So I am stuck on a problem of a quadratic module.
Let $x^2 + 2y^2 – 2xy – 2 \ge k (x + 2y) \;\forall\; x, y \in \mathbb{R}$ then find the number of integral values of k.
Since it it's the part of quadratics module and the calculus part is introduced in latter modules.
So I reject any possibility of derivatives coming in. Moreover only concept related to quadratic equation in two variables in the module is the condition of resolving general quadratic equation in x and y into two linear factors.
My attempt:
Converting it into general form: $$f(x,y)=ax^2+2hxy+by^2+2gx+2fy+c$$
$$(1)x^2+2(-1)xy+(2)y^2+2(\frac{-k}{2})x+2(-k)y+(-2)\ge0$$
$$a=1 , h=-1 , b=2 , g=-\frac{k}{2} , f=-k , c=-2$$
Condition for linear factors: $$abc+2fgh-af^2-bg^2-ch^2=0$$
On Substituting: $$k^2=-\frac{4}{5}$$
which is not true for any real number
$\implies$no linear factors
Now a quadratic in x: $$x^2-(2y+k)x+2y^2-2ky-2\ge0$$
We need to have coefficient of $x^2\gt0$ and Discriminant of given quadratic $D\lt0$
$$D=(4)y^2+(-4k)y+(9k^2-8)<0$$
$$(2y-k)^2+(8k^2-8)<0$$
$$k\in\phi$$
which means there is no value of k which can satisfy the relation.
Please confirm the solution as I have no answer key at the moment.
Pardon me for any mistakes.
Edit: The answer key says that there are "$0$" integral values of k satisfying the required condition.
Best Answer
I did not read your method but the conclusion is the same with mine: whatever the value of $k\in\mathbb R$, the difference $$x^2+2y^2–2xy–2-k(x+2y)=\left(x-y-\frac k2\right)^2+\left(y-\frac{3k}2\right)^2-\frac{4+5k^2}2$$ cannot be $\ge0$ for all real numbers $x,y$. Its minimum value is $-\frac{4+5k^2}2<0.$