Chapter 2 – Relations And Functions

Class 11 Mathematics · 43 questions · 0 with answers

Solved examples

example-1Multiple choice

Let A = {1, 2, 3, 4} and B = {5, 7, 9}. Determine

(i)A × B
(ii)B × A
(iii)Is A × B = B × A ?
(iv)Is n (A × B) = n (B × A) ? Solution Since A = {1, 2, 3, 4} and B = {5, 7, 9}. Therefore, (i) A × B = {(1, 5), (1, 7), (1, 9), (2, 5), (2, 7), (2, 9), (3, 5), (3, 7), (3, 9), (4, 5), (4, 7), (4, 9)} (ii) B × A = {(5, 1), (5, 2), (5, 3), (5, 4), (7, 1), (7, 2), (7, 3), (7, 4), (9, 1), (9, 2), (9, 3), (9, 4)} (iii) No, A × B ≠ B × A. Since A × B and B × A do not have exactly the same ordered pairs. (iv) n (A × B) = n (A) × n (B) = 4 × 3 = 12 RELATIONS AND FUNCTIONS 23 n (B × A) = n (B) × n (A) = 4 × 3 = 12 Hence n (A × B) = n (B × A)

example-2Multiple choice

Find x and y if:

(i)(4x + 3, y) = (3x + 5, – 2)
(ii)(x – y, x + y) = (6, 10) Solution (i) Since (4x + 3, y) = (3x + 5, – 2), so 4x + 3 = 3x + 5 or x=2 and y=–2 (ii) x – y = 6 x + y = 10 ∴ 2x = 16 or x=8 8–y=6 ∴ y=2

example-3Long answer

If A = {2, 4, 6, 9} and B = {4, 6, 18, 27, 54}, a ∈ A, b ∈ B, find the set of ordered pairs such that 'a' is factor of 'b' and a < b. Solution Since A = {2, 4, 6, 9} B = {4, 6, 18, 27, 54}, we have to find a set of ordered pairs (a, b) such that a is factor of b and a < b. Since 2 is a factor of 4 and 2 < 4. So (2, 4) is one such ordered pair. Similarly, (2, 6), (2, 18), (2, 54) are other such ordered pairs. Thus the required set of ordered pairs is {(2, 4), (2, 6), (2, 18), (2, 54), (6, 18), (6, 54,), (9, 18), (9, 27), (9, 54)}.

example-4Short answer

Find the domain and range of the relation R given by R = {(x, y) : y = x +; where x, y ∈ N and x < 6}. Solution When x = 1, y = 7 ∈ N, so (1, 7) ∈ R. Again for, x = 2 . y = 2+ = 2 + 3 = 5 ∈ N, so (2, 5) ∈ R. Again for x = 3, y = 3 + = 3 + 2 = 5 ∈ N, (3, 5) ∈ R. Similarly for x = 4

example-5Multiple choice

Is the following relation a function? Justify your answer

(i)R1 = {(2, 3), ( , 0), (2, 7), (– 4, 6)}
(ii)R2 = {(x, | x |) | x is a real number} Solution Since (2, 3) and (2, 7) ∈ R1 ⇒ R1 (2) = 3 and R1 (2) = 7 So R1 (2) does not have a unique image. Thus R1 is not a function.
(iii)R2 = {(x, | x |) / x ∈R} For every x ∈ R there will be unique image as | x | ∈ R. Therefore R2 is a function.

example-6Short answer

Find the domain for which the functions f (x) = 2x2 – 1 and g (x) = 1 – 3x are equal. Solution For f (x) = g (x) ⇒ 2x2 – 1 = 1 – 3x ⇒ 2x2 + 3x – 2 = 0 ⇒ 2x2 + 4x – x – 2 = 0 ⇒ 2x (x + 2) – 1 (x + 2) = 0 ⇒ (2x – 1) (x + 2) = 0 Thus domain for which the function f (x) = g (x) is ,–2 .

example-7Multiple choice

Find the domain of each of the following functions.

(i)f ( x) = 2
(ii)f (x) = [x] + x x + 3x + 2 RELATIONS AND FUNCTIONS 25 Solution g ( x) (i) f is a rational function of the form , where g (x) = x and h (x) = x2 + 3x + 2. h ( x) Now h (x) ≠ 0 ⇒ x2 + 3x + 2 ≠ 0 ⇒ (x + 1) (x + 2) ≠ 0 and hence domain of the given function is R – {– 1, – 2}. (ii) f (x) = [x] + x,i.e., f (x) = h (x) + g (x) where h (x) = [x] and g (x) = x The domain of h = R and the domain of g = R. Therefore Domain of f = R

example-8Multiple choice

Find the range of the following functions given by x−4

(i)
(ii)16 – x 2 x−4 Solution x−4 = 1, x > 4 x−4 x−4 (i) f (x) = = x−4 − ( x − 4) = −1, x < 4 x−4 x−4 Thus the range of = {1, –1}. x−4 (ii) The domain of f, where f (x) = 16 − x 2 is given by [– 4, 4]. For the range, let y = 16 − x 2 then y2 = 16 – x2 or x2 = 16 – y2 Since x ∈ [– 4, 4] Thus range of f = [0, 4]

example-9Short answer

Redefine the function which is given by f (x) = x −1 + 1 + x , – 2 ≤ x ≤ 2 Solution f (x) = x −1 + 1 + x , – 2 ≤ x ≤ 2 – x +1 −1 − x, – 2 ≤ x < –1 = – x +1 + x +1, –1≤ x < 1 x −1 + 1 + x, 1 ≤ x ≤ 2 – 2 x , – 2 ≤ x < –1 = 2, –1 ≤ x < 1 2 x,1 ≤ x ≤ 2

example-10Short answer

Find the domain of the function f given by f (x) = [ x] –[ x] – 6 Solution Given that f (x) = , f is defined if [x]2 – [x] – 6 > 0. [ x] –[ x] – 6 or ([x]–3) ([x] + 2) > 0, ⇒ [x] < – 2 or [x] > 3 ⇒ x<–2 or x≥4 Hence Domain = ( – ∞ , – 2) ∪ [4, ∞ ).

example-11Multiple choice

The domain of the function f defined by f (x) = is x− x

(A)R
(B)R +
(C)R –
(D)None of these Solution The correct answer is (D). Given that f (x) = x− x x – x = 0 if x ≥0 where x– x = 2x if x <0 RELATIONS AND FUNCTIONS 27 Thus is not defined for any x ∈ R. x− x Hence f is not defined for any x ∈ R, i.e. Domain of f is none of the given options. 1 1

example-12Multiple choice

If f (x) = x3 − 3 , then f (x) + f ( ) is equal to x x

(A)2x 3
(B)2
(C)0
(D)1 x3 Solution The correct choice is C. Since f (x) = x3 – x3 1 1 1 1 f = 3 − = 3 – x3 x x 1 x x3 1 3 1 1 3 Hence, f (x) + f = x − 3 + 3 –x =0 x x x

example-13Fill in the blanks

Let A and B be any two sets such that n(B) = p, n(A) = q then the total number of functions f : A → B is equal to __________. Solution Any element of set A, say xi can be connected with the element of set B in p ways. Hence, there are exactly pq functions.

example-14Fill in the blanks

Let f and g be two functions given by f = {(2, 4), (5, 6), (8, – 1), (10, – 3)} g = {(2, 5), (7, 1), (8, 4), (10, 13), (11, – 5)} then. Domain of f + g is __________ Solution Since Domain of f = Df = {2, 5, 8, 10} and Domain of g = Dg = {2, 7, 8, 10, 11}, therefore the domain of f + g = {x | x ∈ D f ∩ Dg} = {2, 8, 10}

Questions

Q6Short answer

6 y=4+ ∉ N and for x = 5 , y = 5 + ∉N 4 5 Thus R = {(1, 7), (2, 5), (3, 5)}, where Domain of R = {1, 2, 3} Range of R = {7, 5}

Q1Multiple choice

Let A = {–1, 2, 3} and B = {1, 3}. Determine

(i)A × B
(ii)B × A
(iii)B × B
(iv)A × A

Q2Short answer

If P = {x : x < 3, x ∈ N}, Q = {x : x ≤ 2, x ∈ W}. Find (P ∪ Q) × (P ∩ Q), where W is the set of whole numbers.

Q3Multiple choice

If A = {x : x ∈ W, x < 2} B = {x : x ∈ N, 1 < x < 5} C = {3, 5} find

(i)A × (B ∩ C)
(ii)A × (B ∪ C)

Q4Multiple choice

In each of the following cases, find a and b.

(i)(2a + b, a – b) = (8, 3)
(ii), a – 2b = (0, 6 + b)

Q5Multiple choice

Given A = {1, 2, 3, 4, 5}, S = {(x, y) : x ∈ A, y ∈ A}. Find the ordered pairs which satisfy the conditions given below:

(i)x + y = 5
(ii)x + y < 5
(iii)x + y > 8 2 2

Q6Short answer

Given R = {(x, y) : x, y ∈ W, x + y = 25}. Find the domain and Range of R.

Q7Short answer

If R1 = {(x, y) | y = 2x + 7, where x ∈ R and – 5 ≤ x ≤ 5} is a relation. Then find the domain and Range of R1.

Q8Short answer

If R2 = {(x, y) | x and y are integers and x2 + y2 = 64} is a relation. Then find R2.

Q9Short answer

If R3 = {(x, x ) | x is a real number} is a relation. Then find domain and range of R3.

Q10Multiple choice

Is the given relation a function? Give reasons for your answer.

(i)h = {(4, 6), (3, 9), (– 11, 6), (3, 11)}
(ii)f = {(x, x) | x is a real number}  1 
(iii)g =  n ,  |n is a positive integer   n 
(iv)s = {(n, n2) | n is a positive integer} (v) t = {(x, 3) | x is a real number.

Q11Multiple choice

If f and g are real functions defined by f (x) = x2 + 7 and g (x) = 3x + 5, find each of the following

(a)f (3) + g (– 5)
(b)f × g (14)
(c)f (– 2) + g (– 1)
(d)f (t) – f (– 2) f (t ) − f (5) (e) , if t ≠ 5 t −5 RELATIONS AND FUNCTIONS 29

Q12Multiple choice

Let f and g be real functions defined by f (x) = 2x + 1 and g (x) = 4x – 7.

(a)For what real numbers x, f (x) = g (x)?
(b)For what real numbers x, f (x) < g (x)?

Q13Multiple choice

If f and g are two real valued functions defined as f (x) = 2x + 1, g (x) = x2 + 1, then find.

(i)f + g
(ii)f – g
(iii)f g
(iv)

Q14Short answer

Express the following functions as set of ordered pairs and determine their range. f : X → R, f (x) = x3 + 1, where X = {–1, 0, 3, 9, 7}

Q15Short answer

Find the values of x for which the functions f (x) = 3x2 – 1 and g (x) = 3 + x are equal

Q16Long answer

Is g = {(1, 1), (2, 3), (3, 5), (4, 7)} a function? Justify. If this is described by the relation, g (x) = αx + β, then what values should be assigned to α and β?

Q17Multiple choice

Find the domain of each of the following functions given by 1 1

(i)f ( x) =
(ii)f ( x) = 1 − cos x x+ x x3 − x + 3
(iii)f (x) = x x
(iv)f (x) = x 2 −1 3x (v) f (x) = 2x −8

Q18Multiple choice

Find the range of the following functions given by

(i)f (x) = 2
(ii)f (x) = 1 – x − 2 2– x
(iii)f (x) = x − 3
(iv)f (x) = 1 + 3 cos2x (Hint : – 1 ≤ cos 2x ≤ 1 ⇒ – 3 ≤ 3 cos 2x ≤ 3 ⇒ –2 ≤ 1 + 3cos 2x ≤ 4)

Q19Long answer

Redefine the function f (x) = x − 2 + 2 + x , – 3 ≤ x ≤ 3 x −1

Q20Multiple choice

If f (x) = , then show that x +1 1 1 −1

(i)f = – f (x)
(ii)f − = x x f ( x)

Q21Multiple choice

Let f (x) = x and g (x) = x be two functions defined in the domain R+ ∪ {0}. Find

(i)(f + g) (x)
(ii)(f – g) (x)
(iii)(fg) (x)
(iv)( x)

Q22Long answer

Find the domain and Range of the function f (x) = . x −5 ax − b

Q23Long answer

If f (x) = y = , then prove that f (y) = x. cx − a

Q24Multiple choice

Let n

(A)= m, and n
(B)= n. Then the total number of non-empty relations that can be defined from A to B is (A) mn (B) nm – 1
(C)mn – 1
(D)2mn – 1

Q25Multiple choice

If [x]2 – 5 [x] + 6 = 0, where [ . ] denote the greatest integer function, then

(A)x ∈ [3, 4]
(B)x ∈ (2, 3]
(C)x ∈ [2, 3]
(D)x ∈ [2, 4)

Q26Multiple choice

Range of f (x) = is 1− 2cos x 1 1

(A),1
(B)−1, 3 3 1 1
(C)(– ∞, –1] ∪ ,∞
(D)− ,1 3 3 RELATIONS AND FUNCTIONS 31

Q27Multiple choice

Let f (x) = 1 + x 2 , then

(A)f (xy) = f (x) . f (y)
(B)f (xy) ≥ f (x) . f (y)
(C)f (xy) ≤ f (x) . f (y)
(D)None of these [Hint : find f (xy) = 1+ x 2 y 2 , f (x) . f (y) = 1+ x 2 y 2 + x 2 + y 2 ]

Q28Multiple choice

Domain of a 2 − x 2 (a > 0) is

(A)(– a, a)
(B)[– a, a]
(C)[0, a]
(D)(– a, 0]

Q29Multiple choice

If f (x) = ax + b, where a and b are integers, f (–1) = – 5 and f (3) = 3, then a and b are equal to

(A)a = – 3, b = –1
(B)a = 2, b = – 3
(C)a = 0, b = 2
(D)a = 2, b = 3

Q30Multiple choice

The domain of the function f defined by f (x) = 4− x + is equal to x 2 −1

(A)(– ∞, – 1) ∪ (1, 4]
(B)(– ∞, – 1] ∪ (1, 4]
(C)(– ∞, – 1) ∪ [1, 4]
(D)(– ∞, – 1) ∪ [1, 4) 4− x

Q31Multiple choice

The domain and range of the real function f defined by f (x) = is given by x−4

(A)Domain = R, Range = {–1, 1}
(B)Domain = R – {1}, Range = R
(C)Domain = R – {4}, Range = {– 1}
(D)Domain = R – {– 4}, Range = {–1, 1}

Q32Multiple choice

The domain and range of real function f defined by f (x) = x − 1 is given by

(A)Domain = (1, ∞), Range = (0, ∞)
(B)Domain = [1, ∞), Range = (0, ∞)
(C)Domain = [1, ∞), Range = [0, ∞)
(D)Domain = [1, ∞), Range = [0, ∞) x 2 + 2 x +1

Q33Multiple choice

The domain of the function f given by f (x) = x2 – x – 6

(A)R – {3, – 2}
(B)R – {–3, 2}
(C)R – [3, – 2]
(D)R – (3, – 2)

Q34Multiple choice

The domain and range of the function f given by f (x) = 2 – x − 5 is

(A)Domain = R+, Range = ( – ∞, 1]
(B)Domain = R, Range = ( – ∞, 2]
(C)Domain = R, Range = (– ∞, 2)
(D)Domain = R+, Range = (– ∞, 2]

Q35Multiple choice

The domain for which the functions defined by f (x) = 3x2 – 1 and g (x) = 3 + x are equal is 4 4

(A)−1,
(B)−1, 3 3 4 4
(C)−1,
(D)−1, 3 3 Fill in the blanks :

Q36Fill in the blanks

Let f and g be two real functions given by f = {(0, 1), (2, 0), (3, – 4), (4, 2), (5, 1)} g = {(1, 0), (2, 2), (3, – 1), (4, 4), (5, 3)} then the domain of f . g is given by _________.

Q37Multiple choice

Let f = {(2, 4), (5, 6), (8, – 1), (10, – 3)} g = {(2, 5), (7, 1), (8, 4), (10, 13), (11, 5)} be two real functions. Then, match the following : 4 −1 −3 (a) f – g

(i)2, , 8, , 10, 5 4 13 (b) f + g
(ii){( 2, 20 ) , (8, − 4 ) , (10, − 39)} (c) f . g
(iii){( 2, − 1) , (8, − 5) , (10, − 16 )} RELATIONS AND FUNCTIONS 33 (d) g
(iv){(2, 9), (8, 3), (10, 10)} State True or False for the following statements given in Exercises 38 to 42 :

Q38Multiple choice

The ordered pair (5, 2) belongs to the relation R = {(x, y) : y = x – 5, x, y ∈ Z}

Q39Multiple choice

If P = {1, 2}, then P × P × P = {(1, 1, 1), (2, 2, 2), (1, 2, 2), (2, 1, 1)}

Q40Multiple choice

If A = {1, 2, 3}, B = {3, 4} and C = {4, 5, 6}, then (A × B) ∪ (A × C) = {(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6)}. 1 −14

Q41Multiple choice

If (x – 2, y + 5) = − 2, are two equal ordered pairs, then x = 4, y = 3 3

Q42Multiple choice

If A × B = {(a, x), (a, y), (b, x), (b, y)}, then A = {a, b}, B = {x, y}