Math Problem Statement

17. (15) 已知函数f(x)=sin2x+sinx+cos2x.17\text{. (}15分)\text{ 已知函数}f\left(x\right)=\sin^2x+\sin x+\cos^2x\text{.} (1)(6)求函数f(x)的对称中心.\text{(}1)\left(6分\right)求函数f\left(x\right)的对称中心\text{.} (2)(9)若函数f(x)在区间[0, nπ2]上恰好有2024个零点, 求实数n的取值范围.\left(2\right)\,\left(9分\right)\,若函数f\left(x\right)在区间\left\lbrack0,\text{ }\frac{n\pi}{2}\right\rbrack上恰好有2024个零点\text{, 求实数}n的取值范围\text{.}

Solution

Let's solve the problem step by step.

Part (1): Finding the Center of Symmetry of f(x)f(x)

Given: f(x)=sin2x+sinx+cos2xf(x) = \sin^2 x + \sin x + \cos^2 x

First, simplify the function f(x)f(x): cos2x+sin2x=1\cos^2 x + \sin^2 x = 1

Thus, f(x)=1+sinxf(x) = 1 + \sin x

A function g(x)g(x) has a center of symmetry at x=ax = a if g(a+x)=g(ax)g(a + x) = g(a - x).

For f(x)=1+sinxf(x) = 1 + \sin x, we need to check if it has a center of symmetry.

Let aa be the center of symmetry. Then: 1+sin(a+x)=1+sin(ax)1 + \sin(a + x) = 1 + \sin(a - x)

Since sin\sin is an odd function: sin(a+x)=sin(ax)\sin(a + x) = \sin(a - x)

For this equality to hold true, aa must be a point where sin\sin is symmetrical, i.e., sin(a)=0\sin(a) = 0.

Thus, the center of symmetry aa is: a=kπ,where kZa = k\pi, \, \text{where } k \in \mathbb{Z}

Therefore, the function f(x)=1+sinxf(x) = 1 + \sin x has centers of symmetry at: x=kπ,where kZx = k\pi, \, \text{where } k \in \mathbb{Z}

Part (2): Finding the Range of nn

Given: f(x)=1+sinxf(x) = 1 + \sin x

We need to find the number of zeros of f(x)f(x) in the interval [0,nπ2]\left[ 0, \frac{n\pi}{2} \right].

The zeros of f(x)=1+sinxf(x) = 1 + \sin x occur when: 1+sinx=01 + \sin x = 0 sinx=1\sin x = -1

The sine function sinx=1\sin x = -1 at: x=3π2+2kπ,where kZx = \frac{3\pi}{2} + 2k\pi, \, \text{where } k \in \mathbb{Z}

We need 2024 zeros in the interval: [0,nπ2]\left[ 0, \frac{n\pi}{2} \right]

Let’s count the number of such zeros: 3π2+2kπnπ2\frac{3\pi}{2} + 2k\pi \leq \frac{n\pi}{2} 3+4kn3 + 4k \leq n 4kn34k \leq n - 3 kn34k \leq \frac{n - 3}{4}

Given there are 2024 zeros: k+1=2024k + 1 = 2024 k=2023k = 2023

Thus: n342023\frac{n - 3}{4} \geq 2023 n38092n - 3 \geq 8092 n8095n \geq 8095

Therefore, the value of nn must be: n8095n \geq 8095

Summary

  1. The centers of symmetry of the function f(x)=1+sinxf(x) = 1 + \sin x are x=kπx = k\pi, where kZk \in \mathbb{Z}.
  2. The value of nn such that the function f(x)f(x) has exactly 2024 zeros in the interval [0,nπ2]\left[ 0, \frac{n\pi}{2} \right] is: n8095n \geq 8095

Do you want further details or have any questions?

Here are some related questions you might want to consider:

  1. How do you determine the zeros of a trigonometric function?
  2. What are the general properties of sine and cosine functions?
  3. How do you solve trigonometric equations within a given interval?
  4. What is the significance of symmetry in trigonometric functions?
  5. How can you determine the period of a sine or cosine function?

Tip: When working with trigonometric functions, always remember their periodic nature and key values within one period. This can greatly simplify solving equations involving these functions.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Function Analysis

Formulas

f(x) = sin^2(x) + sin(x) + cos^2(x)
sin(x) = -1

Theorems

Properties of sine and cosine functions
Center of symmetry

Suitable Grade Level

Grades 10-12