Chapter 1 Relations and Functions
- Let * be a binary operation on the set Q of rational numbers as follows: (i) a * b = a – b Operation * is on the set Q. (i) Defined as a * b = a – b (a) Now b * a = b – a. But a – b ≠ b – a ∴ a * b ≠ b * a ∴ Operation * is not comm... 1
- Show that none of the operations given above has identity. The binary operation * on set Q is (i) defined as a * b = a – b. For idenity element e, a * e = e * a = a. But a * e = a – e # a and e * a =... 2
- Let A = N x N and * be the binary operation on A defined by (a, b) * (c, d) = (a + c,b+d) A = N x N Binary operation * is defined as (a, b) * (c, d) = (a + c, b + d). (a) Now, (c, d) * (a, b) = (c + a, d + b) = (a + c, b + d) ⇒ (a... 1
- State whether the following statements are true or false. Justify. (i) For any arbitrary binary operation * on a set N (i) A binary operation on N is defined as a * a = a ∀a ∈ N. Here, operation * is not defined. ∴ Given statement is false. (ii) * is a binary... 2
- Consider a binary operation * on N defined as a * b = a³ +b³. Choose the correct answer: (B) Is * commutative but not associative The binary operation * on set N is defined as a * b = a³ +b³ Now, b * a = b³ + a³ = a³ +b³. ⇒ a * b... 1
- Let f : {1. 3 4} → {1,2, 5} and g: {1, 2, 5} → {1,3} be given by f = {(1, 2), (3, 5), (4,1)} f= {(1,2), (3,5), (4,1)} g = {(I, 3), (2, 3), (5, 1)} (f) = 2, g(2) = 3 ⇒ gof(1) = 3 f(3) = 5, g(5) = 1 ⇒ gof(3) = 1 f(4) = 1, g(1) = 3 ⇒ go... 1
- Find gof and fog, if \(x^{\frac{1}{3}}\) (i) f(x) = |x|, and g(x) = |5x – 2|. (i) f(x) = |x|, and g(x) = |5x – 2| gof(x) = g[f(x)]=g|x| = |5|x|-2|. fog(x) = f[g(x)] = f(|5x- 2|) =||5x-2|| = |5x – 2|. (ii) f(x) = 8x³ an... 1
- State with reasons whether each of the following functions has a inverse or not: (i) f : {1, 2, 3, 4} → {10} with f= {(1, 10), (2, 10), (3, 10), (4, 10)}. (ii) g: {5, 6, 7, 8} → {1, 2, 3, 4} with g = {(5, 4), (6, 3), (7,... 1
- Consider f : R → R given by f(x) = 4x + 3. Show that f is invertible. Find the inverse of f. f : R → R given by f(x) = 4x + 3. f(x1) = 4x1 + 3, f(x2) = 4x2 + 3 If f(x1) = f(x2), then 4x1 + 3 = 4x2 + 3 or 4x1 = 4x2 ⇒ x1 = x2. ⇒ f is o... 1
- Consider f : R+ → [4, ∝) given by f(x) = x² + 4. Show that f is invertible with the inverse of f-1 of f f(x1) = x²1 + 4 and f(x2) = x²2 + 4 So, f(x1) = f(x2) ⇒ x²1 + 4 = x²2 + 4 or x²1 = x²2 ⇒ x1 = x2 As x ∈ R, ∴ x > 0, x²1 = x²2 ⇒ x1 = x2 ⇒ f... 1
- Let f : X → Y be an invertible function. Show that f has a unique inverse. If f is invertible, then gof(x) = Ix and fog(y) = Iy. ⇒ f is one-one and onto. Let there be two inverses g1 and g2 ∴ fog1 (y)= Iy, f0g2(y) =... 1
- Consider f : {1, 2, 3} → {a, b, c] given by f(1) = a, f(2) – b, f(3) = c. Find f-1 and show that (f-1)-1 = f. f : {1, 2, 3} → {a, b, c} so that f(1) = a, f(2) = b, f(3) = c. Now, let X = {1, 2, 3}, Y = {a, b, c). ∴ f : X → Y ∴ f-1 : Y → X such that f... 1
- Let f : X → Y be an invertible function. Show that the inverse of f-1 is f, i.e., (f-1)-1 = f. f : X → Y is an invertible function. ∴ f is one-one and onto ⇒g : Y → X, where g is also one-one and onto such that gof(x) = Ix and fog(y) =... 1
- If f : R → R be given by f(x) = (3-x³)\(\left(3-x^{3}\right)^{\frac{1}{3}}\), when fof(x) is Correct answer is (C) x 2
- Let f : R – (- \(\frac { 4 }{ 3 }\)) → R be a function, defined as f(x) = \(\frac { 4x }{ 3x+4 }\) f : R – -4 / 3 → R is a function defined by f(x) = 4x / 3x+4 Let y = 4x / 3x+4 . ⇒ 3xy + 4y = 4x or x(4 – 3y) = 4y ∴ x = 4y / 4-3y ∴ f-1(4y)... 1
- Show that the function f: R → R, defined by f (x) = \(\frac{1}{x}\) is one-one onto (i) f(x) = 1/x , If f(x1) = f(x2) ⇒ frac1x₁ = frac1x₂ ⇒ x1 = x2 Each x ∈ R, has a unique image in codomain. ⇒ f is one-one. (ii) For each y... 1
- Check the injectivity and surjectivity of the following functions: (i) f : N → N is given by (i) f : N → N given by f(x) = x2. (a) f(x1) = f(x2) ⇒ x₁² = x₂² ⇒ x1 = x2. ∴ f is one-one, i.e., it is injective. (b) There are some members... 1
- Prove that greatest integer function f : R → R, given by f(x) = [x], is neither one-one nor onto f : R → R given by f(x) = [x]. (a) f(1.2) = 1, f(1.5) = 1 ⇒ f is not one-one. (b) All the images of x ∈ R belonging to its domain have integ... 1
- Show that the Modulus Function f : R → R, given by f(x) = [x], is neither one-one nor onto, where | x | is x f : R → R given by f(x) = | x |. (a) f(-1) = |-1| = 1, f(1) = |1| = 1. ⇒ – 1 and 1 have the same image. ∴ f is not one-one. (b) No negative... 1
- Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let f = {1, 4), (2, 5), (3, 6)} be a function from A to B. A = (1, 2, 3}, B = {4, 5, 6, 7} and f = {(1, 4), (2, 5), (3, 6)}. Every member of A has a unique image in B as shown in the figure. ∴ f is o... 1
- In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer. (i) f : R → R defined by f(x) = 3 – 4x. f(x1) = 3 – 4x1, f(x2) = 3 – 4x2 (a) f(x1) = f(x2) ⇒ 3 – 4x1 = 3 – 4x2 ∴ x1 = x2 This shows that f i... 1
- Let A and B be two sets. Show that f : A × B → B × A such that f(a, b) = (b, a) is bijective function. We have f : (A × B) → (B × A) such that f(a, b) = (b, a). (a) ∴ f{a1, b1) = (b1, a1), f(a2, b2) = (b2, a2) and f(a1, b1) = f(a2, b2) ⇒ (b1,1... 2
- Let A = R-{3} and B = R-{1}. consider the function f: A → B defined by f (x) = \(\frac{x-2}{x-3}\). f : A → B, where A = R- 3, B = R – 1, f is defined by (x1 – 2)(x2 – 3) = (x2 – 2)(x1 – 3) or x1x2 – 3x1 – 2x2 + 6 = x1x2 – 2x1 – 3x2 + 6 i.e... 1
- 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. (D) f is neither one-one nor onto https://www.youtube.com/watch?v=RSbkdpekHC0 1