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Functions Exercise 1b Solutions

Education, TS Inter Maths 1A / Nayini Satyanarayana Reddy

Functions Exercise 1b Solutions Functions Exercise 1b The famous mathematician ” Lejeune Dirichlet” defined a function. Function: A variable is a symbol which represents any one of a set of numbers, if two variables x and y so related that whenever a value is assigned to x there is autometically assigned by some rule or correspondence a value to y, then we say y is a function of x. Chapter 1 Functions Exercise 1b Solutions for inter first year students, prepared by Mathematics expert of www.basicsinmaths.com Functions Exercise 1b Exercise 1(b) Solutions I. 1.If f (x) = ex and g(x) = , then show that f og = gof and f-1 = g-1 Sol: Given f (x) = ex and g(x) = (fog) (x) = f (g (x)) = f ( ) = = x ————- (1) (gof) (x) = g (f (x)) = g (ex) = = x = x ————- (2) From (1) and (2) f og = gof let y = f(x) x = f-1(y) y = ex x = f-1 (x) = let g(x) = z x = g-1(z) z = x = ez g-1 (x) = ex 2. If f (y) = , g (y) = then show that fog(y) = y Sol: Given f (y) = , g (y) = Now fog(y) = f(g(y)) ∴ fog(y) = y 3. If f: R ⟶ R, g: R ⟶ R is defined by f(x) = 2×2 + 3 ang g (x) = 3x – 2 then find (i) (fog) (x) (ii) (gof) (x) (iii) (fof) (0) (iv) go(fof)(3) Sol: Given f: R ⟶ R is, g: R ⟶ R is defined by f(x) = 2×2 + 3 ang g (x) = 3x – 2 (i) (fog) (x) = f(g(x)) = f(3x – 2) = 2 (3x – 2)2 + 3 = 2 (9×2 – 12x + 4) + 3 = 18 x2 – 24x + 8 + 3 = 18 x2 – 24x + 11 ∴ (fog) (x) = 18 x2 – 24x + 11 (ii) (gof) (x) = g (f (x)) = g (2×2 + 3) = 3(2×2 + 3) – 2 = 6×2 + 9 – 2 = 6 x2 + 7 ∴ (gof) (x) = 6 x2 + 7 PDF Files || Inter Maths 1A &1B || (New) 6th maths notes|| TS 6 th class Maths Concept TS 10th class maths concept (E/M) (iii) (fof) (0) = f (f (0)) = f (2(0)2 + 3) = f (2(0) + 3) = f (3) = 2 (3)2 + 3 = 2 (9) + 3 = 18 + 3 = 21 (iv) go(fof) (3) = go (f (f (3))) = go (f (2 (3)2 + 3)) = go (f (21)) = g (f (21)) = g (2 (21)2 + 3)) = g (2 (441) + 3)) = g (882 + 3) = g (885) = 3 (885) – 2 = 2655 – 2 = 2653 ∴ go(fof) (3) = 2653 Functions Exercise 1b Ts Inter Maths IA Concept 4. If f: R ⟶ R; g: R ⟶ R is defined by f(x) = 3x – 1, g (x) = x2 + 1, then find (i) (fof) (x2 + 1) (ii) (fog) (2) (iii) (gof) (2a – 3) Sol: Given f: R ⟶ R; g: R ⟶ R is defined by f(x) = 3x – 1, g (x) = x2 + 1 (i) (fof) (x2 + 1) = f (f (x2 + 1)) = f (3 (x2 + 1)– 1) = f (3×2 + 3– 1) = f (3×2 + 2) = 3(3×2 + 2) – 1 = 9×2 + 6 – 1 = 9×2 + 5 (ii) (fog) (2) = f (g (2)) = f (22 + 1) =f (4 + 1) = f (5) = 3(5) – 1 = 15 – 1 = 14 (iii) (gof) (2a – 3) = g (f (2a – 3)) = g (3(2a – 3) – 1) = g (6a – 9 – 1) = g (6a – 10) = (6a – 10)2 + 1 = 362 – 120a + 100 + 1 = 362 – 120a + 101 5. If f(x) = and g(x) = for all x ∈ (0, ∞) then find (gof) (x) Sol: Given f(x) = and g(x) = for all x ∈ (0, ∞) (gof) (x) = g (f (x)) = g ( ) = ∴ (gof) (x) = 6. If f(x) = 2x – 1 and g(x) = for all x ∈ R then find (gof) (x) Sol: Given f(x) = 2x – 1 and g(x) = for all x ∈ R ∴ gof(x) = x 7. If f (x) = 2, g (x) = x2 and h(x) = 2x ∀ x ∈ R, then find (fo(goh)) (x) Sol: Given f (x) = 2, g (x) = x2 and h(x) = 2x ∀ x ∈ R (fo(goh)) (x) = fo (g (h (x)) = fo g(2x) = f (g(2x)) = f((2x)2) = f(4×2) = 2 ∴ (fo(goh)) (x) = 2 Functions Exercise 1b Solutions 8. Find the inverse of the following functions (i) a, b ∈ R, f: R ⟶ R defined by f(x) = ax + b, (a ≠ 0) Given a, b ∈ R, f: R ⟶ R defined by f(x) = ax + b Let y = f(x) x = f-1(y) now y =

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chapter 1 Functions Exercise 1a Solutions

Education, TS Inter Maths 1A / Nayini Satyanarayana Reddy

Functions Exercise 1a Solutions Functions Exercise 1a The famous mathematician ” Lejeune Dirichlet” defined a function. Function: A variable is a symbol which represents any one of a set of numbers, if two variables x and y so related that whenever a value is assigned to x there is autometically assigned by some rule or correspondence a value to y, then we say y is a function of x. Chapter 1 Functions Exercise 1a Solutions for inter first year students, prepared by Mathematics expert of www.basicsinmaths.com Exercise 1a I. 1. If the function f is defined by then find the values of (i) f (3) (ii) f (0) (iii) f (– 1.5) (iv) f (2) + f (– 2) (v) f (– 5 ) Sol: Given Domain of f(x) is (– 3, ∞) (i) f (3) 3 lies in the interval x > 1 ⟹ f(x) = x + 2 f(3) = 3 + 2 = 5 ∴ f (3) = 5 (ii) f (0) 0 lies in interval – 1 ≤ x ≤ 1 ⟹f(x) = 2 ∴ f (0) = 2 (iii) f (– 1.5) – 1.5 lies in interval – 3 < x < – 1 ⟹f(x) = x – 1 f (– 1.5) = – 1.5 – 1 = – 2.5 ∴ f (– 1.5) = – 2.5 (iv) f (2) + f (– 2) 2 lies in the interval x > 1 ⟹ f (x) = x + 2 f (3) = 2 + 2 = 4 f (2) = 4 – 2 lies in interval – 3 < x < – 1 ⟹f(x) = x – 1 f (– 2) = – 2 – 1 = – 3 f (– 2) = – 2 – 1 = – 3 now f (2) + f (– 2) = 4 – 3 = 1 ∴ f (2) + f (– 2) = 1 (v) f (– 5) since domain of f(x) is (– 3, ∞) f (– 5) is not defined 2. If f: R – {0} ⟶ R is defined by f(x) = , then show that f (x) + f (1/x) = 0 Sol: Given f: R – {0} ⟶ R is defined by f(x) = f (1/x) = Now f (x) + f (1/x) = ∴ f (x) + f (1/x) = 0 3. If f: R ⟶ R is defined by f(x) = , then show that f (tan θ) = cos 2θ Sol: Given f: R ⟶ R is defined by f(x) = f (tan θ) = = cos 2θ 4. If f: R – {±1} ⟶ R is defined by f(x) = , then show that f = 2 f (x) Sol: Given f: R – {±1} ⟶ R is defined by f(x) = 5. If A = {– 2, – 1, 0, 1, 2} and f: A ⟶ B is a surjection (onto function) defined by f(x) = x2 + x + 1, then find B Sol: Given A = {– 2, – 1, 0, 1, 2} and f: A ⟶ B is a surjection defined by f(x) = x2 + x + 1 f(– 2) = (–2)2 + (–2) + 1 = 4 – 2 + 1 = 3 f(– 1) = (–1)2 + (–1) + 1 = 1 – 1 + 1 = 1 f(0) = (0)2 + (0) + 1 = 0 + 0 + 1 = 1 f(1) = (1)2 + (1) + 1 =1 +1 + 1 = 3 f( 2) = (2)2 + (2) + 1 = 4 + 2 + 1 = 4 ∴ B = {1, 3, 7} TS 10th class maths concept (E/M) Functions Exercise 1a 6. If A = {1, 2, 3,4} and f: A ⟶ B is a surjection defined by f(x) = , then find range of f. Sol: Given A = {1, 2, 3,4} and f: A ⟶ B is a surjection defined by f(x) = 7. If f (x + y) = f (xy) ∀ x, y ∈ R, then prove that f is a constant function Sol: Given f (x + y) = f (xy) ∀ x, y ∈ R Let x = 0 and y = 0 f (0 + 0) = f (0 × 0) = f (0) f (1) = f (0 + 1) = f (0 × 1) = f (0) f (2) = f (1 + 1) = f (1 × 1) = f (1) = f (0) f (3) = f (1 + 2) = f (1 × 2) = f (2) = f(0) Similarly, f(4) = 0 f(5) = 0 and so on. ∴ f is a constant function PDF Files || Inter Maths 1A &1B || (New) 6th maths notes|| TS 6 th class Maths Concept II. 1. If A = {x/ – 1 ≤ x ≤ 1}, f(x) = x2, g(x) = x3, which of the following are surjections? (i) f: A ⟶ A (ii) g: A ⟶ A Sol: (i) Given A = {x/ – 1 ≤ x ≤ 1}, f(x) = x2 A = {– 1, 0, 1}; f: A ⟶ A f (x) = x2 f (– 1) = (– 1)2 = 1 f (0) = (0)2 = 0 f (1) = (1)2 = 1 ∵ range is not equal to co domain of f f is nor a surjection (ii) A = {x/ – 1 ≤ x ≤ 1}, g (x) = x3 A = {– 1, 0, 1}; g: A ⟶ A g (x) = x3 g (– 1) = (– 1)3 =

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Ts Inter Maths IA Concept

TS Inter Maths, TS Inter Maths 1A / Nayini Satyanarayana Reddy

1. Functoins

Set: A collection of well-defined objects is called set. Ordered pair: Two elements a and b listed in a specific order form. An ordered pair denoted by (a, b). Cartesian product: Let A and B are two non- empty sets. The Cartesian product of A and B is denoted by A × B and is defined as set of all ordered pairs (a, b) where a ϵ A and b ϵB cartesion product Rel

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