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

   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

      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) = , then show that f (x) + f (1/x) = 0

Sol:

Given f: R – {0} ⟶ R is defined by f(x) =

       f (1/x)  =

Now

f (x) + f (1/x)  =

∴ f (x) + f (1/x)  = 0

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

Sol:

    Given f: R ⟶ R is defined by f(x) =

       f (tan θ) =

                    = cos 2θ  

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

Sol:

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

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}

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Functions Exercise 1a

6. If A = {1, 2, 3,4} and f: A ⟶ B is a surjection defined by f(x) = , then find range of f.

Sol:

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

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) =

      let x₁, x₂ ∈ R

     f(x₁) = f(x₂)

      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 =

3y = 2x + 1

3y – 1 = 2x

⟹ x =  ∈ R

Now

∴ f is surjection

f is injection and surjection

∴ f is a bijection

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

let x₁, x₂ ∈ R

     f(x₁) = f(x₂)

       x₁ = x₂

x₁, x₂ ∈ R, f(x₁) = f(x₂) ⟹ x₁ = x₂

∴ f is an injection

Let y = f(x)

       y = 2^x

 x =   ∈ (0, ∞)

Now f(x) = 2^x

                  =

                  = y

∴ f is surjection

f is injection and surjection

∴ f is a bijection

(iii) f: (0, ∞) ⟶ R defined by f(x) =

  let x₁, x₂ ∈ (0, ∞)

     f(x₁) = f(x₂)

       x₁ = x₂

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

∴ f is an injection

Let y = f(x)

       y =

 x = e^y ∈ (0, ∞)

 Now f(x) =

                  =

                  = 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)

       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 =  ∈ [0, ∞)

 Now f(x) = x²

                   =

                  = 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)

       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 = ∈ R

 Now f(x) = x²

                   =

                  = 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

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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) = , 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) =

                                                       = 2 f (x) f (y)

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

and hence reduce the value of

Sol:

Given the function f: R ⟶ R defined by f(x) =

∴ 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)

  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)

  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 = 0

Sol

Given f(x) = cos (log x)

 =

                             cos (log x) cos (log y) – [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) – [2cos (log x) cos (log y)]

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

          ∴    = 0

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