Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 3.0 Junior Find the answer

Given proposition p: The graph of the equation $\frac {x^{2}}{4-m}+ \frac {y^{2}}{m}=1$ is an ellipse with foci on the x-axis; proposition q: "For all $x \in \mathbb{R}, x^2+2mx+1>0$ "; proposition S: "There exists $x \in \mathbb{R}, mx^2+2mx+2-m=0$ ."
(1) If proposition S is true, find the range of the real number $m$ ;
(2) If $p \lor q$ is true, and $\lnot q$ is true, find the range of the real number $m$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Solution: (1) Since proposition S is true,
When $m=0$ , $2=0$ , which is not possible,
When $m \neq 0$ , $\Delta = (2m)^2-4m(2-m) \geq 0$ ,
Therefore, $\text{m0 m>0 4-m>0}$ $, solving this gives$ 0<m<2$,
If q is true, it implies $(2m)^2-4<0$ which gives $-1<m<1$ ,
Since $p \lor q$ is true, and $\lnot q$ is true,
Therefore, p is true and q is false,
Thus, $\begin{cases} \overset{0<m<2}{m\leq -1\text{ or }m\geq 1}\end{cases}$ , solving this gives $\boxed{1 \leq m < 2}$ .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.