As curvas ${C_1}$ e ${C_2}$ da figura s\u00e3o as representa\u00e7\u00f5es gr\u00e1ficas das fun\u00e7\u00f5es $f$ e $g$ definidas, em $\\left[ {0,2\\pi } \\right]$, respetivamente, por:<\/p>\n
$$\\begin{array}{*{20}{c}}
\n{f(x) = \\operatorname{sen} x}&{}&{\\text{e}}&{}&{g(x) = \\operatorname{sen} 2x}
\n\\end{array}$$<\/p>\n
b) $f(x) + g(x) \\geqslant 0$<\/p>\n c) $f(x) \\times g(x) < 0$<\/li>\n b) c)![\"\"](\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/04\/12pag130-14.jpg\" "\\"Gr\u00e1fico\\"")
<\/a><\/p>\n\n
\n
\n{f(x) = g(x)}& \\Leftrightarrow &{\\begin{array}{*{20}{l}}
\n{\\operatorname{sen} x = \\operatorname{sen} 2x}& \\wedge &{0 \\leqslant x \\leqslant 2\\pi }
\n\\end{array}} \\\\
\n{}& \\Leftrightarrow &{\\begin{array}{*{20}{l}}
\n{\\begin{array}{*{20}{l}}
\n{2x = x + 2k\\pi }& \\vee &{2x = \\pi\u00a0 – x + 2k\\pi ,k \\in \\mathbb{Z}}
\n\\end{array}}& \\wedge &{0 \\leqslant x \\leqslant 2\\pi }
\n\\end{array}} \\\\
\n{}& \\Leftrightarrow &{\\begin{array}{*{20}{l}}
\n{\\begin{array}{*{20}{l}}
\n{x = 2k\\pi }& \\vee &{x = \\frac{\\pi }{3} + \\frac{{2k\\pi }}{3},k \\in \\mathbb{Z}}
\n\\end{array}}& \\wedge &{0 \\leqslant x \\leqslant 2\\pi }
\n\\end{array}} \\\\
\n{}& \\Leftrightarrow &{x \\in \\left\\{ {0,\\frac{\\pi }{3},\\pi ,\\frac{{5\\pi }}{3},2\\pi } \\right\\}}
\n\\end{array}$$ as coordenadas dos pontos de intersec\u00e7\u00e3o das duas curvas s\u00e3o:
\n$$\\begin{array}{*{20}{l}}
\n{O\\left( {0,0} \\right)}&;&{A\\left( {\\frac{\\pi }{3},\\operatorname{sen} \\frac{\\pi }{3}} \\right) = \\left( {\\frac{\\pi }{3},\\frac{{\\sqrt 3 }}{2}} \\right)}&;&{B\\left( {\\pi ,0} \\right)}&;&{C\\left( {\\frac{{5\\pi }}{3}, – \\frac{{\\sqrt 3 }}{2}} \\right)}&{\\text{e}}&{D\\left( {2\\pi ,0} \\right)}
\n\\end{array}$$
\n<\/a>\u00ad<\/li>\n
\n$$\\begin{array}{*{20}{l}}
\n{f(x) – g(x) \\geqslant 0}& \\Leftrightarrow &{x \\in \\left[ {\\frac{\\pi }{3},\\pi } \\right] \\cup \\left[ {\\frac{{5\\pi }}{3},2\\pi } \\right]}
\n\\end{array}$$<\/p>\n
\n$$\\begin{array}{*{20}{l}}
\n{f(x) + g(x) \\geqslant 0}& \\Leftrightarrow &{f(x) \\geqslant\u00a0 – g(x)} \\\\
\n{}& \\Leftrightarrow &{x \\in \\left[ {0,\\pi\u00a0 – \\frac{\\pi }{3}} \\right] \\cup \\left[ {\\pi ,\\pi\u00a0 + \\frac{\\pi }{3}} \\right] \\cup \\left\\{ {2\\pi } \\right\\}} \\\\
\n{}& \\Leftrightarrow &{x \\in \\left[ {0,\\frac{{2\\pi }}{3}} \\right] \\cup \\left[ {\\pi ,\\frac{{4\\pi }}{3}} \\right] \\cup \\left\\{ {2\\pi } \\right\\}}
\n\\end{array}$$<\/p>\n
\n$$\\begin{array}{*{20}{l}}
\n{f(x) \\times g(x) < 0}& \\Leftrightarrow &{x \\in \\left] {\\frac{\\pi }{2},\\pi } \\right[ \\cup \\left] {\\pi ,\\frac{{3\\pi }}{2}} \\right[}
\n\\end{array}$$\u00ad<\/li>\n
\n{g(\\frac{\\pi }{6}) = \\operatorname{sen} \\left( {\\frac{{2\\pi }}{3}} \\right) = \\frac{{\\sqrt 3 }}{2}}&;&{g(\\frac{\\pi }{4}) = \\operatorname{sen} \\left( {\\frac{\\pi }{2}} \\right) = 1}&{\\text{e}}&{g(\\frac{{2\\pi }}{3}) = \\operatorname{sen} \\left( {\\frac{{4\\pi }}{3}} \\right) =\u00a0 – \\frac{{\\sqrt 3 }}{2}}
\n\\end{array}$$ conclui-se que o contradom\u00ednio\u00a0da restri\u00e7\u00e3o da fun\u00e7\u00e3o $g$ ao intervalo $\\left] {\\frac{\\pi }{6},\\frac{{2\\pi }}{3}} \\right]$ \u00e9 $D’ = \\left[ { – \\frac{{\\sqrt 3 }}{2},1} \\right]$.<\/li>\n<\/ol>\n