# 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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