The problem provides a universal set $U$, a set $A$, a function $f(x)$, and a function $P(y)$. The universal set $U$ is defined as the set of natural numbers less than 8, i.e., $U = \{x: x \in \mathbb{N} \text{ and } x < 8\}$. The set $A$ is given as $A = \{1, 3, 5\}$. The function $f(x)$ is defined as $f(x) = x^4 + 3x^3 + kx^2 - 3x - 4 + k$. The function $P(y)$ is defined as $P(y) = \frac{1+y^2+y^4}{y^2}$. The problem asks us to: a) Find $A^c$, where $A^c$ is the complement of $A$ with respect to $U$. b) Find the value of $k$ for which $f(-2) = 0$. c) Show that $\sqrt{P(1)}$ is an irrational number.

1. Problem Description

The problem provides a universal set $U$, a set $A$, a function $f(x)$, and a function $P(y)$.

The universal set $U$ is defined as the set of natural numbers less than 8, i.e., $U = \{x: x \in \mathbb{N} \text{ and } x < 8\}$. The set $A$ is given as $A = \{1, 3, 5\}$. The function $f(x)$ is defined as $f(x) = x^4 + 3x^3 + kx^2 - 3x - 4 + k$. The function $P(y)$ is defined as $P(y) = \frac{1+y^2+y^4}{y^2}$.

The problem asks us to:

a) Find $A^c$, where $A^c$ is the complement of $A$ with respect to $U$.

b) Find the value of $k$ for which $f(-2) = 0$.

c) Show that $\sqrt{P(1)}$ is an irrational number.

2. Solution Steps

a) Finding $A^c$:

Since $U = \{x: x \in \mathbb{N} \text{ and } x < 8\}$, we have $U = \{1, 2, 3, 4, 5, 6, 7\}$.

The set $A$ is given as $A = \{1, 3, 5\}$.

The complement of $A$ with respect to $U$, denoted as $A^c$, is the set of elements in $U$ that are not in $A$. Therefore, $A^c = \{2, 4, 6, 7\}$.

b) Finding the value of $k$ such that $f(-2) = 0$:

We are given the function $f(x) = x^4 + 3x^3 + kx^2 - 3x - 4 + k$. We need to find $k$ such that $f(-2) = 0$.

Substitute $x = -2$ into the function:

$f(-2) = (-2)^4 + 3(-2)^3 + k(-2)^2 - 3(-2) - 4 + k$

$f(-2) = 16 + 3(-8) + 4k + 6 - 4 + k$

$f(-2) = 16 - 24 + 4k + 6 - 4 + k$

$f(-2) = -6 + 5k$

Since $f(-2) = 0$, we have $-6 + 5k = 0$.

$5k = 6$

$k = \frac{6}{5}$

c) Showing that $\sqrt{P(1)}$ is an irrational number:

We are given the function $P(y) = \frac{1+y^2+y^4}{y^2}$. We need to show that $\sqrt{P(1)}$ is irrational.

First, find $P(1)$:

$P(1) = \frac{1+1^2+1^4}{1^2} = \frac{1+1+1}{1} = \frac{3}{1} = 3$.

Now, we need to show that $\sqrt{P(1)} = \sqrt{3}$ is irrational.

Assume, for the sake of contradiction, that $\sqrt{3}$ is rational. Then, $\sqrt{3}$ can be expressed as a fraction $\frac{a}{b}$, where $a$ and $b$ are integers with no common factors (i.e., the fraction is in its simplest form), and $b \neq 0$.

So, $\sqrt{3} = \frac{a}{b}$. Squaring both sides, we get $3 = \frac{a^2}{b^2}$, which implies $a^2 = 3b^2$.

This means that $a^2$ is a multiple of

3. If $a^2$ is a multiple of 3, then $a$ must also be a multiple of

3. Therefore, we can write $a = 3k$ for some integer $k$.

Substituting $a = 3k$ into the equation $a^2 = 3b^2$, we get $(3k)^2 = 3b^2$, which simplifies to $9k^2 = 3b^2$. Dividing both sides by 3, we get $3k^2 = b^2$.

This means that $b^2$ is a multiple of

3. Therefore, $b$ must also be a multiple of

3. Since both $a$ and $b$ are multiples of 3, they have a common factor of 3, which contradicts our initial assumption that $a$ and $b$ have no common factors. Therefore, our assumption that $\sqrt{3}$ is rational must be false.

Thus, $\sqrt{3}$ is irrational, which means $\sqrt{P(1)}$ is irrational.

3. Final Answer

a) $A^c = \{2, 4, 6, 7\}$

b) $k = \frac{6}{5}$

c) $\sqrt{P(1)} = \sqrt{3}$, which is an irrational number.