    ♦ ∠MOP = (2π - x)
    ♦ ∠MOQ = (2π - x)
• Since these two angles are same, the points P, Q and M are on the same vertical line.
• The x axis bisects the base PQ of the triangle OPQ
• Thus we get: QM = PM = b
18. So the coordinates of Q are: (a, b)
• But x coordinate is cosine and y coordinate is sine.
• So we get:
sin (-x) = b and cos (-x) = a
19. For the +ve angle x^c, the coordinates of P are: (a, -b)
• But x coordinate is cosine and y coordinate is sine.
• So we get:
sin x = -b and cos x = a
20. comparing the results in (18) and (19), we get:
sin (-x) = -sin x and cos (-x) = cos x

---

21. We have considered the position of the -ve angle in all the four quadrants.
• We get the same result in all the four steps (5), (10), (15) and (20)
• So we can write the first and second identities:
Identity 1: sin (-x) = -sin x
Identity 2: cos (-x) = cos x

---

Next we will derive the third identity. It is related to the sum of two angles. It can be derived in 7 steps:
1. In fig.3.36 below, the green circle is a unit circle.

Fig.3.36

• Beginning from the +ve side of the x axis, the red ray rotated by x^c in the anticlockwise direction.
• Beginning from the +ve side of the x axis, the magenta ray rotated by (x+y)^c in the anticlockwise direction.
    ♦ So the angle between the red and magenta rays is y^c
• Beginning from the +ve side of the x axis, the blue ray rotated by y^c in the clockwise direction.
2. Important points and their coordinates:
• P₁ is the tip of the red ray.
    ♦ It's coordinates will be: (cos x, sin x)
• P₂ is the tip of the magenta ray.
    ♦ It's coordinates will be: (cos (x+y), sin (x+y))
• P₃ is the tip of the blue ray.
    ♦ It's coordinates will be: (cos (-y), sin (-y))
• P₄ is the point where the unit circle cuts the +ve side of the x axis.
    ♦ It's coordinates will be: (1, 0)
3. Using the five points O, P₁, P₂, P₃ and P₄, we can form two triangles:
P₁OP₃ and P₂OP₄
These two triangles are congruent. It can be proved in 5 steps:
(i) ∠ P₂OP₃ = (2π - x - y - y) = (2π - x - 2y)
(ii) In the triangle P₁OP₃,
∠ P₁OP₃ = [y + ∠ P₂OP₃] = [y + (2π - x - 2y)] = [2π - x - y]
(iii) In the triangle P₂OP₄,
∠ P₂OP₄ = [∠ P₂OP₃ + y] = [(2π - x - 2y) + y] = [2π - x - y]
(iv) So the same angle [2π - x - y] is present in both the triangles.
(v) In triangle P₁OP₃, this [2π - x - y] angle is included in between two radii.
• In triangle P₂OP₄ also, this [2π - x - y] angle is included in between two radii.
• So it is a SAS congruence.
4. Since the two triangles are congruent, we get: P₁P₃ = P₂P₄
5. Lengths of the these two sides can be calculated using distance formula.

• The distance between two points A (x1, y1) and B (x2, y2) is given by: AB2 = [(x2 - x1)2 + (y2 - y1)2] • Details can be seen here.

◼ Let us first calculate P₁P₃. It can be written in 3 steps:
(i) First we write the coordinates:
    ♦ Coordinates of P₁ are: (cos x, sin x)
    ♦ Coordinates of P₃ are: (cos (-y), sin (-y))
(ii) Now we apply the distance formula:
(P₁P₃)² = [(cos (-y) - cos x)² + (sin (-y) - sin x)²]
(iii) Using identities 1 and 2, we have: sin (-y) = -sin y and cos (-y) = cos y
• So the result in (ii) becomes:
(P₁P₃)² = [(cos y - cos x)² + (-sin y - sin x)²]
⇒ (P₁P₃)² = cos²y - 2 cos x cos y + cos²x + (-sin y)² - 2 (-sin y) (sin x) + sin²x
⇒ (P₁P₃)² = cos²y - 2 cos x cos y + cos²x + sin²y + 2 sin y sin x + sin²x
⇒ (P₁P₃)² = 2 - 2 cos x cos y + 2 sin y sin x
[∵ cos²y + sin²y  = 1 and cos²x + sin²x = 1]
◼ Next let us calculate P₂P₄. It can be written in 2 steps:
(i) First we write the coordinates:
    ♦ Coordinates of P₂ are: (cos (x+y), sin (x+y))
    ♦ Coordinates of P₄ are: (1, 0)
(ii) Now we apply the distance formula:
(P₂P₄)² = [(1 - cos (x+y))² + (0 - sin (x+y))²]
⇒ (P₂P₄)² = [1 - 2 cos (x+y) + cos²(x+y) + 0² - 2 0 sin (x+y) + sin²(x+y)]
⇒ (P₂P₄)² = [1 - 2 cos (x+y) + cos²(x+y) + sin²(x+y)]
⇒ (P₂P₄)² = [2 - 2 cos (x+y)]
[∵ cos²(x+y) + sin²(x+y) = 1]
6. From (4), the two lengths P₁P₃ and P₂P₄ are equal. So their squares must also be equal.
• Equating the squares, we get:
2 - 2 cos x cos y + 2 sin y sin x = 2 - 2 cos (x+y)
⇒ -2 cos x cos y + 2 sin y sin x = -2 cos (x+y)
⇒ -cos x cos y + sin y sin x = -cos (x+y)
⇒ cos x cos y - sin y sin x = cos (x+y)
7. Thus we get the third identity:
Identity 3: cos (x+y) = cos x cos y - sin x sin y

---

Next we will derive the fourth identity. It is related to the difference of two angles. It can be derived in 3 steps:
1. We have the Identity 3: cos (x+y) = cos x cos y - sin x sin y
2. Put (-y) in place of y. We get:
cos (x+(-y)) = cos x cos (-y) - sin x sin (-y)   
⇒ cos (x-y) = cos x cos y - sin x × -sin y
[∵ From first and second identities, cos (-y) = cos y and sin (-y) = -sin y]   
3. Thus we get the fourth identity:
Identity 4: cos (x-y) = cos x cos y + sin x sin y

---

• Identities 3 and 4 follow a similar pattern. The only difference is in the sign.
• So we can write them together as:
Identities 3 & 4: cos (x±y) = cos x cos y ∓ sin x sin y.
• On the left side, we have ± and on the right side, we have ∓.
   ♦ If on the left side we apply the top sign '+', on the right side also, we must apply the top sign '-'
   ♦ If on the left side we apply the bottom sign '-', on the right side also, we must apply the bottom sign '+'.

---

Next we will derive the fifth identity. It can be derived in 3 steps:
1. We have the Identity 4: cos (x-y) = cos x cos y + sin x sin y
2. put x = $\frac{\pi}{2}$ and y = x. We get:
$\cos (\frac{\pi}{2} - x)=\cos \frac{\pi}{2} \sin x + \sin \frac{\pi}{2} \sin x$
$\cos (\frac{\pi}{2} - x)=0 \times \sin x + 1 \times \sin x$
3. Thus we get the fifth identity:
Identity 5: $\cos (\frac{\pi}{2} - x)= \sin x$

---

Next we will derive the sixth identity. It can be derived in 3 steps:
1. Using the identity 5, we can write:
$\cos \left[\frac{\pi}{2}-\left(\frac{\pi}{2} - x\right)\right]= \sin \left(\frac{\pi}{2} - x\right)$
2. But $\cos \left[\frac{\pi}{2}-\left(\frac{\pi}{2} - x\right)\right]$ is cos x
So the result in (1) becomes: $\cos x = \sin \left(\frac{\pi}{2} - x\right)$
3. Thus we get the sixth identity:
Identity 6: $ \sin \left(\frac{\pi}{2} - x\right)=\cos x$

---

Next we will derive the seventh identity. It can be derived in 4 steps:
1. Using the identity 5, we can write:
$\cos \left[\frac{\pi}{2}-(x+y)\right]= \sin (x+y)$
2. The left side can be rearranged. We get:
$\cos \left[\left(\frac{\pi}{2}-x\right)-y \right]= \sin (x+y)$
3. Using identity 4, we can expand the left side. We get:
$\cos \left(\frac{\pi}{2}-x\right) \, \cos y + \sin \left(\frac{\pi}{2}-x\right) \, \sin y = \sin (x+y)$
⇒ $\sin x \, \cos y + \cos x \sin y = \sin (x+y)$
4. Thus we get the seventh identity:
Identity 7: $\sin (x+y)=\sin x \, \cos y + \cos x \sin y$

---

Next we will derive the eighth identity. It can be derived in 2 steps:
1. In identity 7, put -y in place of y. We get:
$\sin (x+(-y))=\sin x \, \cos (-y) + \cos x \sin (-y) $  
⇒ $\sin (x-y)=\sin x \, \cos y + \cos x \times -\sin y $  
⇒ $\sin (x-y)=\sin x \, \cos y - \cos x \sin y $
2. Thus we get the eithth identity:
Identity 8: $\sin (x-y)=\sin x \, \cos y - \cos x \sin y $

---

• Identities 7 and 8 follow a similar pattern. The only difference is in the sign.
• So we can write them together as:
Identities 7 & 8: sin (x±y) = sin x cos y ± cos x sin y.
• On the left side, we have ± and on the right side, we have ±.
   ♦ If on the left side we apply the top sign '+', on the right side also, we must apply the top sign '+'
   ♦ If on the left side we apply the bottom sign '-', on the right side also, we must apply the bottom sign '-'.

---

Next we will derive the ninth identity. Here we will derive 8 simple identities from 9(a) to 9(h). They can be derived in steps:
1. Consider the Identity 3: cos (x+y) = cos x cos y - sin x sin y.
• We put $\frac{\pi}{2}$ in place of x and x in place of y. We get:
$\cos (\frac{\pi}{2}+x) = \cos \frac{\pi}{2}\, \cos x - \sin \, \frac{\pi}{2} \sin x$
⇒ $\cos (\frac{\pi}{2}+x) =0 \times \cos x - 1 \times \sin x$
⇒ $\cos (\frac{\pi}{2}+x) =- \sin x$
Identity 9(a): $\cos (\frac{\pi}{2}+x) =- \sin x$

2. Consider the Identity 7: $\sin (x+y)=\sin x \, \cos y + \cos x \sin y$
• We put $\frac{\pi}{2}$ in place of x and x in place of y. We get:
$\sin (\frac{\pi}{2}+x) = \sin \frac{\pi}{2}\, \cos x + \cos \, \frac{\pi}{2} \sin x$
⇒ $\sin (\frac{\pi}{2}+x) =1 \times \cos x + 0 \times \sin x$
⇒ $\sin (\frac{\pi}{2}+x) = \cos x$
Identity 9(b): $\sin (\frac{\pi}{2}+x) = \cos x$

3. Consider the Identity 4: cos (x-y) = cos x cos y + sin x sin y.
• We put π in place of x and x in place of y. We get:
cos (π - x) = cos π cos x + sin π sin x.
⇒ cos (π - x) = -1 × cos x + 0 × sin x.
⇒ cos (π - x) = - cos x
Identity 9(c): cos (π - x) = - cos x

4. Consider the Identity 8: $\sin (x-y)=\sin x \, \cos y - \cos x \sin y $
• We put π in place of x and x in place of y. We get:
sin (π - x) = sin π cos x - cos π sin x
⇒ sin (π - x) = 0 × cos x - (-1)  × sin x
⇒ sin (π - x) = sin x
Identity 9(d): sin (π - x) = sin x

5. Consider the Identity 3: cos (x+y) = cos x cos y - sin x sin y.
• We put π in place of x and x in place of y. We get:
cos (π + x) = cos π cos x - sin π sin x
⇒ cos (π + x) = -1 × cos x - 0 × sin x
⇒ cos (π + x) = -cos x
Identity 9(e): cos (π + x) = -cos x 

6. Consider the Identity 3: sin (x+y) = sin x cos y + cos x sin y.
• We put π in place of x and x in place of y. We get:
sin (π + x) = sin π cos x + cos π sin x.
⇒ sin (π + x) = 0 × cos x + (-1) × sin x
⇒ sin (π + x) = -sin x
Identity 9(f): sin (π + x) = -sin x

7. Consider the Identity 4: cos (x-y) = cos x cos y + sin x sin y.
• We put 2π in place of x and x in place of y. We get:
cos (2π - x) = cos 2π cos x + sin 2π sin x.
⇒ cos (2π - x) = +1 × cos x + 0 × sin x.
⇒ cos (2π - x) = cos x
Identity 9(g): cos (2π - x) = cos x.

8. Consider the Identity 8: $\sin (x-y)=\sin x \, \cos y - \cos x \sin y $
• We put 2π in place of x and x in place of y. We get:
sin (2π - x) = sin 2π cos x - cos 2π sin x
⇒ sin (2π - x) = 0 × cos x - 1  × sin x
⇒ sin (2π - x) = -sin x
Identity 9(h): sin (2π - x) = sin x

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• The above identities from 9(a) to 9(h) are related to sin x and cos x. Using them, we can derive similar identities for tan x, cot x, sec x and csc x.
• Let us see an example:
$\tan (\frac{\pi}{2}+x)=\frac{\sin (\frac{\pi}{2}+x)}{\cos (\frac{\pi}{2}+x)}=\frac{\cos x}{\sin x}=\tan x$
• Thus we get the identity: $\tan (\frac{\pi}{2}+x)=\tan x$

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

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