Tiêu đề 12 Maths AR Ch 1.pdf ✅

Class 12 Maths Chapter 1 RELATIONS & FUNCTIONS Assertion and Reason Questions Directions: Each of these questions contains two statements, Assertion and Reason. Each of these questions also has four alternative choices, only one of which is the correct answer. You have to select one of the codes (a), (b), (c) and (d) given below. (a) Assertion is correct, reason is correct; reason is a correct explanation for assertion. (b) Assertion is correct, reason is correct; reason is not a correct explanation for assertion (c) Assertion is correct, reason is incorrect (d) Assertion is incorrect, reason is correct. 1. Assertion: f: R → R defined by f(x) = sin x is a bijection. Reason: If f is both one-one and onto it is bijection. 2. Assertion: f: R → R is a function defined by f(x) = 2x+1 3 . Then f −1 (x) = 3x−1 2 . Reason: f(x) is not a bijection. 3. Assertion: If f is even function, g is odd function, then f g , (g ≠ 0) is an odd function. Reason: If f(−x) = −f(x) for every x of its domain, then f(x) is called an odd function and if f(−x) = f(x) for every x of its domain, then f(x) is called an even function. 4. Assertion: Let L be the set of all lines in a plane and R be the relations in L defined as R = {(L1, L2 ): L1 is perpendicular to L2 }. This relation is not equivalence relation. Reason: A relation is said to be equivalence relation if it is reflexive, symmetric and transitive. 5. Assertion: If f(x) is odd function and g(x) is even function, then f(x) + g(x) is neither even nor odd. Reason: f(x) = { f(x), f(x) is even −f(x), f(x) is odd
6. Assertion: If f: R → R and g: R → R be two mappings such that f(x) = sin x and g(x) = x 2 , then fog ≠ gof. Reason: (fog) x = f(x)g(x) = (gof) x 7. Assertion: If the relation R defined in A = {1,2,3} by aRb, if |a 2 − b 2 | ≤ 5, then R −1 = R Reason: For above relation, domain of R −1 = Range of R. 8. Assertion: Let A = {−1,1,2,3} and B = {1,4,9}, where f: A → B given by f(x) = x 2 , then f is a many-one function. Reason: If x1 ≠ x2 ⇒ f(x1 ) ≠ f(x2 ), for every x1, x2 ∈ domain, then f is one-one or else many-one. 9. Assertion: The function f: R → R given by f(x) = x 3 is injective. Reason: The function f: X → Y is injective, if f(x) = f(y) ⇒ x = y for all x, y ∈ X 10. Assertion: The binary operation ∗: R × R → R given by a ∗ b → a + 2 b is associative. Reason: A binary operation*: A × A → A is said to be associative, if (a ∗ b) ∗ c = a ∗ ( b ∗ c) for all a, b, c ∈ A. 11. Let f(x) = (x + 1) 2 − 1, x ≥ −1 Assertion: The set {x: f(x) = f −1 (x) = {0, −1} Reason: f is a bijection. 12. Assertion: Let f: R → R be defined by f(x) = 1 x , then f is one-one and onto. Reason: x = 0 does not belong to the domain of f. 13. Assertion: Division is a binary operation on the set of natural numbers. Reason: 5 ÷ 4 = 1.25 is not a natural number. 14. Assertion: The binary operation subtraction on the set of real numbers is not commutative. Reason: If a and b are two real numbers, then in general a − b ≠ b − a
Answers 1. (d) 2. (c) 3. (a) 4. (a) 5. (a ) 6. (c) 7. (b) 8. (a) 9. (a) 10. (d) 11. (c) 12. (d ) 13. (d) 14. (a)