Algebra Exam Paper

Q2

Let the roots are $k-d,k,k+d$ such that $d > 0$.

By Vieta's formula, we get $(k-d)+(k)+(k+d) = -3$, which gives $k = -1$.

Therefore, the roots are $(-1-d, -1, -1+d$.

Also, we get $(-1-d)(-1)(-1+d) = -b$ and $(-1-d)(-1) + (-1)(-1+d) + (-1+d)(-1-d) = a$.

$1-d^{2} = b$ and $1+d+1-d+1-d^{2} = a$.

$1-d^{2} = b$ and $3-d^{2} = a$.

Therefore, $(a,b) = (3-d^{2},1-d^{2})$ for all positive integers $d$.

My life is complicated. This is the original solution if I didn't state $d > 0$.

Therefore, $a-b = 2$.

Substitute back to equation we get

$x^{3}+3x^{2}+(b+2)x+b$.

We know that $-1$ is the root of equation. We can factor $(x+1)$ out of the polynomial.

$(x+1)(x^{2}+2x+b)$.

If $x^{2}+2x+b = 0$ has 2 other solutions, we get

$\Delta = 2^{2} - 4\times 1\times b > 0$.

$b < 1$

5) Choose arbitrary primes $p_1$ and $p_2$. Then, $f(p_1)+f(p_1p_2) = f(p_2)$ and $f(p_2)+f(p_1p_2) = f(p_1)$. This gives that $f(p_1) = f(p_2)$ and $f(p_1p_2) = 0$. Let $f(p) = c$ for arbitrary prime $p$.

Let $n$ be an arbitrary positive composite number with prime factorization $p_1p_2\cdots p_k$, where the $p_i$ are not necessarily unique. Then, we propose that $f(n) = (2-k)c$. We will show this by induction. From the formula given, $f(n) = f(p_2\cdots p_k)-f(p_1)$. First, note that if $k = 2$, $f(n) = 0$ and if $k = 3$, $f(n) = -f(p_1) = -c$. Let $n_i$ have $i$ non-unique prime factors. Then, if $f(n_k) = (2-k)c$, we have from our equation that $f(n_{k+1}) = (2-k)c-c = (2-(k+1))c$. Our induction is complete.

Note that $2014 = 2\times 19\times 53$, giving $f(2014^2) = -4c$; $2015 = 5\times 13\times 31$, giving $f(2015^3) = -7c$; and $2016 = 2^5\cdot 3^2\cdot 7$, giving $f(2016^5) = -38c$. Thus, $f(2014^2)+f(2015^3)+f(2016^5) = -49c$.

We have $f(2^{2014})+f(3^{2015})+f(5^{2016}) = -2012c-2013c-2014c = 2013$, which gives $c = -1/3$. Thus, $f(2014^2)+f(2015^3)+f(2016^5) = 49/3$.

What age group is this aimed at?

Question 4: Using Cauchy-Schwarz, the inequality is proven trivially.

Using the equality case for C-S, this implies that a^2 = b^2 = (c^2)/4 = (d^2)/16 implying that a = b = c/2 = d/4. Performing the substitutions gives a + b + c + d = 8.

4a) does not have a unique answer. The equality is satisfied by any $a$, $b$, $c$, and $d$ that satisfy $a = b = d/4$ and $c = d/2$ as long as $d \neq 0$, which gives $a+b+c+d = 2d$, which can obviously take on any positive value within the constraints of the problem.

2) Let $P(x) = P(x+c) = P(x+2c) = 0$. Then, expand to find that $\frac{P(x+2c)-P(x)}{2} = 3cx^2+(6c^2+6c)x+(4c^3+6c^2+ac) = 0$ and $P(x+c)-P(x) = 3cx^2+(3c^2+6c)x+(c^3+3c^2+ac) = 0$. Then, $[P(x+c)-P(x)]-\left[\frac{P(x+2c)-P(x)}{2}\right] = 3c^2x+(3c^3+3c^2) = 0$, which gives $x = -1-c$. Thus, the three roots in arithmetic sequence are $x = -1-c$, $x+c = -1$, and $x+2c = -1+c$. From Vieta's formulas, this gives $a = (-1-c)(-1)+(-1-c)(-1+c)+(-1)(-1+c) = 3-c^2$ and $b = -(-1-c)(-1)(-1+c) = 1-c^2$ for integer $c$.

3a) If we let $y = 0$, we have $f(0)+x\cdot f(0) = x\cdot f(x)$, or $f(x) = \frac{f(0)+x\cdot f(0)}{x}$ for $x \neq 0$. If we go back to our original equation and also let $x = 0$, we find that $f(0) = 0$. Thus, $f(x) = 0$ for all $x \in \mathbb{R}$.

3b) Since the value of $f(0)$ is undefined, we take our previous solution of $f(x) = \frac{f(0)+x\cdot f(0)}{x}$ and substitute $c$ for $f(0)$ to get $f(x) = c\frac{x+1}{x}$. In 3a), we must consider that this $f$ does not map $0$ to a single value in $\mathbb{R}$ except in the case that $f(0) = 0$, but as $0 \notin A$, we avoid that problem here.