NCERT MISCELLANEOUS PROBLEMS

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    Miscellaneous Problems

    Question 1: Determine whether or not each of the definition of given below gives a binary operation.
    In the event that * is not a binary operation, give justification for this.
    Solution :
    It is given that f: R → R is defined as f(x) = 10x + 7.
    For one – one
    Let f(x) = f(y), where x, y ∈ R.
    ⇒ 10x + 7 = 10y + 7
    ⇒ x = y
    ∴ f is a one – one function.
    For onto
    For y ∈ R, let y = 10x + 7.
    ⇒ 𝑥 =y−7/ 10    ∈ 𝑹
    Therefore, for any y ∈ R, there exists 𝑥 =y−7 / 10   ∈ 𝑹 such that
    𝑓(𝑥) = 𝑓 (y−7/ 10) = 10 (y−7 /10) + 7 = 𝑦 − 7 + 7 = 𝑦
    ∴ f is onto.
    Therefore, f is one – one and onto.
    Thus, f is an invertible function.
    Let us define 𝑔: R → R as 𝑔(𝑦) = y−7 /10
    Now, we have
    𝑔𝑜𝑓(𝑥) = 𝑔(𝑓(𝑥)) = 𝑔(10𝑥 + 7) =(10x+7)−7/ 10=  10x /10= x
    and
    𝑓𝑜𝑔(𝑦) = 𝑓(𝑔(𝑦)) = 𝑓 (y−7 /10) = 10 ( y−7 /10)+7 = 𝑦 − 7 + 7 = 𝑦
    ∴ 𝑔𝑜𝑓 = I𝑅 𝑎𝑛𝑑 𝑓𝑜𝑔 = IR.                                                                                             Hence, the required function 𝑔: R → R is defined as 𝑔(𝑦) = y−7/ 10.
    Question 2:
    Let f: W → W be defined as f(n) = n − 1, if is odd and f(n) = n + 1, if n is even. Show that f is invertible. Find the inverse of f. Here, W is the set of all whole numbers.            Solution :
    It is given that:
    f: W → W is defined as 𝑓(𝑛) = {n−1, if n is odd                n+1, if n is even
    For one – one
    Let f(n) = f(m).
    It can be observed that if n is odd and m is even, then we will have n − 1 = m + 1.
    ⇒ n − m = 2
    However, this is impossible.
    Similarly, the possibility of n being even and m being odd can also be ignored under a similar argument.
    ∴ Both n and m must be either odd or even. Now, if both n and m are odd,
    Then, we have
    f(n) = f(m)
    ⇒ n − 1 = m – 1
    ⇒ n = m
    Again, if both n and m are even,
    Then, we have
    f(n) = f(m)
    ⇒ n + 1 = m + 1
    ⇒ n = m
    ∴ f is one – one.
    For onto
    It is clear that any odd number 2r + 1 in co-domain N is the image of 2r in domain N and any even number 2r in co-domain N is the image of 2r + 1 in domain N.
    ∴ f is onto.
    Hence, f is an invertible function.
    Let us define 𝑔: W → W as 𝑔(𝑚) = {m+1, if m is even              m−1, if m is odd
    Now, when n is odd
    𝑔𝑜𝑓(𝑛) = 𝑔(𝑓(𝑛)) = 𝑔(𝑛 − 1) = 𝑛 − 1 + 1 = 𝑛 and
    When n is even
    𝑔𝑜𝑓(𝑛) = 𝑔(𝑓(𝑛)) = 𝑔(𝑛 + 1) = 𝑛 + 1 − 1 = 𝑛                                                                Similarly,
    When m is odd
    𝑓𝑜𝑔(𝑚) = 𝑓(𝑔(𝑚)) = 𝑓(𝑚 − 1) = 𝑚 − 1 + 1 = 𝑚 and
    When m is even
    𝑓𝑜𝑔(𝑚) = 𝑓(𝑔(𝑚)) = 𝑓(𝑚 + 1) = 𝑚 + 1 − 1 = 𝑚
    ∴ 𝑔𝑜𝑓 = IW and 𝑓𝑜𝑔 = IW
    Thus, f is invertible and the inverse of f is given by f ‾¹= 𝑔, which is the same as f.
    Hence, the inverse of f is f itself
    Question 3:
    If f: R → R is defined by f(x) = x² − 3x + 2, find f(f(x)).
    Solution :
    It is given that f: R → R is defined as f(x) = x² − 3x + 2.
    𝑓(𝑓(𝑥)) = 𝑓(𝑥² − 3𝑥 + 2)
    = (𝑥² − 3𝑥 + 2)2 − 3(𝑥² − 3𝑥 + 2) + 2
    = (𝑥^4 + 9𝑥² + 4 − 6𝑥³ − 12𝑥 + 4𝑥²) + (−3𝑥² + 9𝑥 − 6) + 2
    = 𝑥^4 − 6𝑥³ + 10𝑥² – 3x
    Question 4:
    Show that function f: R → {x ∈ R: −1 < x < 1} defined by f(x) = 𝑥1+|𝑥|′ x ∈ R is one – one and onto function.
    Solution :
    It is given that f: R → {x ∈ R: −1 < x < 1} is defined as f(x) = 𝑥/ 1+|𝑥|′ x ∈ R.
    For one – one
    Suppose f(x) = f(y), where x, y ∈ R.
    ⇒ x /1+|x|= y /1+|y|
    It can be observed that if x is positive and y is negative,
    Then, we have x1+x=y1+y ⇒ 2𝑥𝑦 = 𝑥 − 𝑦
    Since, x is positive and y is negative
    x > y ⇒ x − y > 0
    But, 2xy is negative.
    Then 2𝑥𝑦 ≠ 𝑥 − 𝑦
    Thus, the case of x being positive and y being negative can be ruled out.
    Under a similar argument, x being negative and y being positive can also be ruled out.
    ∴ x and y have to be either positive or negative.                                                           When x and y are both positive, we have
    𝑓(𝑥) = 𝑓(𝑦) ⇒ x / 1+x=y / 1+y ⇒ 𝑥 + 𝑥𝑦 = 𝑦 + 𝑥𝑦 ⇒ 𝑥 = 𝑦
    When x and y are both negative, we have
    𝑓(𝑥) = 𝑓(𝑦) ⇒ x /1− x=y / 1− y ⇒ 𝑥 − 𝑥𝑦 = 𝑦 − 𝑥𝑦 ⇒ 𝑥 = 𝑦
    ∴ f is one – one.                                                                                                         For onto
    Now, let y ∈ R such that −1 < y < 1.
    If y is negative, then, there exists 𝑥 = y1+y ∈ R such that
    𝑓(𝑥) = 𝑓 (y / 1+y) = (y / 1+y)1+|y / 1+y|=y/ 1+y / (1+(−y1+y)) =y /( 1+y−y)= y
    If y is positive, then, there exists 𝑥 = y1− y ∈ R such that
    𝑓(𝑥) = 𝑓 (y / 1− y) = (y / 1− y)1+|y / 1− y|=y /1− y1+(−y / 1− y)=y / (1− y+ y)=y
    ∴ f is onto.
    Hence, f is one – one and onto.
    Question 5:
    Show that the function f: R → R given by f(x) = x³ is injective.
    Solution :
    f: R → R is given as f(x) = x³.
    For one – one
    Suppose f(x) = f(y), where x, y ∈ R.
    ⇒ x³ = y³ (1)
    Now, we need to show that x = y.
    Suppose x ≠ y, their cubes will also not be equal.
    ⇒ x³ ≠ y³
    However, this will be a contradiction to (1).
    ∴ x = y Hence, f is injective.
    Question 6:
    Give examples of two functions f: N → Z and 𝑔: Z → Z such that 𝑔of is injective but 𝑔 is not injective.
    (Hint: Consider f(x) = x and 𝑔 (x) = |𝑥|)                                                                      Solution :
    Define f: N → Z as f(x) = x and 𝑔: Z → Z as 𝑔 (x) = |𝑥|.
    We first show that 𝑔 is not injective.
    It can be observed that
    𝑔(−1) =|−1| = 1
    𝑔(1) = |1| = 1
    ∴ 𝑔(−1) = 𝑔(1), but −1 ≠ 1.
    ∴ 𝑔 is not injective.
    Now, 𝑔 of: N → Z is defined as 𝑔𝑜𝑓(𝑥) = 𝑔(𝑓(𝑥)) = 𝑔(𝑥) = |𝑥|.
    Let x, y ∈ N such that 𝑔of(x) = 𝑔of(y).
    ⇒ |𝑥| = |𝑦|
    Since x and y ∈ N, both are positive.
    ∴ |𝑥| = |𝑦| ⇒ 𝑥 = 𝑦
    Hence, gof is injective.
    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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