22 Nov 2020 6:46 pm

Question 1.

Let f = {(x, y) |x,y; ∈ N and y = 2x} be a relation on N. Find the domain, co-domain and range. Is this relation a function?

Answer:

X = {1,2,3,….}

Y = {1,2,3,….}

f = {(1,2) (2, 4) (3, 6) (4, 8) ….}

Domain = {1, 2, 3, 4 ….}

Co – Domain = {1, 2, 3, 4 ….}

Range = {2, 4, 6, 8 }

Yes this relation is a function.

Question 2.

Let X = {3, 4, 6, 8}. Determine whether the relation

R = {(x,f(x)) |x ∈ X, f(x) = x^{2} + 1}

is a function from X to N?

Answer:

f(x) = x^{2} + 1

f(3) = 3^{2} + 1 = 9 + 1 = 10

f(4) = 4^{2} + 1 = 16 + 1 = 17

f(6) = 6^{2} + 1 = 36 + 1 = 37

f(8) = 8^{2} + 1 = 64 + 1 = 65

yes, R is a function from X to N

Question 3.

Given the function f: x → x^{2} – 5x + 6, evaluate

(i) f(-1)

(ii) f(2a)

(iii) f(2)

(iv) f(x – 1)

Solution:

Give the function f: x → x^{2} – 5x + 6.

(i) f(-1) = (-1)^{2} – 5(1) + 6 = 1 + 5 + 6 = 12

(ii) f(2a) = (2a)^{2} – 5(2a) + 6 = 4a^{2} – 10a + 6

(iii) f(2) = 2^{2} – 5(2) + 6 = 4 – 10 + 6 = 0

(iv) f(x – 1) = (x – 1)^{2} – 5(x – 1) + 6

= x^{2} – 2x + 1 – 5x + 5 + 6

= x^{2} – 7x + 12

Question 4.

A graph representing the function f(x) is given in it is clear that f(9) = 2.

**(i) Find the following values of the function**

(a) f(0)

(b) f(7)

(c) f(2)

(d) f(10)**Answer:**

(a) f (0) = 9

(b) f (7) = 6

(c) f (2) = 6

(d) f(10) = 0

(ii) For what value of x is f(x) = 1 ?

Answer:

When f(x) = 1 the value of x is 9.5

(iii) Describe the following

(i) Domain

(ii) Range.

Answer:

Domain = {0, 1, 2, 3,… .10}

= {x / 0 < x < 10, x ∈ R}

Range = {0,1,2,3,4,5,6,7,8,9}

= {x / 0 < x < 9, x ∈ R}

(iv) What is the image of 6 under f?**Answer:**

The image of 6 under f is 5.

Question 5.

Let f (x) = 2x + 5. If x ≠ 0 then find

f(x+2)−f(2)x**Answer:**

f(x) = 2x + 5

f(x + 2) = 2(x + 2) + 5

= 2x + 4 + 5

= 2x + 9

**Question 6.**

A function/is defined by f(x) = 2x – 3

(i) find f(0)+f(1)2

(ii) find x such that f(x) = 0.

(iii) find x such that/ (A:) = x.

(iv) find x such that fix) =/(l – x).

Answer:

(i) f(x) = 2x – 3

f(0) = 2(0) – 3 = -3

f(1) = 2(1) – 3 = 2 – 3 = -1

(ii) f(x) = 0

2x – 3 = 0

2x = 3

x = 32

(iii) f(x) = x

2x – 3 = x

2x – x = 3

x = 3

(iv) f(1 – x) = 2(1 – x) – 3

= 2 – 2x – 3

= – 2x – 1

f(x) = f(1 – x)

2x – 3 = – 2x – 1

2x + 2x = 3 – 1

4x = 2

x = 24 = 12

Question 7.

square piece of material, 24 cm on a side, by cutting equal squares from the corners and turning up the sides as shown. Express the volume V of the box as a function of x.

Solution:

After cutting squares we will get a cuboid,

length of the cuboid (l) = 24 – 2x

breadth of the cuboid (b) = 24 – 2x

height of the cuboid (h) = 2x

Volume of the box = Volume of the cuboid

V = (24 – 2x)(24 – 2x) (x)

= (24 – 2x)^{2} (x)

= (576 + 4x^{2} – 96x) x

= 576x + 4x^{3} – 96x^{2}

V = 4x^{3} – 96x^{2} + 576x

V(x) = 4x^{3} – 96x^{2} + 576x

Question 8.

A function f is defined by f(x) = 3 – 2x. Find x such that f(x^{2}) = (f (x))^{2}.

Answer:

f(x) = 3 – 2x

f(x^{2}) = 3 – 2 (x^{2})

= 3 – 2x^{2}

(f (x))^{2} = (3 – 2x)^{2}

= 9 + 4x^{2} – 12x

But f(x^{2}) = (f(x))^{2}

3 – 2 x^{2} = 9 + 4x^{2} – 12x

-2x^{2} – 4x^{2 }+ 12x + 3 – 9 = 0

-6x^{2} + 12x – 6 = 0

(÷ by – 6) ⇒ x^{2} – 2x + 1 = 0

(x – 1) (x – 1) = 0

x – 1 = 0 or x – 1 = 0

x = 1

The value of x = 1

Question 9.

A plane is flying at a speed of 500 km per hour. Express the distance d travelled by the plane as function of time t in hours.

Solution:

Speed = distance covered time taken

⇒ distance = Speed × time

⇒ d = 500 × t [ ∵ time = t hrs]

⇒ d = 500 t

Question 10.

The data in the adjacent table depicts the length of a woman’s forehand and her corresponding height. Based on this data, a student finds a relationship between the height (y) and the forehand length (x) as y = ax + b , where a, b are constants.

(i) Check if this relation is a function.

(ii) Find a and b.

(iii) Find the height of a woman whose forehand length is 40 cm.

(iv) Find the length of forehand of a woman if her height is 53.3 inches.

Length ‘x’ of forehand (in cm) | Height y (in inches) |

35 | 56 |

45 | 65 |

50 | 69.5 |

55 | 74 |

Answer:

The relation is y = 0.9x + 24.5

(i) Yes the relation is a function.

(ii) When compare with y = ax + b

a = 0.9, b = 24.5

(iii) When the forehand length is 40 cm, then height is 60.5 inches.

Hint: y = 0.9x + 24.5

= 0.9 × 40 + 24.5

= 36 + 24.5

= 60.5 feet

(iv) When the height is 53.3 inches, her forehand length is 32 cm

Hint: y = 0.9x + 24.5

53.3 = 0.9x + 24.5

53.3 – 24.5 = 0.9 x

28.8 = 0.9 x

x = 28.80.9

x = 32 cm

A function may be represented by

(a) Set of ordered pairs

(b) Table form

(c) Arrow diagram

(d) Graphical form

Vertical line test

A curve drawn in a graph represents a function, if every vertical line intersects the curve in at most one point.

1. One – One function (injection)

A function f: A → B is called one-one function if distinct elements of A have a distinct image in B.

2. Many – One function

,A function f: A → B is called many-on function if two or more elements of A have same image in B.

3. Onto function (surjection)

A function f: A → B is said to be on to function if the range of f is equal to the co-domain of f.

4. Into function

A function f: A → B is called an into function if there exists at least one element in B which is not the image of any element of A.

5. Bijection

A function f: A → B is both one – one and onto, then f is called a bijection from A to B.

**Horizontal line test**

A function represented in a graph is one – one, if every horizontal line intersects the curve in at most one point.

**Special cases of function**

1. Constant function

A function f: A → B is called a constant function if the range of f contains only one element.

2. Identity function

A function f : A → A defined by f(x) = x for all x ∈ A is called an identity function on A and is denoted by I_{A}.

3. Real valued function

A function f: A → B is called a real valued function if the range of f is a subset of the set of all real numbers R.