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Algebra Difficulty 2.8 Junior Find the answer

Given propositions P: "If b2=acb^2=ac (a,b,cRa, b, c \in \mathbb{R}), then a,b,ca, b, c form a geometric sequence", and Q: "The function f(x)=cos(π2+x)f(x) = \cos\left(\frac{\pi}{2} + x\right) is an odd function", then the true proposition among the following is

Pick one

Solution

For proposition P: If b2=acb^2=ac,
we can assume a=b=c=0a=b=c=0,
which obviously satisfies the condition, but they do not form a geometric sequence,
thus, this proposition is false.
For proposition Q: "The function f(x)=cos(π2+x)=sinxf(x) = \cos\left(\frac{\pi}{2} + x\right) = -\sin x is an odd function",
thus, proposition Q is true.
Therefore, pqp \lor q is a true proposition,
hence, the correct choice is: A\boxed{A}.
By determining the truth of propositions P and Q, we can judge the truth of the compound proposition.
This question examines the judgment of compound propositions, trigonometric functions, and sequence problems, making it a basic question.

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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.