OP Malhotra Class 11 Maths Solutions Chapter 2 Relations and Functions Ex 2(c)

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S Chand Class 11 ICSE Maths Solutions Chapter 2 Relations and Functions Ex 2(c)

Question 1.
Let X = {1, 2, 3, 4}. Determine whether or not each relation is a function from X into X.
(i) f = {(2, 3), (1, 4), (2, 1), (3, 2), (4, 4)}
(ii) g = {(3, 1), (4, 2), (1, 1)}
(iii) h = {(2, 1), (3, 4), (1, 4), (4, 4)}
Solution:
Given X = {1, 2, 3, 4}
(i) Given relation f = {(2, 3), (1, 4), (2, 1), (3, 2), (4, 4)}
Since element 2 ∈ X has two images 1 and 3
∴ each element of X does not have unique image.
Hence f is not a function.

(ii) Given relation g = {(3, 1), (4, 2), (1, 1)}
Clearly each element of X has unique image in x under relation g. Thus relation g is a function.

(iii) Given h = {(2, 1), (3, 4), (1, 4), (4, 4)}
Clearly each element of X has unique image in X. ∴ relation h is a function.

Question 2.
State for each of the following relations whether it is function or not. (Write Yes or No)
(i) {(1, 2), (2, 2), (3, 2), (4, 2)}
(ii) {(x, y): x ∈ A, y ∈ B is surname of x} where A is the set of people in India and B is the set of surnames.
(iii) {(x, y) : x ∈ A, ∈ B, y is the area of a square of side x} where A is the set of measurements of length.
(iv) {(x, j): x ∈ B, y ∈ P, y is passenger on x} where B is the set of buses of a school and P is the set of pupils of some schools.
(v) {(x, y) : x ∈ A, y ∈ B, y is sewn onto y}, where A is the set of buttons and B is the set of shirts.
Solution:
(i) since each element has unique image.
∴ given relation represents a function.

(ii) Since a unique people can have more than one surname. given relation is not a function.

(iii) Given relation = {(x, y): x ∈ A, y ∈ B, y is the area of square with side x} where A is the set of measurements of length.
Clearly y = x²
So each element has unique image.
Thus given relation represents a function.

(iv) Since a bus contains so many pupils of some schools. So one element have so many images.
Hence each element does not have unique image.
∴ given relation is not a function.

(v) Since if a button sewn on one shirt then it can’t be sewn on other shirt.
Thus each element has unique image.
∴ given relation represents a function.

Question 3.
The ordered pairs are represented by the points shown. For each diagram, state whether it represents a relation or a function. Justify your answer.

Solution:
(i) From graph, it is observe that set of ordered pairs be : {(1, 1), (2, 1), (2, 2), (3, 3)}
Clearly it is a subset of A x B and hence represents a relation. But element 2 has two images 1 and 2.
∴ given relation does not represents a function.
As each element does not have unique image.

(ii) From graph, set of ordered pairs be {(1, 1), (2, 2), (3, 3)}
Since no two ordered pairs have the same first component i.e. each element has unique image.
∴ given set represents a function.

(iii) From graph, the set of ordered pair be {(1, 1), (3, 2)} clearly it is a subset of AxB and hence represents a relation
Here the element 2 in domain does not have any image in codomain.
∴ it neither represents a function.

Question 4.
The range and domain of a function f(x) = \(\frac { 3 }{ x }\) + 1 are subsets of A and B respectively, where A = {- \(\frac { 1 }{ 2 }\), 0, \(\frac { 2 }{ 3 }\), \(\frac { 6 }{ 7 }\), 1} and B = {- 5, 0, 4\(\frac { 1 }{ 2 }\), 5, 5 \(\frac { 1 }{ 2 }\)} as ordered pairs.
Solution:

Question 5.
Which of the four statements given below is different from others ?
(a) f : A → B
(b) f : x → f(x)
(c) f is a mapping of A and B
(f) f is a function of A into B
Solution:
In option (a)fis a function from A into B
So option (c) and (d) ares also coincide with option (a).
f : x → f(x) is also defined as function from A to B
Thus all the four statements are same.

Question 6.
A = {- 2, – 1, 1, 2} and f = \(\left\{\left(x, \frac{1}{x}, x \in A\right)\right\}\)
(i) List the domain of f
(ii) List the range of f
(iii) Is f a function ?
Solution:
Given f be a function defined by f(x) = \(\frac { 1 }{ x }\) ∀ x ∈ A where A = {- 2, – 1, 1, 2}

Since each element in domain offhas unique image, i.e. no two ordered pairs have same first component.
∴ f represents a function.

Question 7.
f : x → highest prime factor of x.
(i) Find the range offwhen the domain is {12, 13, 14, 15, 16 and 17}
(ii) State a domain of five integers for which the range is (3).
(iii) A set of positive integers is called S. What can be said about these integers iff(S) = S ?
Solution:
Given f : x → highest prime factor of x
(i) Here factors of 12 are 1, 2, 3, 4, 6, 12
∴ 3 be the highest prime factor of 12.
∴ 3 ∈ R_f
factors of 13 are 1 and 13
∴ 13 be the highest prime factor of 13
∴ 13 ∈ R_f
Since 1, 2, 7, 14 are the factors of 14
∴ 7 be the highest prime factor of 14.
∴ 7 ∈ R_f
Since 1, 3, 5, 15 are the factors of 15.
Here 5 be the highest prime factor of 15
∴ 5 ∈ R_f
Since 1, 2, 4, 8 and 16 are the factors of 16
Here 2 be the highest prime factors of 16
∴ 2 ∈ R_f
Since 1, 17 are factors of 17. Here 17 be the highest prime factor of 17. ∴ 17 ∈ R_f
∴ range (f) = {2, 3. 5, 7, 13, 17}

(ii) given R_f = 3 i.e. highest prime factor of each element of domain offbe 3.
Since 3 be the highest prime factor of 3, 6, 9, 12, 18. [since highest prime factor of 15 be 5]
∴ Required D_f = {3, 6, 9, 12, 18}

(iii) Given f(S) = S
i. e. highest prime factors of S is equal to S which is only possible if S be the set of all prime numbers. Since factors of any prime number p are 1 and p.
So p be the highest prime factor of p.

Question 8.
(i) For x > 3, f(x) = 3x – 2 and for – 2 < x < 2, f(x) = x² – 2, find f(0) + f(4).
(ii) If f : R → R defined bye f(x) = \(\left\{\begin{array}{l} 4 x-1 \text { for } x>4 \\
x^2-2 \text { for }-2 \leq x<4 \\
3 x+4 \text { for } x<-2 \end{array}\right.\) find f(5) + f(0) + f(- 5).
Solution:
(i) Given f(x) = \(\begin{cases}3 x-2 ; & x>3 \\ x^2-2 ; & -2 \leq x \leq 2\end{cases}\)
When x = 0 ; f(x) = x² – 2 ∴ f(0) = 0² – 2 = – 2
When x = 4 > 2 ∴ f(x) = 3x – 2
⇒ f(4) = 3 x 4 – 2 = 10
∴ f(0) + f(4) = – 2 + 10 = 8

(ii) Given f(x) = \(\left\{\begin{array}{l}
4 x-1 ; x>4 \\
x^2-2 ;-2 \leq x<4 \\
3 x+4 ; x<-2 \end{array}\right.\) When x = 5 > 4 ∴ f (x) = 4x – 1 ⇒ f(5) = 20 – 1 = 19
When x = 0 i.e. – 2 < 0 < 4 ∴ f(x) = x² – 2 ⇒ f(0) = 0² – 2 = – 2
When x = – 5 < – 2 ∴ f(x) = 3x + 4 ⇒ f(- 5) = – 15 + 4 = 11
Thus f(5) + f(0) + f(- 5) = 19 – 2 – 11 = 6

Question 9.
What is the fundamental difference between a function and a relation? Let X = {1,2, 3, 4} and Y = {1, 5, 9, 11, 15, 16}. Determine which of the following sets are :
(i) relation
(ii) function
(iii) neither
(a) f₁ = {(x, y) : y = x², x ∈ X, y ∈ Y}
(b) f₂ = {(1, 1), (2, 11), (3, 1), (4, 15)}
(c) f₃ = {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)}
(d) f₄ = {(1, 1), (2, 7), (3, 5)} (Roorkee)
Solution:
If X and Y are non empty sets.
Then any subset of X × Y be a relation from X to Y. Further a function is a special type of relation in which no two ordered pairs have same first component.
(a) Given X = {1, 2, 3, 4} and Y = {1, 5, 9, 11, 15, 16}
Also, given f₁ = {(x, y) : y = x², x ∈ X, y ∈ Y}
When x = 1 ∈ X ⇒ y = 1² = 1 ∈ Y ⇒ (1, 1) ∈ f₁
When x = 2 ∈ X ⇒ y = 2² = 4 ∈ Y ⇒ (2, 4) ∉ f₁
When x = 3 ∈ X ⇒ y = 3² = 9 ∉ Y ⇒ (3, 9) ∈ f₁
When x = 4 ∈ X ⇒ y = 4² = 16 ∈ Y ⇒ (4, 16) ∈ f₁
∴ f₁ = {(1, 1), (3, 9), (4, 16)} clearly each ordered pair is a member of X × Y
∴ f₁ represents a relation from X to Y.
Further an element 2 in X has no image in Y.
∴ f₁ does not represents a function.

(b) Given f₂ = {(1, 1), (2, 11), (3, 1), (4, 15)}
Clearly each ordered pair of f₂ is a member of X x Y. Also no two ordered pairs have same first component.
∴ f₂ represents a function.

(c) Given f₃ = {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)}
Clearly each ordered pair of f₃ be a member of X x Y.
∴ f₃ ⊆ X x Y
∴ f₃ represents a relation from X to Y. Further element 2 in X have two images 9 and 11 in Y.
∴ f₃ does not represents a function.

(d) Given f₄ = {(1, 1), (2, 7), (3,5)}
Further (2, 7), ∉ X x Y
∵ f₄ is neither a relation nor a function.

Question 10.
A certain jet plane has an average speed of 500 km per hour. It can carry sufficient fuel for a 5 hour flight.
(i) Define the relation, as a set, between the distance d (in km) and time t (in hours) for this plane.
(ii) State the domain of the relation.
(iii) State the range of this relation.
(iv) Is this relation a function ?
Solution:
(i) Since speed = \(\frac { distance }{ Time }\) ⇒ 500 = \(\frac { d }{ t }\) ⇒ d = 500t
∴ required relation R = {(t, d) : d = 500t, t ∈ R and t ≤ 5}

(ii) Domain (R) = {t: 0 < t ≤ 5, t ∈ R}

(iii) Range (R) = {d : 0 < d ≤ 2500, d ∈ R} [∵ 0 < t ≤ 5 0 < 500t ≤ 2500 ⇒ 0 < d ≤ 2500]

(iv) Clearly each element in domain of R has unique image i.e. no two ordered pairs have same first component.
∴ given relation represents a function.

Question 11.
The domain of a function is the set of positive integers less than 12. If y find all ordered pairs satisfying the function. Graph the function.
Solution:
Clearly D_f = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}
Given y = f(x) = | X -4|
When x = 1 ⇒ y = | 1 – 4 | = 3 ⇒ (1, 3) ∈ f
When x = 2 ⇒ y = | 2 – 4 | = 2 ⇒ (2, 2) ∈ f
When x = 3 ⇒ y = | 3 – 4 | = 1 ⇒ (3, 1) ∈ f
When x = 4 ⇒ y = | 4 – 4 | = 0 ⇒ (4, 0) ∈ f
When x = 5 ⇒ y = | 5 – 4 | = 1 ⇒ (5, 1) ∈ f
When x = 6 ⇒ y = | 6 – 4 | = 2 ⇒ (6, 2) ∈ f
When x = 7 ⇒ y = | 7 – 4 | = 3 ⇒ (7, 3) ∈ f
When x = 8 ⇒ y = | 8 – 4 | = 4 ⇒ (8, 4) ∈ f
When x = 9 ⇒ y = | 9 – 4 | = 5 ⇒ (9, 5) ∈ f
When x= 10 ⇒ y = | 10 – 4 | = 6 ⇒ (10, 6) ∈ f
When x = 11 ⇒ y = |11 – 4 | = = 7 ⇒ (11, 7) ∈ f
∴ f = {(1, 3), (2, 2), (3, 1), (4, 0), (5, 1), (6, 2), (7, 3), (8, 4), (9, 5), (10, 6), (11, 7)

Question 12.
Let X = {2, 3} and Y = {1, 3, 5}. How many different functions are there from X into Y ?
Solution:
We know that if f: A → B be a function s.t n (A) = m and n (B) = n
Then no. of functions from A to B = [n (B)]^n(A) = n^m
Here X = {2, 3} ∴ n (X) = 2 and Y = {1, 3, 5} ∴ n (Y) = 3
Thus, required no. of functions from X into Y = [n (Y)]^n(x) = 3² = 9