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MTH1020 - Analysis of change - S2 2025

MTH1020 - Analysis of change - S2 2025

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Choose from below the correct response:

"If a function g is an even function, then the composition function $f\circ g$ f\circ g is even for any function f."

This statement is:

True because when g(x)=x^2 we have g(-x) = (-x)^2 which is equal to g(x).

it will depend on the functions f and g.

True because f(g(-x)) = f(g(x)) for all x in the domain of g, as g is an even function.

False because g(-x) = -g(x), thus f(g(-x)) = f(-g(x)) which is only equal to f(g(x)) when f is even.

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Choose the correct response for the following statement:

"For all odd functions g, the composition function $f\circ g$ f\circ g is odd for any function f."

This statement:

Can be proven false using a counterexample.

Can be proven true with a general argument using properties of odd and even functions.

Can be proven true using the odd function g(x)=x^3.

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Choose the correct response for the following statement:

"If f is an even function, then the composition function $f\circ g$ f\circ g will be even for any function g."

This statement:

Can be proven false using a counterexample.

Can be proven true with a general argument using properties of odd and even functions.

Is only true for even functions g.

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Let $g: \mathbb{R} \to \mathbb{R}$ g: \mathbb{R} \to \mathbb{R} be defined by g(x)=3x+2.

Evaluate the proof that g(x) is bijective.

Proof:

Suppose

g(x)=g(y). Then

3x+2=3y+2. By subtracting

from

both sides, we have

3x=3y. Dividing by

.

So

g is injective.

Suppose y is a real number. Then y=3x+2. So 3x=y-2 and hence x=(y-2)/3. So g is surjective.

Thus g is bijective.

Choose the most complete correct answer.

The proof is correct.

The result is false, but the proof is correct.

There is an error in showing

g is injective.

There are errors in showing g is injective and in showing g is surjective.

There are no errors so far, but the proof needs another step.

There is an error in showing g is surjective.

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From the following functions, choose the function which is injective.

$k:\mathbb{R}\rightarrow \mathbb{R}^{+}, k(x) =3e^{x-1}$ k:\mathbb{R}\rightarrow \mathbb{R}^{+}, k(x) =3e^{x-1}

$f:\mathbb{R}\rightarrow \mathbb{R}, f(x) = (x-2)^2+1$ f:\mathbb{R}\rightarrow \mathbb{R}, f(x) = (x-2)^2+1

$h:\mathbb{R}\rightarrow \mathbb{R}, h(x) = x^3-2x^2-x+2$ h:\mathbb{R}\rightarrow \mathbb{R}, h(x) = x^3-2x^2-x+2

$g:\mathbb{R}\rightarrow \mathbb{R}^{+}\cup {0}, g(x) =x^2$ g:\mathbb{R}\rightarrow \mathbb{R}^{+}\cup {0}, g(x) =x^2

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From the following functions, choose the function which is surjective.

$f:R\rightarrow R, f(x) = (x-2)^2+1$ f:R\rightarrow R, f(x) = (x-2)^2+1

$k:R\rightarrow R, k(x) =|x|$ k:R\rightarrow R, k(x) =|x|

$h:R\rightarrow R, h(x) = 3e^{x-1}-2$ h:R\rightarrow R, h(x) = 3e^{x-1}-2

$g:R^{+}\rightarrow R^{+}, g(x) =x^2$ g:R^{+}\rightarrow R^{+}, g(x) =x^2

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The following are lines from a proof by induction. Which line uses the inductive hypothesis?

Statement: For every positive integer n,

$1+2+ \dots + n = \frac{n(n+1)}{2}$ 1+2+ \dots + n = \frac{n(n+1)}{2}

Proof.

Since $1=(1\cdot 2)/2$ 1=(1\cdot 2)/2, the statement is true for n=1.
Assume the statement is true for a positive integer k.
$1+2+\dots+k+(k+1) = k(k+1)/2 +(k+1)$ 1+2+\dots+k+(k+1) = k(k+1)/2 +(k+1)
Now k(k+1)/2+(k+1) = (k+1)(k/2 + 1) = (k+1)(k+2)/2
By the principle of mathematical induction, $1+2+\dots+n = n(n+1)/2$ 1+2+\dots+n = n(n+1)/2 for every positive integer n.

The inductive hypothesis is used in step 4.

The inductive hypothesis is used in step 3.

The inductive hypothesis is used in step 1.

The inductive hypothesis is used in step 2.

The inductive hypothesis is used in step 5.

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Choose the correct response from below.

To prove the following statement:

"An integer n is divisible by 3 if and only if its last two digits sum to an integer that is divisible by 3".

we need to show:

A direct example, such as n = 39, where 3+9=12 which is divisible by 3.

If the sum of the last two digits of an integer n is divisible by 3 then n is also divisible by 3.

If an integer n is divisible by 3 then its last two digits sum to a number that is divisible by 3.

If an integer n is divisible by 3 then its last two digits sum to a number that is divisible by 3. And; If the sum of the last two digits of an integer n is divisible by 3 then n is also divisible by 3.

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