Functions Exercise 1c Solutions ||TS||

Functions Exercise 1c

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

Exercise 1c

Exercise 1(c) Solutions

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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