# Australian School of Excellence 2015 Algebra Exam

• Each question is worth 7 points

• Time allowed is 4 hours

• No books, notes or calculators permitted

1) Find all real numbers $a$ such that there are four different real numbers $x$ satisfying the equation

$x(x+1)^3=(2x+a)(x+a-1).$

2) Find all functions $f:\mathbb{R} \rightarrow \mathbb{R}$ such that for $x, y \in \mathbb{R}$ we have

$f(x f(y) + x) = xy+f(x).$

3) Suppose that $x$, $y$ and $z$ are real numbers satisfying the following three equations.

\begin{aligned} x+y+z &=2\\ x^2 +y^2 + z^2 &=6\\ x^3+y^3+z^3&=8 \end{aligned}

Find all possible values of $x^{2015} + y^{2015} + z^{2015}$.

4) Suppose that a function $f:\mathbb{R} \rightarrow \mathbb{R}$ has the property that

$f(x)^2 \leq f(y)$

for every $x > y$.

Prove that $0 \leq f(x) \leq 1$ for all $x \in \mathbb{R}$.

5) Prove the following inequality for all $a, b, c > 0$.

$\dfrac {a^2(a^3 + b^3)}{a^2 + ab + b^2} +\dfrac {b^2(b^3 + c^3)}{b^2 + bc + c^2} +\dfrac {c^2(c^3 + a^3)}{c^2 + ca + a^2} \geq 2abc$ Note by Sharky Kesa
4 years, 5 months ago

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3). let $s_1,s_2,s_3$ be the symmetric sums. let $p_n$ denote the nth power sum. then by newtons sum: $\begin{array}{c}a p_1=s_1=2\\ p_2=s_1p_1-2s_2=4-2s_2=6\Longrightarrow s_2=-1\\ p_3=s_1p_2-s_2p_1+3s_3=14+3s_3=8\Longrightarrow s_3=-2\end{array}$ so, x,y,z are roots of $x^3-2x^2-x+2=0$ $x^3-x-2x^2+2=x(x^2-1)-2(x^2-1)=(x-2)(x-1)(x+1)=0$ since the expression is symmetric, order doesnt matter. we have $(x,y,z)=(2,1,-1)$. put this in to get $x^{2015}+y^{2015}+z^{2015}=\boxed{2^{2015}}$ this is the only value possible. are we suppose to expand $2^{2015}$?

- 4 years, 5 months ago

Do you think we do? Of course not! Nice solution. 7 out of 7.

- 4 years, 5 months ago

Is the answer to the second question is $f(x)=0$

- 4 years, 5 months ago

Is the answer to the first question

$\large{a \in \left(\frac{1}{24} (27-\sqrt{3}),\frac{1}{24}(27+\sqrt{3}) \right)}$

- 4 years, 5 months ago

- 4 years, 5 months ago

As my answers are wrong , proof would be useless :(

- 4 years, 5 months ago

Although for the first question, expand the equation and we get a quartic polynomial.

For roots to be real and distinct its derivative must have three distinct roots

$4x^3+6x^2+2x+3(a-1)$.....$(1)$ And for that to happen , we again take its derivative, we get

$12x^2+12x+2=0$

For real and distinct roots

$f(x_{1}) f(x_{2})<0$ where $x_{1,2}=\frac{-3 \pm \sqrt{3}}{4}$ in $(1)$

and then i found interval for $a$

- 4 years, 5 months ago

Now plz tell me if this is correct?

$\large{a \in \left(\frac{1}{432} (153-29 \sqrt{57}),\frac{1}{432}(153+29 \sqrt{57}) \right)}$

- 4 years, 5 months ago

- 4 years, 5 months ago

Putting $y=0$ we get $f(xf(0)+x)=f(x))$ $xf(0)+x=x$ $\boxed{f(0)=0}.........(1)$ Now we assume that there exists $f(y)=-1$ Then putting it in original functional equation we get, $f(0)=xy+f(x)$ $-xy=f(x)........(2)$ now put $x=y$ $f(y)=-y^{2}$ But $f(y)=-1$ Therefore $y=1$ or $y=-1$. Now putting these values in $(2)$ we get $f(x)=x$ or $f(x)=-x$ . since above two functions satisfy the given functional equation and $(1)$ Hence they are the solutions of the given functional equation.

- 4 years, 5 months ago

You would get 3 out of 7 for this proof. In line 2, how do you know $f$ is injective?

- 4 years, 5 months ago

Answer is right I guess (function)

- 4 years, 5 months ago

Yeah it is. But you have done some assumptions.

- 4 years, 5 months ago

Proving that the function is injective is very easy. For two reals a and b let f(a)=f(b). We take two cases. In the first case put y=a and in second case put y=b. Now note that in both cases the LHS is the same. Therefore comparing the RHS you obtain that a=b. Hence function is injective

- 4 years, 5 months ago

Using $AM-GM$ $a^{2}+b^{2}\geq2ab$ $\frac{3(a^{2}+b^{2})}{2}\geq(a^{2}+ab+b^{2})$ This implies $\frac{a^{2}(a^{3}+b^{3})}{a^{2}+ab+b^{2}}\geq\frac{2a^{2}(a^{3}+b^{3})}{3(a^{2}+b^{2})}$ Using Generalized Power mean inequality We get $\frac{a^{2}(a^{3}+b^{3})}{a^{2}+ab+b^{2}}\geq\frac{2a^{2}(\sqrt{\frac{a^{2}+b^{2}}{2}})}{3}$ Now by $RMS-AM$ inequality we get $\frac{a^{2}(a^{3}+b^{3})}{a^{2}+ab+b^{2}}\geq\frac{a^{2}(a+b)}{3}\geq\frac{2a^{2}\sqrt{ab}}{3}$ Therefore $\sum_{cyc}^{a,b,c}\frac{a^{2}(a^{3}+b^{3})}{a^{2}+ab+b^{2}}\geq\sum_{cyc}^{a,b,c}\frac{2a^{2}\sqrt{ab}}{3}$ Now using simple $AM-GM$ on the RHS of the above inequality we get $\sum_{cyc}^{a,b,c}\frac{a^{2}(a^{3}+b^{3})}{a^{2}+ab+b^{2}}\geq\sum_{cyc}^{a,b,c}\frac{2a^{2}\sqrt{ab}}{3}\geq2abc$

$\sum_{cyc}^{a,b,c}\frac{a^{2}(a^{3}+b^{3})}{a^{2}+ab+b^{2}}\geq2abc$ Hence proved .

- 4 years, 5 months ago

Could you provide the solutions to all questions please? So that we can reference and improve. Thanks!

- 4 years, 5 months ago

@Sharky Kesa : For question 1 my solution is as follows : The given equation simplifies to x^4+3.x^3+x^2+(3-3a)x +(a-a^2)=0. Let the roots be p,q,r,s. Now, sum of roots two at a time=1, sum of roots three at a time =(3a-3) and all four roots at a time=(a-a^2)

Now we use the identity : (pqr+pqs+prs+qrs)^2=(p^2.q^2.r^2 + .........) + 2pqrs(pq+pr+ps+qr+qs+rs) Thus we obtain the inequality : (pqr+pqs+prs+qrs)^2 >= 3pqrs(pq+pr+ps+qr+qs+rs) Substituting the values we obtain a condition : 4a^2-7a+3 >=0. Solving that we obtain that a should lie outside [3/4,1].

Please tell me whether the answer and the solution is correct or wrong.

- 4 years, 5 months ago