Exercise 1.2
The Exercise 1.2 deals with functions, sets and relation , injection , surjection and certain other properties of sets.
The exercise 1.2 also include one-one and onto functions. The Exercise 1.2 provides a lot of working problems to make you more familiar and through with the concepts..
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QUESTIONS:
Question 1: Show that the function f: 𝐑∗ → 𝐑∗ defined by 𝑓(𝑥) = 1/𝑥 is one-one and onto, where 𝐑∗ is the set of all non-zero real numbers. Is the result true, if the domain 𝐑∗ is replaced by N with co-domain being same as 𝐑∗? |
Solution 1: It is given that f: R* → 𝑅∗ is defined by f(x) = 1/𝑥 For one – one: Let x, y ∈ R∗ such that f(x) = f(y) ⇒ 1/𝑋=1/𝑌 ⇒ x = y ∴ f is one – one. For onto: It is clear that for y ∈ R∗ , there exists x = 1/𝑦 ∈ R∗ [as y ≠ 0] such that f(x) = 1/(1/𝑦) = y ∴ f is onto. Thus, the given function f is one – one and onto. Now, consider function g: N → R∗ defined by g(x) = 1/x We have, 𝑔(𝑥1) = 𝑔(𝑥2) ⇒ =1/X1=1/X2 ⇒ 𝑥1 = 𝑥2 ∴ g is one – one. Further, it is clear that g is not onto as for 1.2 ∈ = R∗ there does not exit any x in N such that g(x) = 1/1.2 . Hence, function g is one-one but not onto. |
Question 2: Check the injectivity and surjectivity of the following functions: |
(i) f: N → N given by f(x) = x² Sol. It is seen that for x, y ∈ N, f(x) = f(y) ⇒ x² = y² ⇒ x = y. ∴ f is injective. Now, 2 ∈ N. But, there does not exist any x in N such that f(x) = x² = 2. ∴ f is not surjective. Hence, function f is injective but not surjective. (ii) f: Z → Z is given by f(x) = x² It is seen that f(−1) = f(1) = 1, but −1 ≠ 1. ∴ f is not injective. Now, −2 ∈ Z. But, there does not exist any element x ∈ Z such that f(x) = −2 or x² = −2. ∴ f is not surjective. Hence, function f is neither injective nor surjective. |
(iii)f: R → R is given by f(x) = x² It is seen that f(−1) = f(1) = 1, but −1 ≠ 1. ∴ f is not injective. Now, −2 ∈ R. But, there does not exist any element x ∈ R such that f(x) = −2 or x² = −2. ∴ f is not surjective. Hence, function f is neither injective nor surjective. (iv) f: N → N given by f(x) = x³ It is seen that for x, y ∈ N, f(x) = f(y) ⇒ x³ = y³ ⇒ x = y. ∴ f is injective. Now, 2 ∈ N. But, there does not exist any element x ∈ N such that f(x) = 2 or x³ = 2. ∴ f is not surjective Hence, function f is injective but not surjective. (v) f: Z → Z is given by f(x) = x³ It is seen that for x, y ∈ Z, f(x) = f(y) ⇒ x³ = y³ ⇒ x = y. ∴ f is injective. Now, 2 ∈ Z. But, there does not exist any element x ∈ Z such that f(x) = 2 or x³ = 2. ∴ f is not surjective. Hence, function f is injective but not surjective. |
Question 3: Prove that the Greatest Integer Function f: R → R given by f(x) = [x], is neither one – one nor onto, where [x] denotes the greatest integer less than or equal to x. Solution : f: R → R is given by, f(x) = [x] It is seen that f(1.2) = [1.2] = 1, f(1.9) = [1.9] = 1. ∴ f(1.2) = f(1.9), but 1.2 ≠ 1.9. ∴ f is not one – one. Now, consider 0.7 ∈ R. It is known that f(x) = [x] is always an integer. Thus, there does not exist any element x ∈ R such that f(x) = 0.7. ∴ f is not onto. Hence, the greatest integer function is neither one – one nor onto. |
Question 4: In Show that the Modulus Function f: R → R given by 𝑓(𝑥) = |𝑥|, is neither one – one nor onto, where |𝑥| is x, if x is positive or 0 and |X| is − x, if x is negative. Solution : f: R → R is given by f(x) = |x| = { {X if X≥0, −X if x<0 It is clear that f(-1) = |-1| = 1 and f(1) = |1| = 1 ∴ f(−1) = f(1), but −1 ≠ 1. ∴ f is not one – one. Now, consider −1 ∈ R. It is known that f(x) = |𝑥| is always non-negative. Thus, there does not exist any element x in domain R such that f(x) = |𝑥| = −1. ∴ f is not onto. Hence, the modulus function is neither one-one nor onto. |
Question 5: Show that the Signum Function f: R → R, given by 𝑓(𝑥) = {1 if X>0 0, if X=0 −1, if X<0 is neither one-one nor onto. Solution: f: R → R is given by f(x) == {1 if X>0 0, if X=0 −1, if X<0 It is seen that f(1) = f(2) = 1, but 1 ≠ 2. ∴ f is not one – one. Now, as f(x) takes only 3 values (1, 0, or −1) for the element −2 in co-domain R, there does not exist any x in domain R such that f(x) = −2. ∴ f is not onto. Hence, the Signum function is neither one – one nor onto |
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Question 8: Let A and B be sets. Show that f: A × B → B × A such that (a, b) = (b, a) is bijective function. Solution: f: A × B → B × A is defined as f(a, b) = (b, a). Let (a1, b1 ), (a2, b2 ) ∈ A × B such that f(a1, b1 ) = f(a2, b2 ) ⇒ (b1, a1 ) = (b2, a2 ) ⇒ b1 = b2 and a1 = a2 ⇒ (a1, b1 ) = (a2, b2 ) ∴ f is one – one. Now, let (b, a) ∈ B × A be any element. Then, there exists (a, b) ∈ A × B such that f(a, b) = (b, a). [By definition of f ] ∴ f is onto. Hence, f is bijective. |
Question 9: Let f: N → N be defined by f(n) = {n+1/2 ,if n is odd n/2,if n is even for all n ∈ N State whether the function f is bijective. Justify your answer. Solution: f: N → N is defined as f(n) == {n+1/2 ,if n is odd n/2,if n is even for all n ∈ N It can be observed that: f(1) = 1+1/2 = 1 and f(2) = 2/2 = 1 [By definition of f(n)] f(1) = f(2), where 1 ≠ 2 ∴ f is not one-one. Consider a natural number (n) in co-domain N. Case I: n is odd ∴ n = 2r + 1 for some r ∈ N. Then, there exists 4r + 1∈ N such that f(4r + 1) = 4𝑟+1 + 1/2= 2r + 1 Case II: n is even ∴ n = 2r for some r ∈ N. Then, there exists 4r ∈ N such that f(4r) = 4𝑟/2 = 2r. ∴ f is onto. Hence, f is not a bijective function. |
Question 10: Let A = R − {3} and B = R − {1}. Consider the function f: A → B defined by (𝑥) = (𝑥−2/𝑥−3) . Is f one-one and onto? Justify your answer. Solution: A = R − {3}, B = R − {1} and f: A → B defined by f(x) = (𝑋−2/𝑋− 3 ) Let x, y ∈ A such that f(x) = f(y) ⇒ X−2/(X− 3) = Y−2/(Y− 3) ⇒ (x – 2)(y – 3) = (y – 2)(x – 3) ⇒ xy – 3x – 2y + 6 = xy – 2x – 3y + 6 ⇒ – 3x – 2y = – 2x – 3y ⇒ x = y ∴ f is one-one. Let y ∈ B = R − {1}. Then, y ≠ 1. The function f is onto if there exists x ∈ A such that f(x) = y. Now, f(x) = y ⇒ X−2/(X− 3) = y ⇒ x – 2 = xy – 3y ⇒ x(1 – y) = – 3y + 2 ⇒ x = 2− 3y/1−y ∈ A [y ≠ 1] Thus, for any y ∈ B, there exists 2− 3y1−y ∈ A such that f (2− 3y/(1−y)) = (2−3𝑦/(1−𝑦))−2(2−3𝑦/(1−𝑦))−3 = 2−3y−2 + 2y/(2−3y−3 + 3y )= −𝑦/−1=𝑦 ∴ f is onto. Hence, function f is one – one and onto. |
Question 11: Let f: R → R be defined as f(x) = x4 . Choose the correct answer. (A) f is one-one onto (B) f is many-one onto (C) f is one-one but not onto (D) f is neither one-one nor onto Solution : f: R → R is defined as f(x) = xˆ4. Let x, y ∈ R such that f(x) = f(y). ⇒ xˆ4 = yˆ4 ⇒ x = ± y ∴ f(x) = f(y) does not imply that x = y. For example f(1) = f(–1) = 1 ∴ f is not one-one. Consider an element 2 in co-domain R. It is clear that there does not exist any x in domain R such that f(x) = 2. ∴ f is not onto. Hence, function f is neither one – one nor onto. The correct answer is D. |
Question 12: Let f: R → R be defined as f(x) = 3x. Choose the correct answer. (A) f is one – one onto (B) f is many – one onto (C) f is one – one but not onto (D) f is neither one – one nor onto Solution: f: R → R is defined as f(x) = 3x. Let x, y ∈ R such that f(x) = f(y). ⇒ 3x = 3y ⇒ x = y ∴f is one-one. Also, for any real number (y) in co-domain R, there exists 𝑦/3 in R such that f (y/3) = 3 (y/3) = y ∴ f is onto. Hence, function f is one – one and onto. The correct answer is A. |