RowQ
The Vault
RowQ
The Vault
CBSE Class 11 Maths · 11 questions · 26 marks
A relation is nothing more than a rule that pairs elements of one set with elements of another, and a function is the special, well-behaved case where each input gets exactly one output. This chapter builds that idea from ordered pairs and the Cartesian product upward, then introduces the standard functions — modulus, signum, greatest integer, polynomial — whose graphs you will see again all year. Getting comfortable with domain and range now saves you a lot of trouble in limits and calculus later.
If n(A) = 4 and n(B) = 3, then the number of relations from A to B is:
Answer
The number of relations is 4096. A relation from A to B is any subset of A × B, and n(A × B) = 4 × 3 = 12. A set with 12 elements has 2¹² subsets, so the number of relations is 2¹² = 4096. The option 12 is only the number of ordered pairs, not the number of relations.
The range of the real function f(x) = |x - 4| is:
Answer
The range is [0, ∞). The modulus of any real quantity is never negative, so f(x) ≥ 0 for every x. The value 0 is attained at x = 4, and as x moves away from 4 in either direction |x - 4| grows without bound. Hence every non-negative real number occurs as an output and the range is [0, ∞). The set [4, ∞) confuses the input value 4 with the output values.
The domain of the real function f(x) = 1/√(x - 5) is:
Answer
The domain is (5, ∞). For the square root to be defined we need x - 5 ≥ 0, and because the root sits in the denominator it must also be non-zero, so x - 5 > 0. That gives x > 5, i.e. the interval (5, ∞). The option [5, ∞) wrongly includes x = 5, where the denominator becomes 0 and f is undefined.
If [x] denotes the greatest integer function, then [-3.4] + [2.7] equals:
Answer
The value is -2. By definition [x] is the greatest integer that is not greater than x. For -3.4 the integers below it are -4, -5, ..., and the greatest of these is -4, so [-3.4] = -4 (not -3, which is larger than -3.4). For 2.7 we get [2.7] = 2. Adding, -4 + 2 = -2.
Assertion (A): The relation R = {(1, 4), (1, 6), (2, 5)} from A = {1, 2} to B = {4, 5, 6} is a function. Reason (R): A relation from A to B is a function if every element of A has exactly one image in B.
Answer
A is false but R is true. Reason R is the correct definition of a function. Applying it to R, the element 1 of A appears in two ordered pairs, (1, 4) and (1, 6), so 1 has two different images. That violates the requirement of exactly one image. Therefore the given relation is not a function and Assertion A is false, while Reason R remains a true statement.
If A = {2, 5} and B = {1, 3, 4}, write A × B in roster form and state n(A × B).
Answer
A × B contains every ordered pair whose first component comes from A and second from B. Pairing 2 with each element of B gives (2, 1), (2, 3), (2, 4). Pairing 5 with each element of B gives (5, 1), (5, 3), (5, 4). So A × B = {(2, 1), (2, 3), (2, 4), (5, 1), (5, 3), (5, 4)}. n(A × B) = n(A) × n(B) = 2 × 3 = 6, which matches the six pairs listed.
For f(x) = x² - 3x + 2, find f(0) and f(-2), and determine all real values of x for which f(x) = 0.
Answer
f(0) = 0² - 3(0) + 2 = 2. f(-2) = (-2)² - 3(-2) + 2 = 4 + 6 + 2 = 12. For the zeros, solve x² - 3x + 2 = 0. Splitting the middle term: x² - 2x - x + 2 = x(x - 2) - 1(x - 2) = (x - 1)(x - 2). Setting (x - 1)(x - 2) = 0 gives x = 1 or x = 2. So f(0) = 2, f(-2) = 12, and f(x) = 0 at x = 1 and x = 2.
If (2a + 1, 3) = (7, b - 4), find the values of a and b.
Answer
Two ordered pairs are equal only when the corresponding components are equal. Comparing first components: 2a + 1 = 7, so 2a = 6 and a = 3. Comparing second components: 3 = b - 4, so b = 7. Hence a = 3 and b = 7.
Let R = {(x, y) : x ∈ N, y ∈ N, y = 2x + 1 and x ≤ 5}. (a) Write R in roster form. (b) State its domain and range. (c) Is R a function from {1, 2, 3, 4, 5} to N? Justify your answer. (d) Find the number of relations that can be defined from {1, 2, 3, 4, 5} to the range of R.
Answer
(a) Substitute x = 1, 2, 3, 4, 5 into y = 2x + 1. x = 1 gives y = 3; x = 2 gives y = 5; x = 3 gives y = 7; x = 4 gives y = 9; x = 5 gives y = 11. So R = {(1, 3), (2, 5), (3, 7), (4, 9), (5, 11)}. (b) Domain = set of first components = {1, 2, 3, 4, 5}. Range = set of second components = {3, 5, 7, 9, 11}. The codomain here is N, and the range is a subset of it. (c) Yes, R is a function. Each of the five elements 1, 2, 3, 4, 5 appears as the first component of exactly one ordered pair, so every input has exactly one image, which is the definition of a function. (d) The domain set has 5 elements and the range set has 5 elements, so the Cartesian product has 5 × 5 = 25 ordered pairs. The number of relations is the number of subsets of that product, namely 2²⁵ = 33554432.
Let f(x) = 3x - 4 and g(x) = x² + 1 be real functions. (a) Find (f + g)(x) and (f - g)(x). (b) Evaluate (fg)(2). (c) Write (f/g)(x) and state its domain. (d) Show that there is no real x for which f(x) = g(x).
Answer
(a) (f + g)(x) = (3x - 4) + (x² + 1) = x² + 3x - 3. (f - g)(x) = (3x - 4) - (x² + 1) = -x² + 3x - 5. (b) f(2) = 3(2) - 4 = 2 and g(2) = 2² + 1 = 5. So (fg)(2) = f(2) × g(2) = 2 × 5 = 10. (c) (f/g)(x) = (3x - 4)/(x² + 1). The quotient is defined wherever g(x) ≠ 0. Since x² ≥ 0 for all real x, we always have x² + 1 ≥ 1 > 0, so g never vanishes. Hence the domain is all of R. (d) Setting f(x) = g(x) gives 3x - 4 = x² + 1, i.e. x² - 3x + 5 = 0. The discriminant is D = (-3)² - 4(1)(5) = 9 - 20 = -11. Since D < 0, the quadratic has no real roots, so there is no real x with f(x) = g(x).
Read the following and answer the questions that follow: A city cab service charges a fixed booking fee of ₹40 for every ride plus ₹14 for each kilometre travelled. If x denotes the distance in kilometres, the total fare in rupees is modelled by the real function f(x) = 40 + 14x, where x ≥ 0. (a) State the domain and the range of f. (b) Find the fare for a 9 km ride. (c) A passenger paid ₹250. How far did the cab travel? (d) Two passengers were charged different fares. Can they have travelled the same distance? Justify using the nature of f.
Answer
(a) Distance cannot be negative, so the domain is [0, ∞). At x = 0 the fare is ₹40, and the fare increases without bound as x increases, so the range is [40, ∞). (b) f(9) = 40 + 14(9) = 40 + 126 = 166. The fare is ₹166. (c) Solve 40 + 14x = 250, so 14x = 210 and x = 15. The cab travelled 15 km. (d) No. f is a linear function with non-zero slope 14, so it is one-one: if f(x₁) = f(x₂) then 40 + 14x₁ = 40 + 14x₂, giving 14x₁ = 14x₂ and x₁ = x₂. Equal distances therefore force equal fares, so different fares must come from different distances.
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