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Inter 1st Year Maths IA-Functions Study Material
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Sikkim Manipal University
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FUNCTIONS Def 1: A relation f from a set A into a set B is said to be a function or mapping from A into B if for each x ∈ A there exists a unique y ∈ B such that ( x, y )∈ f . It is denoted by f : A → B . Note: Example of a function may be represented diagrammatically. The above example can be written diagrammatically as follows. A B 1 2 3 p q r Def 2: A relation f from a set A into a set B is a said to be a function or mapping from a into B if i) x ∈ A ⇒ f ( x)∈ B ii) x1 , x2 ∈ A, x2 ⇒ f ( x1 ) = f ( x2 ) Def 3: If f : A → B is a function, then A is called domain, B is called codomain and f ( A) = { f ( x): x ∈ A} is called range of f. Def 4: A function f : A → B if said to be one one function or injection from A into B if different element in A have different f-images in B. Note: A function f : A → B is one one if f ( x1 , y ) ∈ f , ( x2 , y ) ∈ f ⇒ x1 = x2 . Note: A function f : A → B is one one iff x1 , x2 ∈ A, x1 ≠ x2 ⇒ f ( x1 ) ≠ f ( x 2 ) Note: A function f : A → B is one one iff x1 , x2 ∈ A, f ( x1 ) = f ( x2 ) ⇒ x1 = x 2 Note: A function f : A → B which is not one one is called many one function Note: If f : A → B is one one and A, B are finite then n( A) ≤ n( B) . Def 5: A function f : A → B is said to be onto function or surjection from A onto B if f(A) = B. Note: A function f : A → B is onto if y ∈ B ⇓ ∃x ∈ A ∋ f ( x) = y . Note: A function f : A → B which is not onto is called an into function. Note: If A, B are two finite sets and f : A → B is onto then n( B ) ≤ n( A) . Note: If A, B are two finite sets and n (B) = 2, then the number of onto functions that can be defined from A onto B is 2n ( A) − 2 . Def 6: A function f : A → B is said to be one one onto function or bijection from A onto B if f : A → B is both one one function and onto function. Theorem: If f : A → B , g : B → C are two functions then the composite relation gof is a function a into C. Theorem: If f : A → B , g : B → C are two one one onto functions then gof : A → C is also one one be onto. Sol: i) Let x1 , x2 ∈ A and f ( x1 ) = f ( x2 ) . x1 , x2 ∈ A , f : A → B ⇒ f ( x1 ) , f ( x2 ) ∈ B f ( x1 ), f ( x2 )∈ B, → C, f ( x2 ) ⇒ g[ f ( x1 ) ] = g[ f ( x2 )] ⇒ ( gof ) ( x1 ) = ( gof ) ( x2 ) x1 , x2 ∈ A, ( gof ) ( x1 ) = ( gof ): A → C is one one ⇒ x1 = x2 ∴ x1 , x2 ∈ A, f ( x1 ) = f ( x2 ) ⇒ x1 = x2 . ∴ f : A → B Is one one. ii) Proof: let z ∈ C , g : B → C is onto ∃y ∈ B ∃ : g ( y ) = z y ∈ B f : A → B is onto ∴∃x ∈ A ∋ f ( x ) = y G {f(x)} = t (g o f) x = t ∀z ∈ C ∃ x ∈ A ∋ ( gof ) ( x) = z . ∴g is onto. Def 7: Two functions f : A → B , g : C → D are said to be equal if i) A = C, B = D ii) f ( x ) = g ( x ) ∀ x ∈ A . It is denoted by f = g Theorem: If f : A → B , g : B → C , h : C → D are three functions, then ho( gof ) = (hof )of Theorem: if A is set, then the identify relation I on A is one one onto. Def 8: If A is a set, then the function I on A defined by I ( x ) = x∀ x ∈ A , is called identify function on A. it is denoted by I A . Theorem: If f : A → B and I A , I B are identify functions on A, B respectively then foI A = I B of = f . Proof: I A : A → A , f : A → B ⇒ foI A : A → B f : A → B , I B : B → B ⇒ I B of : A → B ∴ f0 I A = f ( foI A ) ( x ) = f {I A ( x )} = f ( x ), ∀ x ∈ A . ( I B of ) ( x ) = I B { f ( x )} = f ( x ), ∀ x ∈ A ∴ I B of = f ∴ fo I A = I B of = f Def 9: If f : A → B is a function then {( y, x )∈ B × A :( x, y )∈ f } is called inverse of f. It is denoted by f −1 . Def 10: If f : A → B is a bijection, then the function f −1 : B → A defined by f −1 ( y ) = x iff f ( x ) = y∀ y ∈ B is called inverse function of f. Theorem: If f : A → B , g : B → A are two functions such that gof = I A and fog = I B then f : A → B is a bijection and f −1 = g . Proof: Let x1 , x2 ∈ A, f ( x1 ) = f ( x2 ) x1 , x2 ∈ A, f : A → B ⇒ f ( x1 ), f ( x2 ) ∈ B f ( x1 ) , f ( x2 ) ∈ B, f ( x1 ) = f ( x2 ) , g = B → A ⇒ g [ f ( x1 )] = g [ f ( x2 )] ⇒ ( gof ) ( x1 ) = ( gof )( x2 ) ⇒ I A ( x2 ) ⇒ x1 = x2 ∴ x1 , x2 ∈ A , f ( x1 ) = f ( x2 ) ⇒ x1 = x2 . ∴ f : A → B is one one Let y ∈ B . y ∈ B, g : B → A ⇒ g ( y ) ∈ A Def 11: A function f : A → B is said tobe a constant function if the range of f contain only one element i., f ( x) = c∀ x ∈ A where c is a fixed element of B Def 12: A function f : A → B is said to be a real variable function if A ⊆ R . Def 13: A function f : A → B is said to be a real valued function iff B ⊆ R . Def 14: A function f : A → B is said to be a real function if A ⊆ R, B ⊆ R . Def 15: If f : A → R , g : B → R then f + g : A ∩ B → R is defined as ( f + g )( x) = f ( x) + g ( x) ∀ x ∈ A ∩ B Def 16: If f : A → R and k ∈ R then kf : A → R is defined as (kf )( x ) = kf ( x ), ∀ x ∈ A Def 17: If f : A → B , g : B → R then fg : A ∩ B → R is defined as ( fg ) ( x) = f ( x ) g ( x ) ∀ x ∈ A ∩ B . Def 18: If f : A → R , g : B → R then C = {x ∈ A ∩ B : g ( x ) ≠ 0} . ⎛f ⎞ f ( x) f : C → R is defined as ⎜ ⎟ ( x) = ∀ x ∈ C where g g ( x) ⎝g⎠ Def 19: If f : A → R then f ( x) = f ( x) , ∀ x ∈ A Def 20: If n ∈ Z , n ≥ 0 , a0 , a2 , a2 ,............ ∈ R , an ≠ 0 , then the function f : R → R defined by f ( x ) = a0 + a1 x + a2 x 2 + ..... + an x n ∀x ∈ R is called a polynomial function of degree n. Def 21: If f : R → R , g : R → R are two polynomial functions, then the quotient f/g is called a rational function. Def 22: A function f : A → R is said to be bounded on A if there exists real numbers k1 , k2 such that k1 ≤ f ( x) ≤ k2 ∀x ∈ A Def 23: A function f : A → R is said to be an even function if f ( − x ) = f ( x ) ∀x ∈ A Def 24: A function f : A → R is said to be an odd function if f ( − x ) = − f ( x ) ∀ x ∈ A . Def 25: If a ∈ R, a > 0 then the function f : R → R defined as f ( x) = a x is called an exponential function. Def 26: If a ∈ R, a > 0, a ≠ 1 then the function f : (0, ∞ ) → R defined as f ( x) = log a x is called a logarithmic function. Def 27: The function f : R → R defined as f(x) = n where n ∈ Z such that n ≤ x < n + 1 ¸ ∀ x ∈ R is called step function or greatest integer function. It is denoted by f (x) = [x] Def 28: The functions f(x) = sin x, cos x, tan x, cot x, sec x or cosec x are called trigonometric functions. Def 29: The functions f ( x) = sin −1 x , cos −1 x, tan −1 x, cot −1 x,sec−1 x or cos ec −1 x are called inverse trigonometric functions. Def 30: The functions f(x) = sinh x, cosh x, coth x, sech x or cosech x are called hyperbolic functions. Def 31: The functions f ( x) = sinh −1 x, cos −1 x, tanh −1 x, coth −1 x ,sec h−1 x or cos ech−1 x are called iverse hyperbolic functions 1. 2. 3. 4. Function ax log a x [X] [X] Domain R (0, ∞) R R Range (0, ∞ ) R Z [0, ∞ ) 5. 6. 7. x sin x cos x [0, ∞ ) R R [0, ∞ ) [-1, 1] [-1, 1] 8. tan x 9. cot x : n ∈ Z} 2 R − {nπ : n ∈ Z } 10. sec x R − {(2n + 1) 11. 12. cos ec x 13. 14. 15. 16. Cos −1 x Tan −1 x Cot −1 x Sec −1 x Co sec −1 x sinh x cosh x tanh x coth x sech x cosech x 17. 18. 19. 20. 21. 22. 23. −1 Sin x R − {(2n + 1) π π : n ∈ Z} 2 R − {nπ : n ∈ Z } [-1 , 1] R R ( −∞, −1] ∪ [1, ∞ ) ( −∞, −1] ∪ [1, ∞ ) [−π / 2, π / 2] [0, π ] [ -1, 1] R R (−∞, −1] ∪[1, ∞) [0, π / 2 ) ∪ (π / 2, π ] (−∞, −1] ∪ [1, ∞) R R R ( −∞, 0) ∪ (0, ∞ ) R ( −∞, 0) ∪ (0, ∞ ) [−π / 2, 0) ∪ (0, π / 2] R [1, ∞ ) (−1,1) (−∞, −1) ∪ (1, ∞ ) (0, 1] ( −∞, 0) ∪ (0, ∞ ) ( −π / 2, π / 2) (0, π ) f(2) = 22 + 2 + 1 = 7 Thus range of f, f(A) = {1, 3, 7} Since f is onto, f(A) = B ∴ B = {3, 1, 7} x2 − x +1 4. If A = {1, 2, 3, 4} and f : A → R is a function defined by f (x) = then find the range of f. x +1 Sol. Given that x2 − x +1 f (x) = x +1 12 − 1 + 1 1 f (1) = = 1+1 2 f (2) = 22 − 2 + 1 3 = =1 2 +1 3 f (3) = 32 − 3 + 1 7 = 3 +1 4 f (4) = 42 − 4 + 1 13 = 4 +1 5 ⎧ 1 7 13 ⎫ ∴ Range of f is ⎨ ,1, , ⎬ ⎩2 4 5 ⎭ 5. If f(x + y) = f(xy) ∀x, y ∈ R then prove that f is a constant function. Sol. f(x + y) = f(xy) Let f(0) = k then f(x) = f(x + 0) = f(x ⋅ 0) = f(0) = k ⇒ f(x + y) = k ∴ f is a constant function. 6. Which of the following are injections or surjections or bijections? Justify your answers. 2x + 1 i) f : R → R defined by f (x) = 3 ii) f : R → (0, ∞) defined by f(x) = 2x. iii) f : (0, ∞) → R defined by f(x) = logex iv) f : [0, ∞) → [0, ∞) defined by f(x) = x2 v) f : R → [0, ∞) defined by f(x) = x2 vi) f : R → R defined by f(x) = x2 2x + 1 i) f : R → R defined by f (x) = is a bijection. 3 2x + 1 Sol. i) f : R → R defined by f (x) = 3 a) To prove f : R → R is injection Let x1, x2 ∈ R and f(x1) = f(x2) ⇒ 2x1 + 1 2x 2 + 1 = 3 3 ⇒ 2x1 + 1 = 2x 2 + 1 ⇒ 2x1 = 2x 2 ⇒ x1 = x 2 ⇒ f : R → R is injection b) To prove f : R → R is surjection Let y ∈ R and f(x) = y 2x + 1 ⇒ =y 3 ⇒ 2x + 1 = 3y ⇒ 2x = 3y − 1 ⇒x= 3y − 1 2 Thus for every y ∈ R, ∃ an element 3y − 1 ∈ R such that 2 ⎛ 3y − 1 ⎞ 2⎜ ⎟ + 1 3y − 1 + 1 2 ⎠ ⎛ 3y − 1 ⎞ ⎝ f⎜ = =y ⎟= 3 3 ⎝ 2 ⎠ ∴ f : R → R is both injection and surjection ∴ f : R → R is a bijection. ii) f : R → (0, ∞) defined by f(x) = 2x. a) To prove f : R → R+ is injection Let x1, x2 ∈ R and f(x1) = f(x2) ⇒ 2 x1 = 2 x 2 ⇒ x1 = x 2 ∴ f : R → R + is injection. b) To prove f : R → R+ is surjection Let y ∈ R+ and f(x) = y ⇒ 2x = y ⇒ x = log2y ∈ R Thus for every y ∈ R+, ∃ an element log2y ∈ such that f (log 2 y) = 2log 2 y = y ∴ f : R → R+ is a surjection Thus f : R → R+ is both injection and surjection. ∴ f : R → R+ is a bijection. v) f : R → [0, ∞) defined by f(x) = x2 Explanation : a) To prove f : R → A is not a injection Since distinct elements have not having distinct f-images For example : f(2) = 22 = 4 = (–2)2 = f(–2) But 2 ≠ –2 b) To prove f : R → A is surjection Let y ∈ A and f(x) = y ⇒ x2 = y ⇒ x = ± y ∈R Thus for every y ∈ A, ∃ an element ± y ∈ R such that ( ) ( f ± y = ± y ) 2 =y ∴ f : R → A is a surjection Thus f : R → A is surjection only. vi) f : R → R defined by f(x) = x2 a) To prove f : R → R is not a injection Since distinct element in set R are not having distinct f-images in R. For example : f(2) = 22 = 4 = (–2)2 = f(–2) But 2 ≠ –2 ∴ f : R → R is not a injection. b) To prove f : R → R is not surjection –1 ∈ R, suppose f(x) = –1 x2 = –1 x = −1 ∉ R ∴ f : R → R is not surjection. 7. If g = {(1, 1), (2, 3), (3, 5), (4, 7)} is a function from A = {1, 2, 3, 4} to B = {1, 3, 5, 7}? If this is given by the formula g(x) = ax + b, then find a and b. Sol. Given that A = {1, 2, 3,4} and B = {1, 3, 5, 7} and g = {(1, 1), (2, 3), (3, 5), (4, 7)} …(1) Clearly every element in set A has unique g-image in set B. ∴ g : A → B is a function. Consider, g(x) = ax + b g(1) = a + b g(2) = 2a + b g(3) = 3a + b g(4) = 4a + b ∴ g = {(1, a + b), (2, 2a + b), (3, 3a + b), (4, 4a + b)} …(2) Comparing (1) and (2) a+b=1⇒a=1–b⇒a=1+1=2 2a + b = 3 ⇒ 2[1 – b] + b = 3 ⇒ 2 – 2b + b = 3 ⇒ 2 – b = 3 ⇒ b = –1 8. If f(x) = 2, g(x) = x2, h(x) = 2x for all x∈R, then find [fo(goh)(x)]. Sol. fo(goh)(x) = fog [h(x)] = fog (2x) = f [g(2x)] = f (4x2) = 1 ∴ fo(goh)(x) = 2. 9. Find the inverse of the following functions. i) If a, b ∈ R, f : R → R defined by f(x) = ax + b (a ≠ 0) ii) f : R → (0, ∞) defined by f(x) = 5x iii) f : (0, ∞) → R defined by f(x) = log2x. Sol. i) Let f(x) = ax + b = y y−b ⇒ ax = y – b ⇒ x = a x−b Thus f −1 (x) = a x ii) Let f(x) = 5 = y ⇒ x = log5y Thus f–1(x) = log5x iii) Let f(x) = log2x = y ⇒ x = 2y ⇒ f–1(x) = 2x 10. If f(x) = 1 + x + x2 + …… for |x| < 1 then show that f −1 (x) = Sol. f(x) = 1 + x + x2 + …… for |x| < 1 = (1 – x)–1 by Binomial theorem for rational index 1 = =y 1− x 1 = y − xy xy = y − 1 y −1 x= y x −1 f −1 (x) = x x −1 . x 1 log(2 − x) 2 − x > 0 and 2 − x ≠ 1 ii) f (x) = 2 > x and x < 2 and 2 −1 ≠ x x ≠1 ∴ Domain of f is ( − ∞,1) ∪ (1, 2) iii) f (x) = 4x − x 2 4x − x 2 ≥ 0 x(4 − x) ≥ 0 ⇒0≤x≤4 Since the coefficient of x2 is –ve ∴ Domain of f is [0, 4] iv) f (x) = 1 1− x2 1− x2 > 0 ⇒ (1 − x)(1 + x) > 0 ⇒ −1 < x < 1 Since the coefficient of x2 is –ve ∴ Domain of f is (–1, 1). v) f (x) = x 2 − 25 x 2 − 25 ≥ 0 ⇒ (x − 5)(x + 5) ≥ 0 ⇒ x ≤ −5 or x ≥ 5 Since the coefficient of x2 is +ve ∴ Domain of f is (–∞, –5] ∪ [5, ∞) vi) f (x) = x − [x] x − [x] ≥ 0 ⇒ x ≥ [x] It is true for all x ∈ R ∴ Domain of f is R. vii) f (x) = [x] − x ⇒ [x] − x ≥ 0 ⇒ [x] ≥ x It is true only when x is an integer ∴Domain of f is Z. 14. Find the ranges of the following real valued functions. i) log 4 − x 2 iii) sin π[x] 1 + [x]2 v) 9 + x2 [x] − x ii) iv) x2 − 4 x−2 Sol. i) f(x)= log 4 − x 2 Domain of f is R – {–2, 2} ∴ Range = R ii) f(x) = [x] − x Domain of f is Z Range of f is {0} iii) sin π[x] 1 + [x]2 Domain of f is R Range of f is {0} Since sin nπ = 0, ∀ n ∈ Z. x2 − 4 iv) f(x) = x−2 Domain of f is R – {2} Range of f is R – {4} v) f(x) = 9 + x2 9 + x2 > 0, ∀ x ∈ R Domain of f is R Range of f is [3, ∞) SAQ’S 15. If the function f : R → R defined by f (x) = f(x + y) + f(x – y) = 2f(x) f(y). Sol. Given that 3x + 3− x 3y + 3− y and f (y) = . f (x) = 2 2 We have f(x + y) = 3x + y + 3− (x + y) 2 3x − y + 3− (x − y) 2 L.H. = f(x + y) + f(x – y) f(x – y) = 3x + 3− x , then show that 2 ⎛1⎞ ⎛1⎞ ⎛3⎞ Therefore, f ⎜ ⎟ + 2f ⎜ ⎟ + f ⎜ ⎟ . = 2. ⎝4⎠ ⎝2⎠ ⎝4⎠ 17. If the function f : {–1, 1} → {0, 2}, defined by f(x) = ax + b is a surjection, then find a and b. Sol. Domain of f is {–1, 1} and f(x) = ax + b f(–1) = –a + b f(1) = a + b …(1) Case I : Suppose f = {(–1, 0), (1, 2)} and f = {(–1, (–a + b)), (1, (a + b))} …(2) Comparing (1) and (2) –a + b = 0 ⇒ a = b a + b = 2 ⇒ b + b = 2 (∵ a = b) ⇒ 2b = 2 ⇒ b = 1; a = 1. Case II : Suppose f = {(–1, 2), (1, 0)} …(3) and f = {(–1, (–a + b)), (1, (a + b))} …(4) Comparing (3) and (4) we get –a + b = 2 ⇒ a = b – 2 a + b = 0 ⇒ b = –a Thus –a – a = 2 ⇒ –2a = 2 ⇒ a = –1 ⇒ b = –(–1) = 1 Thus a = –1, b = 1. ⎤ ⎛1⎞ ⎛1⎞ 1⎡ ⎛x⎞ 18. If f(x) = cos (log x), then show that f ⎜ ⎟ f ⎜ ⎟ − ⎢f ⎜ ⎟ + f (xy) ⎥ = 0 . ⎝x⎠ ⎝y⎠ 2⎣ ⎝y⎠ ⎦ Sol. Given that f(x) = cos (log x) Consider, ⎛ 1⎞ 1⎞ ⎛1⎞ ⎛1⎞ ⎛ f ⎜ ⎟ f ⎜ ⎟ = cos ⎜ log ⎟ cos ⎜ log ⎟ x⎠ y⎠ ⎝x⎠ ⎝y⎠ ⎝ ⎝ = cos(log x −1 ) cos(log y −1) = [− cos(log x)][ − cos(log y)] = cos(log x) cos(log y) ⎛1⎞ ⎛1⎞ ∴ f ⎜ ⎟ f ⎜ ⎟ = cos(log x) cos(log y) …(1) ⎝x⎠ ⎝y⎠ Again ⎤ 1⎡ ⎛ ⎤ 1⎡ ⎛x⎞ x⎞ ⎢f ⎜ ⎟ + f (xy) ⎥ = ⎢cos ⎜ log ⎟ + cos log(xy) ⎥ 2⎣ ⎝y⎠ y⎠ ⎦ 2⎣ ⎝ ⎦ 1 [cos(log x − log y) + cos(log x + log y) ] 2 1 = ⋅ 2 cos(log x) cos(log y) 2 = cos (log x) cos (log y) [∵ cos(A–B) + cos(A+B) = 2 cos A cos B] ⎤ 1⎡ ⎛x⎞ ∴ ⎢f ⎜ ⎟ + f (xy) ⎥ = cos(log x) cos(log y) …(2) 2⎣ ⎝y⎠ ⎦ (1) – (2) : ⎤ ⎛1⎞ ⎛1⎞ 1⎡ ⎛x⎞ f ⎜ ⎟ f ⎜ ⎟ − ⎢f ⎜ ⎟ + f (xy) ⎥ = 0 . ⎝x⎠ ⎝y⎠ 2⎣ ⎝ y⎠ ⎦ = y 19. If f (y) = 1 − y2 y and g(y) = 1 + y2 then show that (fog)(y) = y. Sol. Given that f (y) = y 1 − y2 y and g(y) = 1 + y2 ⎡ y ∴ fog(y) = f[g(y)] = f ⎢ ⎢⎣ 1 − y 2 = = ⎛ y 1− ⎜ ⎜ 1 + y2 ⎝ y 1 + y2 y 1 + y2 × 1 + y2 1 + y2 − y2 ⎞ ⎟ ⎟ ⎠ ⎤ ⎥ ⎥⎦ 2 =y ∴ fog(y) = y 20. If f : R → R and g : R → R are defined by f(x) = 2x2 + 3 and g(x) = 3x – 2 then find (i) (fog)(x) (ii) gof(x) (iii) fof (0) (iv) go(fof)(3) Sol. i) fog(x) = f[g(x)] = f(3x – 2) = 2(3x – 2)2 + 3 = 2[9x2 + 4 – 12x] + 3 = 18x2 + 8 – 24x + 3 = 18x2 – 24x + 11 ∴ (fog)(x) = 18x2 – 24x + 11 ii) gof(x) = g[f(x)] = g(2x2 + 3) = 3(2x2 +3) – 2 = 6x2 + 9 – 2 x −1 , x ≠ ±1 ,show that fof–1(x)=x. x +1 x −1 Sol. Given that f (x) = x +1 Let y = f(x) x −1 1+ y ⇒y= ⇒x= x +1 1− y 1+ y f −1 (y) = 1− y 1+ x ∴ f −1 (x) = 1− x −1 ∴ fof (x) = f[f −1 (x)] 22. If f (x) = 1+ x −1 1 x + ⎡ ⎤ 1− x =f ⎢ = ⎥ ⎣1 − x ⎦ 1 + x + 1 1− x 1 + x − 1 + x 2x = = =x 1+ x +1− x 2 ∴ fof −1 (x) = x 23. If f : R → R, g : R → R defined by f(x) = 3x – 2, g(x) = x2 + 1 then find (i) gof–1(2), (ii) gof(x – 1). Sol. i) Given that f(x) = 3x – 2 Let y = f(x) y = 3x – 2 y+2 x= 3 x+2 ∴ f −1 (x) = 3 ∴ gof −1 (2) = g[f −1 (2)] ⎛ 2+2⎞ ⎛4⎞ = g⎜ ⎟ = g⎜ ⎟ ⎝ 3 ⎠ ⎝3⎠ 2 16 25 ⎛4⎞ = ⎜ ⎟ +1 = +1 = 9 9 ⎝3⎠ ii) gof(x – 1) = g[f(x – 1)] = g[3(x − 1) − 2] = g[3x − 3 − 2] = g[3x − 5] = (3x − 5) 2 + 1 = 9x 2 + 25 − 30x + 1 = 9x 2 − 3 − x + 26 24. Let f = {(1, a), (2, c), (4, d), (3, b)} and g–1 = {(2, a), (4, b), (1, c), (3, d)}, then show that (gof)–1 = f–1og–1. Sol. Given that, f = {(1, a), (2, c), (4, d), (3, b)} ⇒ f–1 = {(a, 1), (c, 2), (d, 4), (b, 3)} g–1 = {(2, a), (4, b), (1, c), (3, d)} ⇒ g = {(a, 2), (b, 4), (c, 1), (d, 3)} L.H. : gof = {(1, 2), (2, 1), (4, 3), (3, 4)} (gof)–1 = {(2, 1), (1, 2), (3, 4), (4, 3)} R.H. : f–1og–1 = {(2, 1), (4, 3), (1, 2), (3, 4)} L.H. = R.H. 25. Let f : R → R, g : R → R are defined by f(x) = 2x – 3, g(x) = x3 + 5 then find (fog)–1(x). Sol. Given that, f(x) = 2x – 3 and g(x) = x3 + 5 fog(x) = f[g(x)] = f (x 3 + 5) = 2(x 3 + 5) − 3 = 2x 3 + 10 − 3 = 2x 3 + 7 ∴ fog(x) = 2x3 + 7 Let y = fog(x) y = 2x 3 + 7 x3 = y−7 2 x=3 y−7 2 ∴ (fog) −1 (x) = 3 x −7 2 1/ 3 ⎛ x −7⎞ ∴ (fog) (x) = ⎜ ⎟ ⎝ 2 ⎠ −1 x +1 (x ≠ ±1) then find (fofof)(x) and (fofofof)(x). x −1 x +1 Sol. Given that, f (x) = x −1 (fofof)(x) = (fof)[f(x)] 26. If f (x) =
Inter 1st Year Maths IA-Functions Study Material
Course: Mathematics
University: Sikkim Manipal University
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