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# Algebra Exam Paper

Junior Exam J3

Each problem is worth 7 marks.

Time: 4 hours

No books, notes or calculators permitted

Note: You must answer with proof.

Q1

Let $$a$$, $$b$$, $$x$$, $$y$$ and $$z$$ be integers such that

$0 \leq x \leq a \leq y \leq b \leq z$

$a + b + x + y + z = 2016$

(a) Find the maximum value of $$x + y + z$$. Find all $$x$$, $$y$$ and $$z$$ which satisfy this.

(b) Find the minimum value of $$x + y + z$$. Find all $$x$$, $$y$$ and $$z$$ which satisfy this.

Q2

The polynomial

$P(x) = x^3 + 3x^2 + ax + b$

has 3 distinct integer roots which can be arranged to form an arithmetic progression. Find all $$a$$ and $$b$$ which satisfy this.

Q3

Let $$A$$ denote the set containing all non-zero real numbers.

(a) Find all functions $$f:\mathbb {R} \rightarrow \mathbb {R}$$ such that $$x f(xy) + f(-y) = x f(x)$$.

(b) Find all functions $$f:A \rightarrow \mathbb {R}$$ such that $$x f(xy) + f(-y) = x f(x)$$.

Q4

Assuming $$a$$, $$b$$, $$c$$ and $$d$$ are positive reals:

(a) Find the value of $$a + b + c + d$$ if

$\dfrac {1}{a} + \dfrac {1}{b} + \dfrac {4}{c} + \dfrac {16}{d} = \dfrac {64}{a + b + c + d}$

(b) Prove that

$\dfrac {1}{a} + \dfrac {1}{b} + \dfrac {4}{c} + \dfrac {16}{d} \geq \dfrac {64}{a + b + c + d}$

Q5

$$f:\mathbb {N}^{+} \rightarrow \mathbb {R}$$ is a function such that

$f(p) + f(pn) = f(n)$

where $$n$$ is a number and $$p$$ is a prime number. It is also known that

$f(2^{2014}) + f(3^{2015}) + f(5^{2016}) = 2013$.

Find the value of

$f(2014^2) + f(2015^3) + f(2016^5)$.

Note by Sharky Kesa
2 years ago

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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, $$f(2014^2)+f(2015^3)+f(2016^5) = 49/3$$. · 2 years ago

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. · 2 years ago

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 $$c$$. · 2 years ago

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 years ago

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. · 2 years ago

What age group is this aimed at? · 2 years ago

12 years to 15 years. · 2 years ago

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

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

$$b < 1$$ · 2 years ago

Sorry, forgot to state that the roots were distinct. · 2 years ago

I let the roots are $$k-d,k,k+d$$ such that $$d > 0$$. This will definitely give you distinct roots. · 2 years ago

could u explain how u got a - b = 2? · 2 years ago

Subtract $$3-d^{2} = a$$ and $$1-d^{2} = b$$. · 2 years ago

thanks!! · 2 years ago

Doesn't $$b < 1$$ come directly from the fact that $$b = 1 - d^2$$, since $$d^2$$ can take any value in the interval $$(0,\infty)$$ · 2 years ago