Maths Olympiad Prep

Track / Stage 5 / 101 of 400 #701 of 1964

Problem 701

AIME late
Algebra Difficulty 5.1 Find the answer

(2004 National High School Mathematics League Sichuan Preliminary) Given the inequality m2+(cos⁡2θ−5)m+4sin⁡2θ⩾0m^{2}+\left(\cos ^{2} \theta-5\right) m+4 \sin ^{2} \theta \geqslant 0 always holds, find the range of the real number mm.

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

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Official solution

1. The original expression is equivalent to m2+mcos⁡2θ−5m+4−4cos⁡2θ⩾0m^{2}+m \cos ^{2} \theta-5 m+4-4 \cos ^{2} \theta \geqslant 0, which simplifies to (m−4)cos⁡2θ+m2−5m+4⩾0(m-4) \cos ^{2} \theta+m^{2}-5 m+4 \geqslant 0. Let cos⁡2θ=\cos ^{2} \theta= t∈[0,1],f(t)=(m−4)t+m2−5m+4⩾0t \in[0,1], f(t)=(m-4) t+m^{2}-5 m+4 \geqslant 0, then {f(0)=m2−5m+4⩾0f(1)=m2−4m⩾0⇒m⩾4\left\{\begin{array}{l}f(0)=m^{2}-5 m+4 \geqslant 0 \\ f(1)=m^{2}-4 m \geqslant 0\end{array} \Rightarrow m \geqslant 4\right. or m⩽0m \leqslant 0.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.