The examination

We have $2(x-2)^2=2x^2-8x+8\geq 0$ , with equality for $x=2$ . Thus $3x^2-6x+9\geq x^2+2x+1=(x+1)^2$ and $\sqrt{3x^2-6x+9}\geq x+1$ . The quantity to be minimised is $(x^2-4)^2-x+\sqrt{3x^2-6x+9}+2003\geq (x^2-4)^2+2004$ , with the minimum of $\boxed{2004}$ being attained at $x=2$ .

If we define $g(x) \; = \; \frac{3x}{\sqrt{3x^2 + 6}}$ then $g(1) = 1$ and $g'(x) \; = \; \frac{18}{(3x^2 + 6)^{\frac32}} \; > \; 0$ and hence we deduce that $g(x) < 1$ for $x < 1$ , while $g(x) > 1$ for $x > 1$ . We now note that, if we define $f(x) \; = \; x^4 - 8x^2 - x + \sqrt{3x^2 - 6x + 9} + 2019$ then $f(-x) > f(x)$ for all $x > 0$ , so $f$ is certainly minimized at nonnegative $x$ . But $f'(x) \; = \; 4x^3 - 16x - 1 + \frac{3(x-1)}{\sqrt{3x^2 - 6x + 9}} \; = \; 4x(x^2-4) - 1 + g(x-1)$ so $f'(2) = 0$ and we deduce that $f'(x) < 0$ for $0 < x < 2$ , while $f'(x) > 0$ for $x>2$ . Thus the minimum value of $f(x)$ occurs at $x=2$ , and is therefore $\boxed{2004}$ .

One interesting note, not explanation, is that the answer to this problem is when I, and most of the students who took this test, were born, in 2004.

I noticed your request that calculus not be used; unfortunately, that is the only method I know.

$f=x^4-8 x^2+\sqrt{3 x^2-6 x+9}-x+2019$

$\frac{\partial f}{\partial x}=4 x^3+\frac{6 x-6}{2 \sqrt{3 x^2-6 x+9}}-16 x-1$

$\text{Solve}\left[\frac{\partial f}{\partial x}=0\right] \Longrightarrow \{2.,-1.91467,-0.130657,1.00097\, -1.41505 i,1.00097\, +1.41505 i\}$

$\frac{\partial ^2f}{\partial x^2} @ \{2.,-1.91467,-0.130657,1.00097\, -1.41505 i,1.00097\, +1.41505 i\} \Longrightarrow \\ \{32.6667,28.0934,-15.2116,-15328.-4449.75 i,-15328.+4449.75 i\}$

As only the first two have positive real valuses, they are the local minimums.

$f @ \{2.,-1.91467,-0.130657,1.00097\, -1.41505 i,1.00097\, +1.41505 i\} \Longrightarrow \{2004.,2010.64,2022.13,2019.02\, +29.6511 i,2019.02\, -29.6511 i\}$

Of the first two (the local minimums), the first at $x=2$ where $f(2)=2004$ is the smallest and is therefore the global minimum.