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# Australian School of Excellence 2015 Algebra Exam

• Each question is worth 7 points

• Time allowed is 4 hours

• No books, notes or calculators permitted

• Write full proofs with your answers

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

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
11 months, 2 weeks 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: $\begin{array}a p_1=s_1=2\\ p_2=s_1p_1-2s_2=4-2s_2=6\Longrightarrow s_2=-1\\ p_3=s_1p_2-s_2p_1+3s_3=14+3s_3=8\Longrightarrow s_3=-2\end{array}$ 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 $$2^{2015}$$? · 11 months, 2 weeks ago

Do you think we do? Of course not! Nice solution. 7 out of 7. · 11 months, 2 weeks ago

Is the answer to the second question is $$f(x)=0$$ · 11 months, 2 weeks ago

No. Also, please give proof. · 11 months, 2 weeks ago

Is the answer to the first question

$$\large{a \in \left(\frac{1}{24} (27-\sqrt{3}),\frac{1}{24}(27+\sqrt{3}) \right)}$$ · 11 months, 2 weeks ago

No. Also, please give proof with your answers. · 11 months, 2 weeks ago

Now plz tell me if this is correct?

$$\large{a \in \left(\frac{1}{432} (153-29 \sqrt{57}),\frac{1}{432}(153+29 \sqrt{57}) \right)}$$ · 11 months, 2 weeks 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 $$a$$ · 11 months, 2 weeks ago

As my answers are wrong , proof would be useless :( · 11 months, 2 weeks 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. · 11 months, 1 week ago

Could you provide the solutions to all questions please? So that we can reference and improve. Thanks! · 11 months, 2 weeks 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 . · 11 months, 2 weeks 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. · 11 months, 2 weeks ago

You would get 3 out of 7 for this proof. In line 2, how do you know $$f$$ is injective? · 11 months, 2 weeks 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 · 11 months, 1 week ago

Answer is right I guess (function) · 11 months, 2 weeks ago