NCERT EXERCISE 2.1

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    Chapter 2 – Inverse Trigonometric Functions                                                              Exercise 2.1

    Question 1:
    Find The principal value of sin
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    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.

    Question 8:
    Given a non-empty set X, consider P(X) which is the set of all subsets of X.
    Define the relation R in P(X) as follows:
    For subsets A, B in P(X), ARB if and only if A ⊂ B. Is R an equivalence relation on P(X)? Justify you answer
    Solution :
    Since every set is a subset of itself, A R A for all A ∈ P(X).
    ∴ R is reflexive.
    Let A R B ⇒ A ⊂ B.
    This cannot be implied to B ⊂ A.
    For instance, if A = {1, 2} and B = {1, 2, 3}, then it cannot be implied that B is related to A.
    ∴ R is not symmetric.
    Further, if A R B and B R C, then A ⊂ B and B ⊂ C.
    ⇒ A ⊂ C
    ⇒ A R C
    ∴ R is transitive.
    Hence, R is not an equivalence relation as it is not symmetric

    Question 9:
    Given a non-empty set X, consider the binary operation *: P(X) × P(X) → P(X) given by
    A*B = A∩B∀A,B in P(X) is the power set of X. Show that X is the identity element for this operation and X is the only invertible element in P(X) with respect to the operation*.
    Solution :
    It is given the binary operation *:
    P(X) × P(X) → P(X) given by A * B = A ∩ B ∀ A, B in P(X)
    We know that A∩X = A = X ∩ A for all A ∈ P(X)
    ⇒ A * X = A = X * A for all A ∈ P(X)
    Thus, X is the identity element for the given binary operation *.
    Now, an element A ∈ P(X) is invertible if there exists B ∈ P(X) such that
    A*B = X = B*A [As X is the identity element]
    or
    A∩B = X = B∩A
    This case is possible only when A = X = B.
    Thus, X is the only invertible element in P(X) with respect to the given operation*.
    Hence, the given result is proved.

    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
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