Functions Exercise 1a Solutions

Functions Exercise 1a Solutions

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

Exercise 1a

 if ( function_exists( ‘pgntn_display_pagination’ ) ) pgntn_display_pagination( ‘multipage’ );

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

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 = – 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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