Functions Exercise 1b Solutions

Functions Exercise 1b

Functions Exercise 1b Solutions 

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

Exercise 1b

Exercise 1(b) Solutions

I.

1.If f (x) = ex and g(x) = Functions 1(b) images 1, then show that f og = gof and f-1 = g-1

Sol:

Given     f (x) = ex and g(x) =Functions 1(b) images 1

              (fog) (x) = f (g (x))

                                = f ( Functions 1(b) images 1)

                                =Functions 1(b) images 2

                                = x ————- (1)

            (gof) (x) = g (f (x))

                                = g (ex)

                                =Functions 1(b) images 3

                                = xFunctions 1(b) images 4

                                = x ————- (2)

From (1) and (2)  f og = gof

let y = f(x)

     x = f-1(y)

     y = ex

     x = Functions 1(b) images 5

    f-1 (x) = Functions 1(b) images 1

let g(x) = z

         x = g-1(z)

         z =Functions 1(b) images 1

          x = ez

             g-1 (x)  = ex

2. If f (y) =Functions 1(b) images 6 ,  g (y) =Functions 1(b) images 7 then show that fog(y) = y

Sol:

Given f (y) =Functions 1(b) images 6 ,  g (y) = Functions 1(b) images 7        

Now

     fog(y) = f(g(y))

Functions 1(b) images 8

∴ fog(y) = y

3. If f: R ⟶ R, g: R ⟶ R is defined by f(x) = 2x2 + 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) = 2x2 + 3 ang g (x) = 3x – 2

(i)  (fog) (x) = f(g(x))

                 = f(3x – 2)

                 = 2 (3x – 2)2 + 3

                 = 2 (9x2 – 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 (2x2 + 3)

                 = 3(2x2 + 3) – 2  

                 = 6x2 + 9 – 2

                 = 6 x2 + 7

 ∴ (gof) (x) = 6 x2 + 7

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(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 (3x2 + 3– 1)

                          = f (3x2 + 2)

                          = 3(3x2 + 2) – 1

                          = 9x2 + 6 – 1

                          = 9x2 + 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) = Functions 1(b) images 9 and g(x) =Functions 1(b) images 10 for all x ∈ (0, ∞) then find (gof) (x)

Sol:

Given f(x) = Functions 1(b) images 9 and g(x) = Functions 1(b) images 10 for all x ∈ (0, ∞)

  (gof) (x) = g (f (x))

                    = g (Functions 1(b) images 10 )

                    =Functions 1(b) images 11

               ∴ (gof) (x) =Functions 1(b) images 12

6. If f(x) = 2x – 1 and g(x) =Functions 1(b) images 13 for all x ∈ R then find (gof) (x)

Sol:

   Given f(x) = 2x – 1 and g(x) = Functions 1(b) images 13 for all x ∈ R

  Functions 1(b) images 14           

∴ 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(4x2)

                                   = 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 = ax + b

          ax = y – b

             x =Functions 1(b) images 15

            f-1(y) =Functions 1(b) images 15

     ∴   f-1(x) =Functions 1(b) images 16

(ii) f: R ⟶ (0, ∞) defined by f(x) = 5x

  Let y = f(x)

         x = f-1(y)

  now y = 5x

          x =Functions 1(b) images 17

             f-1(y) =Functions 1(b) images 17

     ∴   f-1(x) =Functions 1(b) images 18

(iii) f: (0, ∞) ⟶ R defined by f(x) =Functions 1(b) images 19

    Let y = f(x)

         x = f-1(y)

   now y =Functions 1(b) images 19

          x = 2y

             f-1(y) = 2y

     ∴   f-1(x) = 2x

9. If f(x) = 1 + x + x2 + … for Functions 1(b) images 20 , then show that f-1 (x) = Functions 1(b) images 21     

Sol:

Given, f(x) = 1 + x + x2 + … for

        1 + x + x2 + … is an infinite G.P

        a = 1, r = x

  S =Functions 1(b) images 22

         =Functions 1(b) images 23

Now

f (x) =Functions 1(b) images 23

Let y = f(x)

         x = f-1(y)

   now y =Functions 1(b) images 23

           1 – x = Functions 1(b) images 24

   x = 1 –Functions 1(b) images 24

       =Functions 1(b) images 25

             f-1(y) =Functions 1(b) images 25

     ∴   f-1(x) =Functions 1(b) images 26

10 . If f: [1, ∞) ⟶ [1, ∞) defined by f (x) = 2x (x – 1) then find f-1 (x)

Sol:

Given f: [1, ∞) ⟶ [1, ∞) defined by f (x) = 2x (x – 1)

Let y = f(x)

         x = f-1(y)

   now y = 2x (x – 1)

            Functions 1(b) images 27= x (x – 1)

    x2 – x =  Functions 1(b) images 27

    x2 – x – Functions 1(b) images 27  = 0

Functions 1(b) images 28

Functions 1(b) images 29

 

Functions Exercise 1b Solutions

II. 

1. If f (x) = Functions 1(b) images 30  , x ≠ ± 1, then verify (fof-1) (x) = x

Sol:

   Given f (x) =Functions 1(b) images 30  , x ≠ ± 1

Let y = f(x)

         x = f-1(y)

   now y =Functions 1(b) images 30

            y (x + 1) = x – 1

            xy + y = x – 1

            1 + y = x – xy

            1 + y = x (1 – y)

             x = Functions 1(b) images 31 

             f-1(y) =Functions 1(b) images 31

     ∴   f-1(x) =Functions 1(b) images 32

Now

        Functions 1(b) images 33

2.  If A = {1, 2, 3}, B = {α, β, γ}, C = {p, q, r} and f: A ⟶ B; g: B ⟶ C is defined by f = {(1, α), (2, γ), (3, β)}, g = {(α, q), (β, r), (γ, p)}, then show that f and g are bijective functions and (gof)-1 = f-1og-1

Sol:

Given A = {1, 2, 3}, B = {α, β, γ}, C = {p, q, r} and f: A ⟶ B; g: B ⟶ C is defined by f = {(1, α), (2, γ), (3, β)}, g = {(α, p), (β, r), (γ, p)}

A = {1, 2, 3}, B = {α, β, γ}

f: A ⟶ B; and f = {(1, α), (2, γ), (3, β)}

 ⟹ f (1) = α; f (2) = γ; f (3) = β

       ⟹    Every element of set A has a unique image in set B

          f is injection (one – one)

range of f = codomain of f

⟹ f is surjection (on to)

∴ f is bijective

B = {α, β, γ}, C = {p, q, r}

g: B ⟶ C is defined by g = {(α, q), (β, r), (γ, p)}

⟹ g (α) =q; g (β) = r; g (γ) = p

       ⟹    Every element of set B has a unique image in set C

          g is injection (one – one)

range of g = codomain of g

⟹ g is surjection (on to)

∴ g is bijective

Now

f = {(1, α), (2, γ), (3, β)}, g = {(α, q), (β, r), (γ, p)}

gof = {(1, q), (2, p), (3, r)

(gof)-1 = {(q, 1), (p, 2), (r, 3)} ———– (1)

f-1 = {(α, 1), (γ, 2), (β, 3)

g-1 = {(q, α), (r, β), (p, γ)}

f-1og-1 = {(q, 1), (p, 2), (r, 3)} ———– (2)

From (1) and (2)

(gof)-1 = f-1og-1

3. If f: R ⟶ R; g: R ⟶ R is defined by f (x) = 3x – 2 and g(x) = x2 + 1          then find
 (i) (gof-1) (2)           (ii) (gof) (x – 1)

 

Sol:

    Given f: R ⟶ R; g: R ⟶ R is defined by f (x) = 3x – 2 and g(x) = x2 + 1

     Let y = f(x)

         x = f-1(y)

   now y = 3x – 2

            3x = y + 2

               x = Functions 1(b) images 34

       f-1(y) =Functions 1(b) images 34

       f-1(x) =Functions 1(b) images 35

Functions 1(b) images 36

∴ (gof-1) (2)   =Functions 1(b) images 37

(ii)  (gof) (x – 1) = g (f (x – 1))

                             = g (3(x – 1) – 2)

                             = g (3x – 3 – 2)

                             = g (3x – 5)

                             = (3x – 5)2 + 1

                             = 9x2 – 30x + 25 + 1

                             = 9x2 – 30x + 26

∴(gof) (x – 1) = 9x2 – 30x + 26

 

4. 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 f = {(1, a), (2, c), (4, d), (3, b)} and g-1 = {(2, a), (4, b), (1, c), (3, d)}

⟹ f -1 = {(a, 1), (c, 2), (d, 4), (b, 3)} and g = {(a, 2), (b, 4), (c, 1), (d, 3)}

Now gof = {(1, 2), (2, 1), (4, 3), (3, 4)}

          (gof)-1 = {(2, 1), (1, 2), (3, 4), (4, 3)} ———(1)

g-1 = {(2, a), (4, b), (1, c), (3, d)} ; f -1 = {(a, 1), (c, 2), (d, 4), (b, 3)}

f-1og-1 = {(2, 1), (1, 2), (3, 4), (4, 3)} ———(2)

from (1) and (2)

(gof)-1 = f-1og-1      

 

5. If f: R ⟶ R; g: R ⟶ R is defined by f (x) = 2x – 3 and g(x) = x3 + 5 then find (fog)-1 (x)

 

Sol:

      Given R ⟶ R; g: R ⟶ R is defined by f (x) = 2x – 3 and g(x) = x3 + 5

       (fog) (x) = f (g(x))

                         = f (x3 + 5)

                          = 2 (x3 + 5) – 3

                         = 2 x3 + 10 – 3 = 2x3 + 7

Let y = (fog) (x)

     ⟹ x = (fog)-1(y)

       y = 2x3 + 7

       2x3 = y – 7

          x3 =Functions 1(b) images 38

          x =Functions 1(b) images 39

(fog)-1(y)=  Functions 1(b) images 39      

(fog)-1(x)=Functions 1(b) images 40        

6. Let f(x) = x2, g(x) = 2x then solve the equation (fog) (x) = (gof) (x)

 

Sol:

  Given f(x) = x2, g(x) = 2x

  (fog) (x) = f(g(x))

                    = f (2x)

                     = (2x)2

                     = 22x

(gof) (x) = g(f(x))

                  = g (x2)

                  =Functions 1(b) images 41

    Now

      (fog) (x) = (gof) (x)

     ⟹ 22x =Functions 1(b) images 41

           2x = x2   [∵ if am = an , then m = n]

           x2 – 2x = 0

         ⟹ x (x – 2) = 0

         ⟹x = 0 or x – 2 = 0

∴ x = 0 or x = 2

7.  If f (x) = , x ≠ ±1 then find (fofof) (x) and (fofofof) (x)

 

Sol:

Given f (x) = Functions 1(b) images 43, x ≠ ±1

Functions 1(b) images 42

Functions 1(b) images 44

 

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Ts Inter Maths IA Concept

TS 10th Class Maths Concept (T/M)

 


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