MathDB

1

Part of 1986 National High School Mathematics League

Problems(2)

Anti-trigonometric Function Problem

Source: 1986 National High School Mathematics League, Exam One, Problem 1

2/24/2020
Let −1<a<0-1<a<0, θ=arcsin⁡a\theta=\arcsin a. Then the solution set to the inequality sin⁡x<a\sin x<a is (A){x∣2nπ+θ<x<(2n+1)π−θ,n∈Z}\text{(A)}\{x|2n\pi+\theta<x<(2n+1)\pi-\theta,n\in\mathbb{Z}\} (B){x∣2nπ−θ<x<(2n+1)π+θ,n∈Z}\text{(B)}\{x|2n\pi-\theta<x<(2n+1)\pi+\theta,n\in\mathbb{Z}\} (C){x∣(2n−1)π+θ<x<2nπ−θ,n∈Z}\text{(C)}\{x|(2n-1)\pi+\theta<x<2n\pi-\theta,n\in\mathbb{Z}\} (D){x∣(2n−1)π−θ<x<2nπ+θ,n∈Z}\text{(D)}\{x|(2n-1)\pi-\theta<x<2n\pi+\theta,n\in\mathbb{Z}\}
functioninequalities
Linear Polynomial

Source: 1986 National High School Mathematics League, Exam Two, Problem 1

2/24/2020
For real numbers a0,a1,⋯ ,an(a0≠a1)a_0,a_1,\cdots,a_n(a_0\neq a_1), we haveai−1+ai+1=2aia_{i-1}+a_{i+1}=2a_i for i=1,2,⋯ ,n−1i=1,2,\cdots,n-1. Prove that P(x)=a0Cn0(1−x)n+a1Cn1x(1−x)n−1+⋯+anCnnxnP(x)=a_0\text{C}_n^0(1-x)^n+a_1\text{C}_n^1x(1-x)^{n-1}+\cdots+a_n\text{C}_n^nx^n is a linear polynomial.
algebrapolynomial