CALCU-Q.1Are the following statements True or Fals

Q.1Are the following statements True or False? (briefly justify your answer) (a) (i)

\emptyset=\{0\}

(ii)

x \in\{x\}

(iii)

\emptyset=\{\emptyset\}

(iv)

\emptyset \in\{\emptyset\}

(b) The following sets have been expressed in set builder notation. Re-express them in list notation. (i)

\{\rho: \rho

is a state capital,

\rho

is in Australia

\}

(ii)

\{x: x=2 n-5

where

n \in \mathbb{N}\}

(iii)

\left\{y: 2 y^{2}=50\right.

where

\left.y \in \mathbb{Z}\right\}

4 marks (iv)

\left\{z: 3 z=n^{2}\right.

where both

n

and

z

are natural numbers

\}

Q. 2 On the set

A=\{1,2,3,4,5\}

we mav define two functions

f: A \longrightarrow A

and

g: A \longrightarrow A

by the

Solution by Steps

step 1

To determine the truth value of the statement

\emptyset = \{0\}

, we compare the empty set to a set containing the number 0

step 2

The empty set has no elements, while the set

\{0\}

has one element, 0. Therefore,

\emptyset \neq \{0\}

and the statement is False

step 3

To determine the truth value of the statement

x \in \{x\}

, we consider any element x and whether it is a member of the set containing only x itself

step 4

Since x is the only element in the set

\{x\}

, it is true that

x \in \{x\}

for any x. Therefore, the statement is True

step 5

To determine the truth value of the statement

\emptyset = \{\emptyset\}

, we compare the empty set to a set containing the empty set

step 6

The empty set has no elements, while the set

\{\emptyset\}

has one element, which is the empty set itself. Therefore,

\emptyset \neq \{\emptyset\}

and the statement is False

step 7

To determine the truth value of the statement

\emptyset \in \{\emptyset\}

, we check if the empty set is an element of the set containing the empty set

step 8

Since the set

\{\emptyset\}

contains the empty set as its only element, it is true that

\emptyset \in \{\emptyset\}

and the statement is True

step 9

To re-express the set builder notation

\{\rho: \rho \text{ is a state capital, } \rho \text{ is in Australia }\}

in list notation, we list the state capitals of Australia

step 10

The state capitals in Australia are Sydney, Melbourne, Brisbane, Perth, Adelaide, Hobart, Darwin, and Canberra. Therefore, the list notation is

\{Sydney, Melbourne, Brisbane, Perth, Adelaide, Hobart, Darwin, Canberra\}

step 11

To re-express the set builder notation

\{x: x = 2n - 5 \text{ where } n \in \mathbb{N}\}

in list notation, we generate values of x for natural numbers n

step 12

The first few natural numbers are 1, 2, 3, ... . Substituting these into the expression

2n - 5

, we get -3, -1, 1, 3, ... . Therefore, the list notation is

\{-3, -1, 1, 3, ...\}

step 13

To re-express the set builder notation

\{y: 2y^2 = 50 \text{ where } y \in \mathbb{Z}\}

in list notation, we solve the equation for y

step 14

Solving

2y^2 = 50

gives

y^2 = 25

, so

y = \pm5

. Since y must be an integer, the list notation is

\{-5, 5\}

step 15

To re-express the set builder notation

\{z: 3z = n^2 \text{ where both } n \text{ and } z \text{ are natural numbers }\}

in list notation, we find values of z that satisfy the equation for natural numbers n

step 16

The first few natural numbers are 1, 2, 3, ... . Squaring these and dividing by 3, we get z values of 1/3, 4/3, 3, ... . However, z must be a natural number, so the only valid z value from these is 3. Therefore, the list notation is

\{3\}

Answer

(a)(i) False, (ii) True, (iii) False, (iv) True (b)(i)

\{Sydney, Melbourne, Brisbane, Perth, Adelaide, Hobart, Darwin, Canberra\}

(ii)

\{-3, -1, 1, 3, ...\}

(iii)

\{-5, 5\}

(iv)

\{3\}

Key Concept

Set theory and set builder notation

Explanation

The truth value of statements involving sets is determined by the definition of the empty set and membership within sets. Set builder notation is a concise way to define a set by specifying a property that its members must satisfy, and it can be converted to list notation by finding all elements that meet the given property.

Q. 2 On the set

A=\{1,2,3,4,5\}

we may define two functions

f: A \longrightarrow A

and

g: A \longrightarrow A

by the table \begin{tabular}{|c||c|c|c|c|c|} \hline

x

& 1 & 2 & 3 & 4 & 5 \\ \hline

f(x)

& 3 & 5 & 3 & 1 & 2 \\ \hline

g(x)

& 4 & 1 & 1 & 2 & 3 \\ \hline

f_{o} g(x)

& & & & & \\ \hline

g_{o} f(x)

& & & & & \\ \hline \end{tabular} Fill in the final two rows by finding the composite functions

f_{o} g

and

g_{o} f

. Q. 3 (a)If

f

and

g

are functions given by

f: A \longrightarrow \mathbb{R}

where

f(x)=\frac{1}{3}\left(\frac{1}{x}+1\right)

g: B \longrightarrow \mathbb{R}

where

g(x)=\frac{1}{3 x-1}

Find (i)The implied maximal domains

A

and

B

for the functions

f

and

g

respectively. (ii)The rule for

f_{o} g

and state its' range. (iii)The rule for

g_{o} f

and state its' range. 6 marks (b)(i)Find the function which is the inverse of

h:\{x \mid x \geq 2\} \longrightarrow \mathbb{R}, h(x)=2-\sqrt{x-2}

(ii)State the domain and range of

h^{-1}(x)

from part (i). (iii)Sketch the graphs of

h(x)

and

h^{-1}(x)

on the SAME set of axes. 7 marks

Solution by Steps

step 1

To find the composite function

f \circ g

, we evaluate

f(g(x))

for each value of

x

in the set

A

step 2

For

x = 1

,

g(1) = 4

and

f(4) = 1

, so

f(g(1)) = 1

step 3

For

x = 2

,

g(2) = 1

and

f(1) = 3

, so

f(g(2)) = 3

step 4

For

x = 3

,

g(3) = 1

and

f(1) = 3

, so

f(g(3)) = 3

step 5

For

x = 4

,

g(4) = 2

and

f(2) = 5

, so

f(g(4)) = 5

step 6

For

x = 5

,

g(5) = 3

and

f(3) = 3

, so

f(g(5)) = 3

step 7

To find the composite function

g \circ f

, we evaluate

g(f(x))

for each value of

x

in the set

A

step 8

For

x = 1

,

f(1) = 3

and

g(3) = 1

, so

g(f(1)) = 1

step 9

For

x = 2

,

f(2) = 5

and

g(5) = 3

, so

g(f(2)) = 3

step 10

For

x = 3

,

f(3) = 3

and

g(3) = 1

, so

g(f(3)) = 1

step 11

For

x = 4

,

f(4) = 1

and

g(1) = 4

, so

g(f(4)) = 4

step 12

For

x = 5

,

f(5) = 2

and

g(2) = 1

, so

g(f(5)) = 1

Answer

f \circ g

for

x = 1, 2, 3, 4, 5

is

1, 3, 3, 5, 3

respectively.

g \circ f

for

x = 1, 2, 3, 4, 5

is

1, 3, 1, 4, 1

respectively.

Key Concept

Composite Functions

Explanation

To find the composite function

f \circ g

or

g \circ f

, we substitute the output of one function into the input of the other and evaluate for each element in the domain.

Solution by Steps

step 1

The maximal domain

A

for

f(x)

is the set of all real numbers except for

x = 0

since

f(x)

involves division by

x

step 2

The maximal domain

B

for

g(x)

is the set of all real numbers except for

x = \frac{1}{3}

since

g(x)

involves division by

3x - 1

step 3

To find the rule for

f \circ g

, we substitute

g(x)

into

f(x)

, giving

f(g(x)) = f\left(\frac{1}{3x-1}\right) = \frac{1}{3}\left(\frac{1}{\frac{1}{3x-1}}+1\right)

step 4

Simplifying

f(g(x))

, we get

f(g(x)) = \frac{1}{3}\left(3x-1+1\right) = x

. The range of

f \circ g

is all real numbers except for

x = \frac{1}{3}

step 5

To find the rule for

g \circ f

, we substitute

f(x)

into

g(x)

, giving

g(f(x)) = g\left(\frac{1}{3}\left(\frac{1}{x}+1\right)\right) = \frac{1}{3\left(\frac{1}{3}\left(\frac{1}{x}+1\right)\right)-1}

step 6

Simplifying

g(f(x))

, we get

g(f(x)) = \frac{1}{\frac{1}{x}+1-1} = x

. The range of

g \circ f

is all real numbers except for

x = 0

Answer

(i) Maximal domain

A

is

\mathbb{R} \setminus \{0\}

, and maximal domain

B

is

\mathbb{R} \setminus \left\{\frac{1}{3}\right\}

. (ii) The rule for

f \circ g

is

x

, and its range is

\mathbb{R} \setminus \left\{\frac{1}{3}\right\}

. (iii) The rule for

g \circ f

is

x

, and its range is

\mathbb{R} \setminus \{0\}

.

Key Concept

Maximal Domains and Composite Functions

Explanation

The maximal domain of a function is the set of all possible inputs for which the function is defined. When composing functions, we substitute one function into another and simplify to find the new rule and its range.

Solution by Steps

step 1

To find the inverse function

h^{-1}(x)

, we set

y = h(x)

and solve for

x

in terms of

y

step 2

We have

y = 2 - \sqrt{x-2}

. To find

h^{-1}(x)

, we swap

x

and

y

to get

x = 2 - \sqrt{y-2}

and solve for

y

step 3

Adding

\sqrt{y-2}

to both sides and then subtracting

x

from both sides gives

\sqrt{y-2} = 2 - x

step 4

Squaring both sides yields

y - 2 = (2 - x)^2

step 5

Adding 2 to both sides gives

y = (2 - x)^2 + 2

. This is the rule for

h^{-1}(x)

step 6

The domain of

h^{-1}(x)

is

[2, \infty)

because

h(x)

is defined for

x \geq 2

step 7

The range of

h^{-1}(x)

is

[0, \infty)

because

h(x)

takes on all values greater than or equal to 0

step 8

To sketch the graphs of

h(x)

and

h^{-1}(x)

, we plot the points and reflect them across the line

y = x

Answer

(i) The inverse function is

h^{-1}(x) = (2 - x)^2 + 2

. (ii) The domain of

h^{-1}(x)

is

[2, \infty)

, and the range is

[0, \infty)

. (iii) The graphs of

h(x)

and

h^{-1}(x)

are reflections of each other across the line

y = x

.

Key Concept

Inverse Functions

Explanation

To find the inverse of a function, we swap the dependent and independent variables and solve for the new dependent variable. The domain and range of the inverse function are the range and domain of the original function, respectively.

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