TOPICS

Miscellaneous

Relations and Functions

**Question-1 :-** The relation f is defined by
The relation g is defined by
Show that f is a function and g is not a function.

It is observed that for 0 ≤ x < 3, f(x) = x² 3 < x ≤ 10, f(x) = 3x Also, at x = 3, f(x) = 3² = 9 or f(x) = 3 × 3 = 9 i.e., at x = 3, f(x) = 9. Therefore, for 0 ≤ x ≤ 10, the images of f(x) are unique. Thus, the given relation is a function. It can be observed that for x = 2, g(x) = 2² = 4 and g(x) = 3 × 2 = 6 Hence, element 2 of the domain of the relation g corresponds to two different images i.e., 4 and 6. Hence, this relation is not a function.

**Question-2 :-** If f(x) = x², find = [f(1.1)-f(1)]/(1.1 - 1).

Given that f(x) = x². [f(1.1) - f(1)]/(1.1 - 1) = [(1.1)² - (1)²]/(1.1 - 1) = (1.21 - 1)/(0.1) = 0.21/0.1 = 2.1

**Question-3 :-** Find the domain of the function f(x) = (x² + 2x + 1)/(x² - 8x + 12).

Given that f(x) = (x² + 2x + 1)/(x² - 8x + 12). (x² - 8x + 12) = x² - 6x - 2x + 12 = x(x - 6) - 2(x - 6) = (x - 2)(x - 6) f(x) = (x² + 2x + 1)/(x - 2)(x - 6) It can be seen that function f is defined for all real numbers except at x = 6 and x = 2. Hence, the domain of f is R – {2, 6}.

**Question-4 :-** Find the domain and the range of the real function f defined by
f (x) = √x - 1

Given that f(x) = f (x) = √x - 1. It can be seen that is defined for (x – 1) ≥ 0. i.e., is defined for x ≥ 1. Therefore, the domain of f is the set of all real numbers greater than or equal to 1 i.e., the domain of f = [1, ∞). As x ≥ 1 ⇒ (x – 1) ≥ 0 ⇒ √x - 1 ≥ 0 Therefore, the range of f is the set of all real numbers greater than or equal to 0 i.e., the range of f = [0, ∞).

**Question-5 :-** Find the domain and the range of the real function f defined by f (x) = |x - 1|.

Given that f(x) = |x - 1|. It is clear that |x – 1| is defined for all real numbers. Domain of f = R Also, for x ∈ R, |x – 1| assumes all real numbers. Hence, the Range of f is the set of all non-negative real numbers.

**Question-6 :-** Let f = {(x, x²/(1 + x²)) : x ∈ R} be a function from R into R. Determine the range
of f.

Given that f(x) = {(x, x²/(1 + x²)) : x ∈ R}. f(x) = {(0, 0), (±0.5, 1/5), (±1, 1/2), (±1.3, 9/13), (±2, 4/5), (3, 9/10),.....} The range of f is the set of all second elements. It can be observed that all these elements are greater than or equal to 0 but less than 1. Range of f = [0, 1)

**Question-7 :-** Let f, g : R → R be defined, respectively by f(x) = x + 1, g(x) = 2x – 3. Find
f + g, f – g and f/g.

Given that f(x) = x + 1, g(x) = 2x – 3. f + g = x + 1 + 2x - 3 = 3x - 2 f - g = x + 1 - 2x + 3 = -x + 4 f/g = (x + 1)/(2x - 3)

**Question-8 :-** Let f = {(1,1), (2,3), (0,–1), (–1, –3)} be a function from Z to Z defined by f(x) = ax + b, for some integers a, b. Determine a, b.

Given that f = {(1,1), (2,3), (0,–1), (–1, –3)} and f(x) = ax + b (1, 1) ∈ f ⇒ f(1) = 1 ⇒ a × 1 + b = 1 ⇒ a + b = 1 (0, –1) ∈ f ⇒ f(0) = –1 ⇒ a × 0 + b = –1 ⇒ b = –1 On substituting b = –1 in a + b = 1, we obtain a + (–1) = 1 ⇒ a = 1 + 1 = 2. Thus, the respective values of a and b are 2 and –1.

**Question-9 :-** Let R be a relation from N to N defined by R = {(a, b) : a, b ∈ N and a = b²}. Are the following true?

(i) (a,a) ∈ R, for all a ∈ N

(ii) (a,b) ∈ R, implies (b,a) ∈ R

(iii) (a,b) ∈ R, (b,c) ∈ R implies (a,c) ∈ R.

Justify your answer in each case.

Given that R = {(a, b) : a, b ∈ N and a = b²}. (i) (a,a) ∈ R, for all a ∈ N It can be seen that 2 ∈ N; However, 2 ≠ 2² = 4. Therefore, the statement '(a, a) ∈ R, for all a ∈ N' is not true. (ii) (a,b) ∈ R, implies (b, a) ∈ R It can be seen that (9, 3) ∈ N because 9, 3 ∈ N and 9 = 3². Now, 3 ≠ 9² = 81; Therefore, (3, 9) ∉ N. Therefore, the statement '(a, b) ∈ R, implies (b, a) ∈ R' is not true. (iii) (a,b) ∈ R, (b, c) ∈ R implies (a, c) ∈ R. It can be seen that (9, 3) ∈ R, (16, 4) ∈ R because 9, 3, 16, 4 ∈ N and 9 = 3² and 16 = 4². Now, 9 ≠ 4² = 16; Therefore, (9, 4) ∉ N Therefore, the statement '(a, b) ∈ R, (b, c) ∈ R implies (a, c) ∈ R' is not true.

**Question-10 :-** Let A ={1,2,3,4}, B = {1,5,9,11,15,16} and f = {(1,5), (2,9), (3,1), (4,5), (2,11)} Are the following true?

(i) f is a relation from A to B

(ii) f is a function from A to B.

Justify your answer in each case.

Given that A ={1,2,3,4}, B = {1,5,9,11,15,16} and f = {(1,5), (2,9), (3,1), (4,5), (2,11)}. A × B = {(1, 1), (1, 5), (1, 9), (1, 11), (1, 15), (1, 16), (2, 1), (2, 5), (2, 9), (2, 11), (2, 15), (2, 16), (3, 1), (3, 5), (3, 9), (3, 11), (3, 15), (3, 16), (4, 1), (4, 5), (4, 9), (4, 11), (4, 15), (4, 16)} (i) f is a relation from A to B A relation from a non-empty set A to a non-empty set B is a subset of the Cartesian product A × B. It is observed that f is a subset of A × B. Thus, f is a relation from A to B. (ii) f is a function from A to B. Since the same first element i.e., 2 corresponds to two different images i.e., 9 and 11, relation f is not a function.

**Question-11 :-** Let f be the subset of Z × Z defined by f = {(ab, a + b) : a, b ∈ Z}. Is f a function from Z to Z? Justify your answer.

Given that f = {(ab, a + b) : a, b ∈ Z}. The relation f is defined as f = {(ab, a + b): a, b ∈ Z} We know that a relation f from a set A to a set B is said to be a function if every element of set A has unique images in set B. Since 2, 6, –2, –6 ∈ Z, (2 × 6, 2 + 6), [(–2) × (–6), (–2) + (–6)) ∈ f i.e., (12, 8), (12, –8) ∈ f It can be seen that the same first element i.e., 12 corresponds to two different images i.e., 8 and –8. Thus, relation f is not a function.

**Question-12 :-** Let A = {9, 10 ,11 ,12 ,13} and let f : A → N be defined by f (n) = the highest prime factor of n. Find the range of f.

Given that A = {9,10,11,12,13} & f(n) = the highest prime factor of n. Prime factor of 9 = 3 Prime factors of 10 = 2, 5 Prime factor of 11 = 11 Prime factors of 12 = 2, 3 Prime factor of 13 = 13 f(9) = The highest prime factor of 9 = 3

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