Chapter 2 – Inverse Trigonometric Functions Exercise 2.1
Question 1: Find The principal value of sin 2 |
Question 7: Given examples of two functions f: N → N and 𝑔: N → N such that 𝑔of is onto but f is not onto. (Hint: Consider f(x) = x + 1 and 𝑔(𝑥) = {x−1,if x>1 and 1, if x=1 Solution : Define f: N → N by f(x) = x + 1 and 𝑔: N → N by 𝑔(𝑥) = {x−1, if x>1 and 1, if x=1 We first show that 𝑔 is not onto. For this, consider element 1 in co-domain N. It is clear that this element is not an image of any of the elements in domain N. ∴ f is not onto. Now, 𝑔of: N → N is defined by 𝑔𝑜𝑓(𝑥) = 𝑔(𝑓(𝑥)) = 𝑔(𝑥 + 1) = 𝑥 + 1 − 1 = 𝑥 [𝑥 ∈ 𝑵 ⇒ 𝑥 + 1 > 1] Then, it is clear that for y ∈ N, there exists x = y ∈ N such that 𝑔of(x) = y. Hence, 𝑔of is onto. |
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Question 9: |
Question 10: Find the number of all onto functions from the set {1, 2, 3, … , n) to itself. Solution : Onto functions from the set {1, 2, 3, … , n} to itself is simply a permutation on n symbols 1, 2, …, n. Thus, the total number of onto maps from {1, 2, … , n} to itself is the same as the total number of permutations on n symbols 1, 2, …, n, which is n. |
Question 11: Let S = {a, b, c} and T = {1, 2, 3}. Find F‾¹ of the following functions F from S to T, if it exists. (i) F = {(a, 3), (b, 2), (c, 1)} (ii) F = {(a, 2), (b, 1), (c, 1)} Solution : S = {a, b, c}, T = {1, 2, 3} (i) F: S → T is defined as F = {(a, 3), (b, 2), (c, 1)} ⇒ F (a) = 3, F (b) = 2, F(c) = 1 Therefore, F‾¹: T → S is given by F‾¹ = {(3, a), (2, b), (1, c)}. (ii) F: S → T is defined as F = {(a, 2), (b, 1), (c, 1)} Since F (b) = F (c) = 1, F is not one – one. Hence, F is not invertible i.e., F‾¹ does not exist |
Question 12: Given a non – empty set X, let *: P(X) × P(X) → P(X) be defined as A * B = (A − B) ∪ (B − A), ∀ A, B ∈ P(X). Show that the empty set Φ is the identity for the operation * and all the elements A of P(X) are invertible with A‾¹ = A. (Hint: (A − Φ) ∪ (Φ − A) = A and (A − A) ∪ (A − A) = A * A = Φ). Solution : It is given that *: P(X) × P(X) → P(X) is defined as A * B = (A − B) ∪ (B − A) ∀ A, B ∈ P(X). Let A ∈ P(X). Then, we have A * Φ = (A − Φ) ∪ (Φ − A) = A ∪ Φ = A Φ * A = (Φ − A) ∪ (A − Φ) = Φ ∪ A = A ∴ A * Φ = A = Φ * A for all A ∈ P(X) Thus, Φ is the identity element for the given operation*. Now, an element A ∈ P(X) will be invertible if there exists B ∈ P(X) such that A * B = Φ = B * A. [As Φ is the identity element] Now, we observed that A * A = ( A – A) ∪ (A – A) = Φ ∪ Φ = Φ for all A ∈ P(X). Hence, all the elements A of P(X) are invertible with A‾¹ = A. |
Question 13: Consider the binary operations *: R × R → R and o: R × R → R defined as 𝑎 ∗ 𝑏 = |𝑎 − 𝑏| and a o b = a, ∀ a, b ∈ R. Show that * is commutative but not associative, o is associative but not commutative. Further, show that ∀ a, b, c ∈ R, a*(b o c) = (a * b) o (a * c). [If it is so, we say that the operation * distributes over the operation o]. Does o distribute over *? Justify your answer. Solution : It is given that *: R × R → and o: R × R → R is defined as 𝑎 ∗ 𝑏 = |𝑎 − 𝑏| and aob = a, ∀ a, b ∈ R For a, b ∈ R, we have 𝑎 ∗ 𝑏 = |𝑎 − 𝑏| and 𝑏 ∗ 𝑎 = |𝑏 − 𝑎| = |−(𝑎 − 𝑏)| = |𝑎 − 𝑏| ∴ a * b = b * a Hence, the operation * is commutative. It can be observed that (1 ∗ 2) ∗ 3 = (|1 − 2|) ∗ 3 = 1 ∗ 3 = |1 − 3| = 2 and 1 ∗ (2 ∗ 3) = 1 ∗ (|2 − 3|) = 1 ∗ 1 = |1 − 1| = 0 ∴ (1 ∗ 2) ∗ 3 ≠ 1 ∗ (2 ∗ 3) where 1, 2, 3 ∈ R. Hence, the operation * is not associative. Now, consider the operation o It can be observed that 1o2 = 1 and 2o1 = 2. ∴ 1o2 ≠ 2o1 where 1, 2 ∈ R. Hence, the operation o is not commutative. Let a, b, c ∈ R. Then, we have (aob)oc = aoc = a and ao(boc) = aob = a ∴ (aob)oc = ao(boc), where a, b, c ∈ R Hence, the operation o is associative. Now, let a, b, c ∈ R, then we have a * (b o c) = a * b =|𝑎 − 𝑏| (a * b) o (a * c) = (|𝑎 − 𝑏|)𝑜(|𝑎 − 𝑐|) = |𝑎 − 𝑏| Hence, a * (b o c) = (a * b) o (a * c). Now, 1 𝑜 (2 ∗ 3) = 1𝑜(|2 − 3|) = 1𝑜1 = 1 (1 o 2) * (1 o 3) = 1 * 1 = |1 − 1| = 0 ∴ 1 o (2 * 3) ≠ (1 o 2) * (1 o 3) where 1, 2, 3 ∈ R Hence, the operation o does not distribute over * |
Question 14: Define a binary operation *on the set {0, 1, 2, 3, 4, 5} as 𝑎 ∗ 𝑏 = {a+b, if a+b<6a+b−6, if a+b≥6 Show that zero is the identity for this operation and each element a ≠ 0 of the set is invertible with 6 − a being the inverse of a. Solution : Let X = {0, 1, 2, 3, 4, 5}. The operation * on X is defined as 𝑎 ∗ 𝑏 = {a+b, if a+b<6 and a+b−6, if a+b≥6 An element e ∈ X is the identity element for the operation *, if 𝑎 ∗ 𝑒 = 𝑎 = 𝑒 ∗ 𝑎 for all 𝑎 ∈ X For 𝑎 ∈ X, we have 𝑎 ∗ 0 = 𝑎 + 0 = 𝑎 [𝑎 ∈ X ⇒ 𝑎 + 0 < 6] 0 ∗ 𝑎 = 0 + 𝑎 = 𝑎 [𝑎 ∈ X ⇒ 0 + 𝑎 < 6] ∴ 𝑎 ∗ 0 = 𝑎 = 0 ∗ 𝑎 for all 𝑎 ∈ X Thus, 0 is the identity element for the given operation *. An element a ∈ X is invertible if there exists b ∈ X such that a * b = 0 = b * a. i.e., {a+b=0=b+a, if a+b<6a+b−6=0=b+a−6, if a+b≥6 ⇒ a = −b or b = 6 − a But, X = {0, 1, 2, 3, 4, 5} and a, b ∈ X. Then, a ≠ −b. ∴ b = 6 − a is the inverse of a for all a ∈ X. Hence, the inverse of an element a ∈X, a ≠ 0 is 6 − a i.e., a−1 = 6 – a |