曲线方程为 $y = -2x^2 - kx - 6 + k$，其中 $k$ 为常数。 求使得该曲线完全位于 x 轴下方的 $k$ 的取值范围。 已知 $2x^2 + bx + c$ 恒为正，写出并化简 $b$ 与 $c$ 之间的关系。解释为什么 $c$ 绝不为负数。

1. State the condition for the curve to lie completely below the x-axis

For the curve $y = -2x^2 - kx - 6 + k$ to lie completely below the x-axis, $y < 0$ for all real $x$. Since the coefficient of $x^2$ is $-2 < 0$, the curve opens downwards. The condition is that the discriminant must be less than zero: $b^2 - 4ac < 0$.

2. Set up the discriminant inequality

$(-k)^2 - 4(-2)(-6 + k) < 0$

3. Expand and simplify the inequality

$k^2 + 8(-6 + k) < 0$
$k^2 + 8k - 48 < 0$

4. Solve the quadratic inequality

Factorising the quadratic expression:
$(k + 12)(k - 4) < 0$

The critical values are $k = -12$ and $k = 4$.
Since the inequality is $< 0$, the solution is the region between the roots:
$-12 < k < 4$

1. Apply the discriminant condition for a quadratic expression to be always positive

For $2x^2 + bx + c$ to be always positive, the coefficient of $x^2$ is positive ($2 > 0$) and the discriminant must be less than zero.
$b^2 - 4(2)(c) < 0$

2. Simplify the inequality to find the relationship between $b$ and $c$

$b^2 - 8c < 0$
$b^2 < 8c$

3. Explain why $c$ is never negative

Since the square of any real number is non-negative, $b^2 \geqslant 0$.
Given $8c > b^2$, it follows that $8c > 0$, which means $c > 0$.
Therefore, $c$ is strictly positive and can never be negative.
(Alternatively, substituting $x = 0$ into the expression gives $2(0)^2 + b(0) + c = c$. Since the expression is always positive, $c > 0$.)