Solve in English please 🙏 All the questions

Setting $P(-1) = 0$:

Setting $P(2) = 0$:

From equation (1), we can express $b$ in terms of $a$:

Substitute this into equation (2):

Now substitute $a=0$ back into $b = -2a$:

So, the values of the constants are $a=0$ and $b=0$.

The expression is $P(x) = ax^3 + bx - a$. With $a=0$ and $b=0$, the expression becomes $P(x) = 0$.
By the Remainder Theorem, the remainder when $P(x)$ is divided by $(x-1)$ is $P(1)$.

Since $b=0$, the remainder is 0.
The values of the constants are $\boxed{a=0, b=0}$ and the remainder is $\boxed{0}$.

Question 1 (ii)

Express as a single fraction $\frac{5}{x^2+2x-3}$, simplifying the numerator.

We need to find two numbers that multiply to $-3$ and add to 2. These numbers are 3 and $-1$.
So, $x^2+2x-3 = (x+3)(x-1)$.

The numerator is already simplified as it is a constant 5. The expression is already a single fraction.
The simplified form of the fraction is $\boxed{\frac{5}{(x+3)(x-1)}}$.

Question 6 (i)

The functions $f$ and $g$ are defined by $f(x) = \frac{3}{x-2}$, $x \in \mathbb{R}, x \neq 2$, and $g(x) = \frac{x-1}{x+2}$, $x \in \mathbb{R}, x \neq -2$.

(a) Find $fg(x)$ and $gf(x)$, stating their domains.

$fg(x) = f(g(x)) = f\left(\frac{x-1}{x+2}\right)$.
Substitute $g(x)$ into $f(x)$:

Simplify the denominator:

So,

The domain of $fg(x)$ requires two conditions:

1. $x$ must be in the domain of $g(x)$, so $x \neq -2$.
2. $g(x)$ must be in the domain of $f(x)$, so $g(x) \neq 2$. $$\frac{x-1}{x+2} \neq 2$$ $$x-1 \neq 2(x+2)$$ $$x-1 \neq 2x+4$$ $$-5 \neq x$$

Combining these, the domain of $fg(x)$ is $x \in \mathbb{R}, x \neq -2, x \neq -5$.

$gf(x) = g(f(x)) = g\left(\frac{3}{x-2}\right)$.
Substitute $f(x)$ into $g(x)$:

Simplify the numerator:

Simplify the denominator:

So,

The domain of $gf(x)$ requires two conditions:

1. $x$ must be in the domain of $f(x)$, so $x \neq 2$.
2. $f(x)$ must be in the domain of $g(x)$, so $f(x) \neq -2$. $$\frac{3}{x-2} \neq -2$$ $$3 \neq -2(x-2)$$ $$3 \neq -2x+4$$ $$2x \neq 1$$ $$x \neq \frac{1}{2}$$

Combining these, the domain of $gf(x)$ is $x \in \mathbb{R}, x \neq 2, x \neq \frac{1}{2}$.

(b) Show that $g$ is not surjective.

A function $g: A \to B$ is surjective if for every element $y$ in the codomain $B$, there exists at least one element $x$ in the domain $A$ such that $g(x) = y$.
The function $g(x) = \frac{x-1}{x+2}$ has a domain of $\mathbb{R} \setminus \{-2\}$. Assuming the codomain is $\mathbb{R}$.

Let $y = g(x)$.

Multiply both sides by $(x+2)$:

Rearrange to solve for $x$:

For $x$ to be defined, the denominator $1-y$ cannot be zero.

This means that there is no value of $x$ in the domain of $g$ that can produce $y=1$. In other words, 1 is not in the range of $g$.
Since the codomain of $g$ is $\mathbb{R}$ (which includes 1), and the range of $g$ does not include all values in the codomain (specifically, it does not include 1), the function $g$ is not surjective.

Question 6 (ii)

A relation $R$ is defined on the set $A = \{1, 2, 3, 4, 5\}$ by $aRb \iff a+b = 2n, n \in \mathbb{N}$. List all the equivalence classes of $A$ under $R$.

The condition $a+b = 2n$ means that the sum $a+b$ must be an even number. This occurs if and only if $a$ and $b$ have the same parity (both even or both odd).

Odd numbers in $A$: $\{1, 3, 5\}$
Even numbers in $A$: $\{2, 4\}$

An equivalence class $[a]$ consists of all elements $x \in A$ such that $aRx$.
For $a=1$:
$[1] = \{x \in A \mid 1Rx\}$. This means $1+x$ must be even, so $x$ must be odd.
The odd numbers in $A$ are $\{1, 3, 5\}$.
So, $[1] = \{1, 3, 5\}$.

For $a=2$:
$[2] = \{x \in A \mid 2Rx\}$. This means $2+x$ must be even, so $x$ must be even.
The even numbers in $A$ are $\{2, 4\}$.
So, $[2] = \{2, 4\}$.

For $a=3$:
$[3] = \{x \in A \mid 3Rx\}$. This means $3+x$ must be even, so $x$ must be odd.
So, $[3] = \{1, 3, 5\}$. This is the same as $[1]$.

For $a=4$:
$[4] = \{x \in A \mid 4Rx\}$. This means $4+x$ must be even, so $x$ must be even.
So, $[4] = \{2, 4\}$. This is the same as $[2]$.

For $a=5$:
$[5] = \{x \in A \mid 5Rx\}$. This means $5+x$ must be even, so $x$ must be odd.
So, $[5] = \{1, 3, 5\}$. This is the same as $[1]$.

The distinct equivalence classes are the sets of numbers with the same parity.
The equivalence classes of $A$ under $R$ are $\boxed{\{1, 3, 5\} \text{ and } \{2, 4\}}$.

Question 7 (i)

This question was solved in a previous turn. The solution was $\boxed{\theta = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \frac{3\pi}{2}}$.

Question 7 (ii)

This question was solved in a previous turn. The solution was $\boxed{\text{Since } \det(M) = -41 \neq 0, \text{ the matrix M is invertible.}}$.