Toán Xét tính chẵn lẻ: 1. f(x)= tanx.sin^2x+sinx 2. f(x)=tan^2x+cosx 05/10/2021 By Adeline Xét tính chẵn lẻ: 1. f(x)= tanx.sin^2x+sinx 2. f(x)=tan^2x+cosx
Đáp án: Ta có: \(\begin{array}{l}1)\,f\left( x \right) = \tan \,x.\sin {\,^2}2x + \sin \,x\\ = \\f\left( { – x} \right) = \tan \left( { – x} \right).{\sin ^2}\left( { – 2x} \right) + \sin \left( { – x} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = – \tan x.{\sin ^2}\left( {2x} \right) – \sin \left( x \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = – \left( {\tan \,x.\sin {\,^2}2x + \sin \,x} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\, = – f\left( x \right)\end{array}\) \( \Rightarrow f\left( x \right)\) là hàm số lẻ. \(\begin{array}{l}2.f(x) = ta{n^2}x + cosx\\ \Rightarrow f\left( { – x} \right) = {\tan ^2}\left( { – x} \right) + cos\,\left( { – x} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \tan {\,^2}x + cos\,x\, = f\left( x \right)\end{array}\) \( \Rightarrow f\left( x \right)\) là hàm số chẵn. Giải thích các bước giải: Trả lời
Đáp án:
Đáp án:
Ta có:
\(\begin{array}{l}1)\,f\left( x \right) = \tan \,x.\sin {\,^2}2x + \sin \,x\\ = \\f\left( { – x} \right) = \tan \left( { – x} \right).{\sin ^2}\left( { – 2x} \right) + \sin \left( { – x} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = – \tan x.{\sin ^2}\left( {2x} \right) – \sin \left( x \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = – \left( {\tan \,x.\sin {\,^2}2x + \sin \,x} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\, = – f\left( x \right)\end{array}\)
\( \Rightarrow f\left( x \right)\) là hàm số lẻ.
\(\begin{array}{l}2.f(x) = ta{n^2}x + cosx\\ \Rightarrow f\left( { – x} \right) = {\tan ^2}\left( { – x} \right) + cos\,\left( { – x} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \tan {\,^2}x + cos\,x\, = f\left( x \right)\end{array}\)
\( \Rightarrow f\left( x \right)\) là hàm số chẵn.
Giải thích các bước giải: