一道很简单的关于函数的高中数学题
\u4e00\u9053\u5f88\u7b80\u5355\u7684\u5173\u4e8e\u51fd\u6570\u7684\u9ad8\u4e2d\u6570\u5b66\u9898sinx\u57280\u5230PI/2\u4e0a\u5355\u8c03\u589e\uff0cPI/2\u5230PI\u5355\u8c03\u51cf\uff0c\u628asinwx\u770b\u6210\u4e00\u4e2a\u6574\u4f53\uff0c\u4e5f\u5c31\u662f\u8bf4wx\u57280\u5230PI/2\u4e0a\u5355\u8c03\u589e\u53ef\u4ee5\u5199\u6210\uff1a
0<=wx<=PI/2;\u53ef\u4ee5\u63a8\u51fa0<=x<=PI/(2w);\u628a\u540e\u8fb9\u8fd9\u90e8\u5206\u4e0ePI/3\u76f8\u7b49\uff0c\u5c31\u53ef\u4ee5\u5f97\u51faw=3/2;
\u5982\u679c\u75283/2\u4ee3\u5165wx\uff0c\u53ef\u4ee5\u770b\u5230sinwx\u5728[PI/3,2PI/3]\u4e0a\u5355\u8c03\u51cf\uff0c\u5f53\u7136\uff0c\u5728[PI/3,PI/2]\u4e5f\u5355\u8c03\u51cf\uff0c\u56e0\u4e3a\u540e\u8005\u88ab\u5305\u542b\u5728\u524d\u8005\u8303\u56f4\u5185\u3002
\u4e0d\u77e5\u9053\u7ed3\u679c\u662f\u4e0d\u662f\u8fd9\u6837\uff0cw=3/2
\u91caPI\u5c31\u662f\u9898\u76ee\u4e2d\u7684"pai".
\u5f53\u7136\uff0c\u8fd9\u4e2a\u9898\u8fd8\u8981\u8003\u8651\u4e09\u89d2\u51fd\u6570\u6574\u6570K\u5728\u5185\uff0c\u53ea\u662f\u5982\u679cw\u662f\u4e00\u4e2a\u786e\u5b9a\u7684\u503c\uff0c\u5e94\u8be5\u5c31\u7528\u8fd9\u79cd\u7b80\u5355\u7684\u65b9\u6cd5\u6765\u89e3\u3002\u518d\u6709\uff0c\u5982\u679c\u7b97\u4e0b\u6765\u7684w\u4ee3\u5165\u539f\u5f0f\u53d1\u73b0\u540e\u9762\u51cf\u533a\u95f4\u5c0f\u4e8ePI/2\uff0c\u90a3\u5c31\u4e0d\u7528\u524d\u9762\u7684\u533a\u95f4\u6c42\u503c\uff0c\u53cd\u8fc7\u6765\u7528\u540e\u9762\u7684\u533a\u95f4\u6c42\u503c\u3002
\u5f00\u53e3\u5411\u4e0a
\u96f6\u70b9\u5728x=1\u4e24\u4fa7
\u6240\u4ee5f(1)<0
1+a²-1+a-2<0
(a+2)(a-1)<0
-2<a<1
把函数图画出来可以知道是一个方波，在正负sin(1)上波动，f(x)<1/2即为方波为负的时候，此时x取值为(2k-1)*2*pi/3<x<(4*pi/3)*k,k为整数
绛旓細绗竴棰樸俧(x1)=x1^2/(2x1+1)锛宖(x2)=x2^2/(2x2+1)浠ｅ叆f(x1)*f(x2)=1锛屽緱[x1^2/(2x1+1)]*[x2^2/(2x2+1)]=1 鍗 x1^2*x2^2=(2x1+1)*(2x2+1)(x1*x2)^2=4(x1*x2)+2(x1+x2)+1 (x1*x2)^2-2(x1*x2)+1=2(x1*x2)+2(x1+x2)+2 (x1*x2-...
绛旓細f(x)=sin(2蟺-x)sin(1.5蟺-x)-鈭3cos²x+鈭3 f(x)=-sinx路-cosx-½鈭3(2cos²x-1)+½鈭3 =½sin2x-½鈭3cos2x+½鈭3 =sin(2x-⅓蟺)+½鈭3 灏2x-⅓蟺鐪嬫垚鏁翠綋:鍗曡皟閫掑鍖洪棿2x-⅓蟺鈭(2k蟺-½蟺,2k...
绛旓細绛旓細x<=1锛宖(x)=2^x x>1锛宖(x)=x^2-2x+2=(x-1)^2+1锛屽紑鍙ｅ悜涓婏紝瀵圭О杞磝=1 鍒嗘鍑芥暟鐨勫浘鍍忚涓嬪浘鎵绀猴紙鍙互缁樺埗绠鍗曠殑鍥惧儚锛塯(x)=f(x)-m=0 f(x)=m瀛樺湪涓や釜涓嶅悓鐨勮В x=1鏃讹細f(1)=2^1=2 x=1+鏃讹紝f(1+)=1^2-2+2=1 鎵浠ワ細1<m<=2 ...
绛旓細3/4蟺•w=k蟺+蟺/2, k鈭圸.w=4K/3+2/3, k鈭圸.鍑芥暟f(x)=cos wx(w>0)鍦ㄥ尯闂淬0锛屜/2銆戜笂鏄崟璋冨嚱鏁,X鈭堛0锛屜/2銆, wx鈭堛0锛寃蟺/2銆.鍘熺偣鍙充晶鍚湁0鐨勪綑寮﹀嚱鏁板崟璋冨尯闂撮暱搴︽渶澶ф椂锛屽尯闂存槸銆0锛屜銆戙傚洜涓哄嚱鏁癴(x)=cos wx(w>0)鍦ㄥ尯闂淬0锛屜/2銆戜笂鏄崟璋冨嚱鏁,...
绛旓細姣斿0 鍟 卤1鍟 卤2鍟婁粈涔堢殑 杩欎釜棰樻秹鍙婂埌0.5鐨勫嶆暟 鎵浠ラ渶甯﹀叆0.5鍊嶆暟 甯﹀叆3/2 锛屽緱3 f锛5/2 锛=5f锛3/2锛夊甫鍏1/2锛屽緱f锛3/2锛=3f锛1/2锛夊甫鍏-1/2,寰 -f锛1/2锛=f锛-1/2锛夎屾鍑芥暟鏄伓鍑芥暟锛屾湁f锛1/2锛= -f锛- 1/2)鎵浠锛1/2)=0 渚濇f锛5/2锛変篃灏=0...
绛旓細杩炵粨DO銆丒F,寤堕暱DO浜F浜庣偣G,鈭礑O鍨傜洿骞冲垎AB 鈭碊O鈯F,鈭燛OG=蟺/2-2x 鈭碘垹EDF鏄垹EOF(鍦嗗績瑙)瀵瑰簲鐨勫渾鍛ㄨ 鈭粹垹EDG=1/2鈭燛OG=蟺/4-x 杩囩偣O浣淥H鈯E 鈭礝E=OD,鈻矱OD涓虹瓑鑵颁笁瑙掑舰 鈭碊H=1/2DE=OD脳cosEDG=cos(蟺/4-x)=鈭2/2(cosx+sinx)DE=鈭2(cosx+sinx)=DF y=4鈭2(...
绛旓細瑙 [1]f(x)鏄鍑芥暟锛屽垯瀹氫箟鍩鍏充簬鍘熺偣瀵圭О锛屽垯a-1+2a=0锛宎=1/3 f(x)=(x^3/3)+bx+b+1 -f(x)=(-x^3/3)-bx-b-1 f(-x)=(-x^3/3)-bx+b+1 f(-x)=-f(x)-bx-b-1=-bx+b+1 b=-1 a=1/3 b=-1 [2]f(x1*x2)=f(x1)+f(x2)浠1=x2=1 f(1)=2f(...
绛旓細濡傛灉鐢3/2浠ｅ叆wx锛屽彲浠ョ湅鍒皊inwx鍦╗PI/3,2PI/3]涓婂崟璋冨噺锛屽綋鐒讹紝鍦╗PI/3,PI/2]涔熷崟璋冨噺锛屽洜涓哄悗鑰呰鍖呭惈鍦ㄥ墠鑰呰寖鍥村唴銆備笉鐭ラ亾缁撴灉鏄笉鏄繖鏍凤紝w=3/2 閲奝I灏辨槸棰樼洰涓殑"pai".褰撶劧锛岃繖涓杩樿鑰冭檻涓夎鍑芥暟鏁存暟K鍦ㄥ唴锛屽彧鏄鏋渨鏄竴涓‘瀹氱殑鍊硷紝搴旇灏辩敤杩欑绠鍗曠殑鏂规硶鏉ヨВ銆傚啀鏈夛紝濡傛灉绠...
绛旓細(1)鍥犱负0<(X-3)/(X+3)<1鏍规嵁鈥滃悓澧炲紓鍑忊濆師鍒,0<m<1鏃,澧炲嚱鏁,m>1鏃,鍑忓嚱鏁板悗涓闂,浣犵殑寮忓瓙鍐欑殑涓嶆竻妤 宸茶禐杩 宸茶俯杩< 浣犲杩欎釜鍥炵瓟鐨勮瘎浠锋槸? 璇勮 鏀惰捣 鍏朵粬绫讳技闂 2020-04-09 涓閬撻珮涓暟瀛﹀嚱鏁棰? 8 2020-04-06 涓閬撻珮涓暟瀛﹀嚱鏁伴? 8 2020-04-24 楂樹腑鏁板鍑芥暟鍥惧儚棰橀毦棰樿繖涓...
绛旓細1. f'(x) = 2x+a+(-1/x)=> 褰搙灞炰簬[1,2]鏃讹紝f'(x)鏄鍑芥暟 => f'(1)<=f'(x)<=f'(2)=> a+1 <= f'(x) <= a+ 7/2 鍑芥暟f(x)鍦╗1,2]涓婃槸鍑忓嚱鏁 => 褰搙灞炰簬[1,2]鏃,f'(x)<=0 => a+ 7/2 <=0 => a鈮-7/2 2. g(x)=f(x)-x^2 = ax-...
