chapter functions inter

Functions Exercise 1c

Functions Exercise 1c Solutions ||TS|| Basics In MAths

Functions Exercise 1c Solutions 

Functions Exercise 1c

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 1c Solutions for inter first year  students, prepared by Mathematics expert of www.basicsinmaths.com

Functions Exercise 1c

 

I.

1.Find the domains of the following real valued functions

 

(i)

f (x) =Functions 1(c) images 1

given function is f (x) =Functions 1(c) images 1

 f (x) is defined when (x2 – 1) (x + 3) ≠ 0

  ⟹ (x2 – 1) ≠ 0 or (x + 3) ≠ 0

  ⟹ (x + 1) (x – 1) ≠ 0 or (x + 3) ≠ 0

  ⟹ x ≠ 1, x ≠ – 1 or x ≠ – 3

∴ Domain of f(x) is R – {– 1, – 3, 1}

(ii)

f (x) = Functions 1(c) images 2

 Given function is f (x) = Functions 1(c) images 2

 f (x) is defined when (x – 1) (x – 2) (x – 3) ≠ 0    

  ⟹ x ≠ 1, x ≠ 2 or x ≠ 3

∴ Domain of f(x) is R – {1, 2, 3}

(iii)

f (x) = Functions 1(c) images 3

Given function is f (x) = Functions 1(c) images 3

f (x) is defined when 2 – x > 0 and 2 – x ≠ 1                        

  ⟹ 2 > x  and  2 – 1 ≠ x                        

⟹ 2 > x  and  x ≠ 1                        

∴ Domain of f(x) is (– ∞, 2) – {1}

(iv)

f (x) = Functions 1(c) images 4

Given function is f (x) = Functions 1(c) images 4

f (x) is defined when x ∈ R      

∴ Domain of f(x) is R

Functions Exercise 1c

(v)

f (x) =Functions 1(c) images 5

Given function is f (x) = Functions 1(c) images 5

f (x) is defined when 4x – x2 ≥ 0      

⟹ x (4 – x) ≥ 0

⟹ x (x – 4) ≤ 0

⟹ (x – 0) (x – 4) ≤ 0

⟹ x ∈ [0, 4]

∴ Domain of f(x) is [0, 4]

(vi) 

f (x) =Functions 1(c) images 6

 Given function is f (x) =Functions 1(c) images 6

 f (x) is defined when 1 – x2 > 0   

 ⟹ x2 – 1 < 0

⟹ (x – 1) (x + 1) < 0

⟹ x ∈ (– 1, 1)

∴ Domain of f(x) is (– 1, 1)

(vii)

f (x) =Functions 1(c) images 7

 Given function is f (x) = Functions 1(c) images 7

  f (x) is defined when x + 1≠ 0   

  ⟹ x ≠ – 1  

  ∴ Domain of f(x) is R – {– 1}

(viii)

f(x) =Functions 1(c) images 8

  Given function is f (x) =Functions 1(c) images 8

  f (x) is defined when x2 – 25 ≥ 0   

  ⟹ (x – 5) (x + 5) ≥ 0  

  ⟹ x ∈ (–∞, –5] ∪ [5, ∞)

 Functions 1(c) images 11

  ⟹ x ∈ R – (– 5, 5)

 ∴ Domain of f(x) is R – (– 5, 5)

Functions Exercise 1c

(ix)

f(x) = Functions 1(c) images 12

Given function is f (x) = Functions 1(c) images 12

f (x) is defined when x – [x] ≥ 0   

  ⟹ x ≥ [x]

 ⟹ x ∈ R   

∴ Domain of f(x) is R  

(x)

f(x) =Functions 1(c) images 13

Given function is f (x) =Functions 1(c) images 13

f (x) is defined when [x] – x ≥ 0   

  ⟹ [x] ≥ x

 ⟹ x ∈ Z   

∴ Domain of f(x) is Z

Functions Exercise 1c

2. find the ranges of the following real valued functions

 

(i)

f(x) =Functions 1(c) images 14

Given function is f (x) = Functions 1(c) images 14

Let y = Functions 1(c) images 14

  ⟹  |4 – x2| = ey

∵ ey > 0 ∀ y ∈ R

∴ Range of f(x) is R

(ii) 

f(x) =Functions 1(c) images 13

Given function is f (x) =Functions 1(c) images 13

f (x) is defined when [x] – x ≥ 0   

  ⟹ [x] ≥ x

 ⟹ x ∈ Z   

Domain of f(x) is Z

Range of f = {0}

(iii) 

f(x) =Functions 1(c) images 15

Given function is f (x) =Functions 1(c) images 15

f (x) is defined when x ∈ R   

             Domain of f(x) is R

           For x ∈ R   [x] is an integer

           Since sin nπ = 0, ∀ n ∈ z

            ⟹ sin π[x] = 0

          ∴ Range of f = {0}

(iv) 

f (x) =Functions 1(c) images 16

           Given function is f (x) = Functions 1(c) images 16

          f (x) is defined when x – 2 ≠ 0

          ⟹ x ≠ 2

         Domain of f(x) is R – {2}

         Let y = Functions 1(c) images 16

            =Functions 1(c) images 17

             = x + 2

       If x = 2 ⟹ y = 2 + 2 = 4

        ∴ Range of f(x) is R – {4}

Functions Exercise 1c

(v) 

f (x) =Functions 1(c) images 18

let y =Functions 1(c) images 18

      y2 = 9 + x2

          x2 = y2 – 9

       x = Functions 1(c) images 19

     it is defined when y2 – 9 ≥ 0

      ⟹ (y – 3) (y + 3) ≥ 0

    y ∈ (– ∞, – 3] ∪ [3, ∞)

but y = Functions 1(c) images 18 ≥ 0

∴ Range of f(x) is [3, ∞)

3. If f and g are real valued functions f(x) = 2x – 1 ang g (x) = x2 then   
    find

Sol:

    Given f and g are real valued functions f(x) = 2x – 1 ang g (x) = x2

(i)

(3f – 2g) (x) = 3 f(x) – 2g (x)

                         = 3 (2x – 1) – 2(x2)

                         = 6x – 3 – 2x2

                          = – 2x2 + 6x – 3                   

∴ (3f – 2g) (x) =– 2x2 + 6x – 3

(ii)

(fg) (x) = f (x) g (x)

         = (2x – 1) (x2)

                = 2x3 + x2   

 ∴ (fg) (x) = 2x3 + x2  

(iii) 

Functions 1(c) images 20   

4. If f = {(1, 2), (2, – 3) (3, – 1)} then find (i) 2f (ii) (fog) 2 + f iii) f2 (iv) Functions 1(c) images 21

Sol:

Given f = {(1, 2), (2, – 3) (3, – 1)}

(i)

(2f) (1) = 2 f (1) = 2 × 2 = 4

(2f) (2) = 2 f (2) = 2 × – 3 = – 6

(2f) (3) = 2 f (3) = 2 × – 1 = – 2

∴ 2f = {(1, 4), (2, – 6) (3, – 2)}    

(ii)

(2 + f) (1) = 2 + f (1) = 2 + 2 = 4

(2 + f) (2) = 2 + f (2) = 2 + (– 3) = – 1

(2 + f) (3) = 2 + f (3) = 2 + (– 1) = 1

∴ 2 + f = {(1, 4), (2, – 1) (3, 1)}

(iii) 

(f2) (1) = [f (1))]2 = 22 = 4

(f2) (2) = [f (2))]2 = (– 3)2 = 9

(f2) (3) = [f (1))]2 = (– 1)2 = 1

∴ f2= {(1, 4), (2, 9) (3, 1)}

(iv) 

Functions 1(c) images 22

II. 

1.Find the domain of the following real valued functions

(i)

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

f(x) is defined when x2 – 3x + 2 ≥ 0       

   x2 – 2x – x + 2 ≥ 0

   x (x – 2) – 1(x – 2) ≥ 0

  (x – 1) (x – 2) ≥ 0

   x ∈ (– ∞, 1] ∪ [2, ∞)

∴ Domain of f(x) is R – (1, 2)

(ii)

f(x) = log (x2 – 4x + 3)

f(x) is defined when x2 – 4x + 3 > 0       

   x2 – 3x – x + 3 > 0

   x (x – 3) – 1(x – 3) > 0

  (x – 1) (x – 3) > 0

   x ∈ (– ∞, 1) ∪ (3, ∞)

∴ Domain of f(x) is R – [1, 3]

(iii)

f(x) =Functions 1(c) images 24

f(x) is defined when 2 + x ≥ 0, 2 – x ≥ 0  and x ≠ 0 

 x ≥ – 2, x ≤ 2 and x ≠ 0   

  – 2 ≤ x ≤ 2 and x ≠ 0   

   x ∈ [–2, 2] – {0}

∴ Domain of f(x) is [–2, 2] – {0}

(iv)

f(x) =Functions 1(c) images 26

f(x) is defined in two cases as follows:

case (i) 4 – x2 ≥ 0 and [x] + 2 > 0    

               x2 – 4 ≤ 0 and [x] > –2 

               (x – 2) (x + 2) ≤ 0 and [x] > –2 

                x ∈ [– 2, 2] and x ∈ [– 1, ∞)

                x ∈ [– 1, 2]

case (ii) 4 – x2 ≤ 0 and [x] + 2 < 0       

                x2 – 4 ≥ 0 and [x] < –2 

               (x – 2) (x + 2) ≥ 0 and [x] < –2 

               x ∈ (– ∞, –2] ∪ [2, ∞) and x ∈ (–∞, –2)

                x ∈ (–∞, –2)

from case (i) and case (ii)

   x ∈ (–∞, –2) ∪ [– 1, 2]

∴ Domain of f(x) is (–∞, –2) ∪ [– 1, 2]

(v)

f(x) = Functions 1(c) images 27

f(x) is defined when Functions 1(c) images 28≥ 0 and x – x2 > o

 x – x2 ≥ (0.3)0   and x2 – x < 0

x – x2 ≥ 1 and x (x – 1) < 0

   x2 –x + 1 ≤ 0 and (x – 0) (x – 1) < 0

 it is true for all x ∈ R and x ∈ (0, 1)

∴ domain of f(x) =  R∩ (0, 1) = (0, 1)

(vi)

f(x) =Functions 1(c) images 29

f(x) is defined when x +|x| ≠ 0       

  |x| ≠ – x      

  |x| ≠ – x      

 ⟹|x| = x      

 ⟹ x > 0   

   x ∈ (0, ∞)

∴ Domain of f(x) is (0, ∞)

2. Prove that the real valued function Functions 1(c) images 30 is an even function

Sol:

Given f(x) = Functions 1(c) images 30

Functions 1(c) images 31 

                         Functions 1(c) images 32

3. Find the domain and range of the following functions

 

(i)

f (x) = Functions 1(c) images 33

Given f(x) =Functions 1(c) images 33

 Since [x] is an integer

  sin π[x] = tan π[x] = 0 ∀ x ∈ R

∴ domain of f(x) is R

and

since tan π[x] = 0

 Range of f(x) = {0}

(ii)

f(x) = Functions 1(c) images 34

Given f (x) = Functions 1(c) images 34

 It is defined when 2 – 3x ≠ 0

⟹ 2 ≠ 3x 

⟹ x ≠ 2/3

∴ Domain of f (x) = R – {2/3} 

Let y = f(x)

        y =Functions 1(c) images 34

  ⟹ y (2 – 3x) = x

          2y – 3xy = x

          2y = x + 3xy  

           2y = x (1 + 3y)

   ⟹ x = Functions 1(c) images 35

    It is defined when 1 + 3y ≠ 0

                                        1 ≠ –3y

                                      y ≠ – 1/3

∴ Range of f (x) = R – {– 1/3} 

(iii) 

f(x) = |x| + | 1 + x|

Given function is f (x) = |x| + |1 + x|

             f (x) is defined for all x ∈ R

           ∴ domain of f(x) = R

        Functions 1(c) images 36

            f (– 3) = |– 3| + |1 – 3|

                        =|– 3| + |– 2|

                       = 3 + 2 = 5 

            f (– 2) = |– 2| + |1 – 2|

                        =|– 2| + |– 1|

                        = 2 + 1 = 3

            f (– 1) = |– 1| + |1 – 1|

                        =|– 1| + |0|

                        = 1 + 0 = 1

            f (0) = |0| + |1 + 0| = 1

            f (1) = |1| + |1 + 1|

                      = 1 + |2|

                      = 1 + 2 = 3

            f (2) = |2| + |1 + 2|

                      = |2| + |3|

                      = 2 + 3 = 5

            f (3) = |3| + |1 + 3|

                      = |3| + |4|

                      = 3 + 4 = 7

∴ Range of f(x) = [1, ∞)


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Functions Exercise 1c Solutions ||TS|| Basics In MAths Read More »

Functions Exercise 1b

Functions Exercise 1b Solutions

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) = 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 10th class maths concept (E/M)

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TS 10th Class Maths Concept (T/M)

 


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Functions Exercise 1a Solutions

chapter 1 Functions Exercise 1a Solutions

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

        Functions 1(a) images 1

   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 Functions 1(a) images 1

      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) = Functions 1(a) images 2, then show that f (x) + f (1/x) = 0

Sol:

Given f: R – {0} ⟶ R is defined by f(x) = Functions 1(a) images 2

       f (1/x)  = Functions 1(a) images 3

Now

f (x) + f (1/x)  = Functions 1(a) images 4

∴ f (x) + f (1/x)  = 0

3. If f: R ⟶ R is defined by f(x) = Functions 1(a) images 5, then show that f (tan θ) = cos 2θ

Sol:

    Given f: R ⟶ R is defined by f(x) =Functions 1(a) images 5

       f (tan θ) =Functions 1(a) images 6

                    = cos 2θ   Functions 1(a) images 7

 

4. If f: R – {±1} ⟶ R is defined by f(x) = Functions 1(a) images 8 , then show that f Functions 1(a) images 9 = 2 f (x)

Sol:

Given f: R – {±1} ⟶ R is defined by f(x) =

            Functions 1(a) images 10

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) = Functions 1(a) images 11, then find range of f.

Sol:

Given A = {1, 2, 3,4} and f: A ⟶ B is a surjection defined by f(x) =Functions 1(a) images 11

    Functions 1(a) images 12

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

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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 = – 1

g (0) = (0)3 = 0

g (1) = (1)3 = 1

∵ range is equal to co domain of f

f is a surjection      

2. Which if the following are injection, surjection or bijection? Justify your answer

(i) f: R ⟶ R defined by f(x) =Functions 1(a) images 13

      let x1, x2 ∈ R

     f(x1) = f(x2)

Functions 1(a) images 14

      2x1 + 1 = 2x2 + 1

      2x1 = 2x2

       x1 = x2

x1, x2 ∈ R, f(x1) = f(x2) ⟹ x1 = x2

∴ f is an injection

Let y = f(x)

       y =Functions 1(a) images 15

3y = 2x + 1

3y – 1 = 2x

⟹ x = Functions 1(a) images 16 ∈ R

Now

         Functions 1(a) images 17           

∴ f is surjection

f is injection and surjection

∴ f is a bijection

 

(ii) f: R ⟶ (0, ∞) defined by f(x) = 2x

let x1, x2 ∈ R

     f(x1) = f(x2)

       Functions 1(a) images 27            

       x1 = x2

x1, x2 ∈ R, f(x1) = f(x2) ⟹ x1 = x2

∴ f is an injection

Let y = f(x)

       y = 2x

 x = Functions 1(a) images 18  ∈ (0, ∞)

 

Now f(x) = 2x

                  = Functions 1(a) images 19

                  = y

∴ f is surjection

f is injection and surjection

∴ f is a bijection

(iii) f: (0, ∞) ⟶ R defined by f(x) =Functions 1(a) images 21

  let x1, x2 ∈ (0, ∞)

     f(x1) = f(x2)

          Functions 1(a) images 20

       x1 = x2

x1, x2 ∈ (0, ∞), f(x1) = f(x2) ⟹ x1 = x2

∴ f is an injection

Let y = f(x)

       y =Functions 1(b) images 1

 x = ey ∈ (0, ∞)

 Now f(x) =Functions 1(a) images 21

                  = Functions 1(a) images 22

                  = y

∴ f is surjection

f is injection and surjection

∴ f is a bijection

(iv) f: [0, ∞) ⟶ [0, ∞) defined by f(x) = x2

  let x1, x2 ∈ [0, ∞)

     f(x1) = f(x2)

     Functions 1(a) images 28           

       x1 = x2 [∵ x1, x2 ∈ [0, ∞)]

 x1, x2 ∈ [0, ∞), f(x1) = f(x2) ⟹ x1 = x2

∴ f is an injection

Let y = f(x)

       y = x2

 x = Functions 1(a) images 23 ∈ [0, ∞)

 Now f(x) = x2

                   =Functions 1(a) images 24

                  = y

∴ f is surjection

f is injection and surjection

∴ f is a bijection

(v) f: R ⟶ [0, ∞) defined by f(x) = x2

  let x1, x2 ∈ R

     f(x1) = f(x2)

        Functions 1(a) images 28 - 1        

       x1 = ± x2 [∵ x1, x2 ∈ R]

 x1, x2 ∈ [0, ∞), f(x1) = f(x2) ⟹ x1 ≠ x2

∴ f is not an injection

Let y = f(x)

       y = x2

 x = Functions 1(a) images 23∈ R

 Now f(x) = x2

                   =Functions 1(a) images 24

                  = y

∴ f is surjection

f is not an injection but surjection

∴ f is not a bijection

(vi) f: R ⟶ R defined by f(x) = x2

  let x1, x2 ∈ R

     f(x1) = f(x2)                

       x1 = ± x2 [∵ x1, x2 ∈ R]

 x1, x2 ∈ [0, ∞), f(x1) = f(x2) ⟹ x1 ≠ x2

∴ f is not an injection

 f (1) = 12 = 1

 f (– 1) = (–1)2 = 1

here ‘– 1’ has no pre image 

∴ f is not a surjection

f is not an injection and not surjection

∴ f is not a bijection

 

Ts Inter Maths IA Concept

hai

 

3. Is 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 A = {1, 2, 3, 4}, B = {1, 3, 5, 7} and g = {(1, 1), (2, 3), (3, 5), (4, 7)}

g (1) = 1; g(2) = 3; g(3) = 5 ; g(4) = 7

here, every element of set A has a unique image in set B

  ∴ g: A ⟶ B is a function

And also given g(x) = ax + b

g (1) = 1

⟹ a (1) + b = 1

       a + b = 1

       b = 1 – a _______________   (1)

g (2) = 3

      a (2) + b = 3

   2a + b = 3

    2a + 1 – a = 3 (from (1))

     a+ 1 = 3

      a = 2

     b = 1 – 2

     b = – 1

∴ a = 2, b = – 1

4. If the function f: R ⟶ R defined by f(x) = Functions 1(a) images 29 , then show that f (x + y) + f (x – y) = 2 f (x) f (y).

Sol:

 Given the function f: R ⟶ R defined by f(x) =Functions 1(a) images 29

Functions 1(a) images 30

                                                       = 2 f (x) f (y)

5. If the function f: R ⟶ R defined by f(x) = Functions 1(a) images 31 , then show that f (1 – x) = 1 – f (x)

and hence reduce the value of Functions 1(a) images 32

Sol:

Given the function f: R ⟶ R defined by f(x) =Functions 1(a) images 31

Functions 1(a) images 33 

∴ f (1 – x) = 1 – f(x)

 

TS 10th Class Maths Concept (T/M)
TS 10th class maths concept (E/M)
 

6. If the function f: {– 1, 1} ⟶ {0, 2}, defined by f(x) = ax + b is a surjection, then find a and b

Sol:

Given the function f: {– 1, 1} ⟶ {0, 2}, defined by f(x) = ax + b is a surjection

Case(i)

         Functions 1(a) images 34 

  If f(– 1) = 0 and f(1) = 2

            a (– 1) + b = 0

           – a + b = 0

              b = a ————(1)

       and

        a (1) + b = 2

         a + b = 2

        a + a = 2         [ from (1)]

         2a = 2

           a = 1

            b = 1

Case (ii)

        Functions 1(a) images 35

  If f(– 1) = 2 and f(1) = 0

            a (– 1) + b = 2

           – a + b =  

              b = 2 + a ————(2)

       and

        a (1) + b = 0

         a + b = 0

        a + 2 + a = 0         [ from (2)]

         2a + 2 = 0

           2a = – 2

            a = – 1

            b = 1

From Case(i) and  Case (ii) a = ±1 and b = 1

7. If f(x) = cos (log x), then show that Functions 1(a) images 36= 0

Sol

Given f(x) = cos (log x)

Functions 1(a) images 37

Functions 1(a) images 38

Functions 1(a) images 36 =

                             cos (log x) cos (log y) – Functions 1(a) images 39 [cos (log x) cos (log y) – sin (log x) sin (log x)

                         + cos (log x) cos (log y) + sin (log x) sin (log x)]

                          = cos (log x) cos (log y) – Functions 1(a) images 39 [2cos (log x) cos (log y)]

                          = cos (log x) cos (log y) – cos (log x) cos (log y)

          ∴ Functions 1(a) images 36   = 0


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