{"id":2222,"date":"2024-06-19T14:56:41","date_gmt":"2024-06-19T05:56:41","guid":{"rendered":"https:\/\/y-page.y.kumamoto-nct.ac.jp\/wp\/Math_23\/?p=2222"},"modified":"2024-07-03T15:17:29","modified_gmt":"2024-07-03T06:17:29","slug":"%e6%95%b0%e5%ad%a6ii_2q%e6%a8%a1%e6%93%ac","status":"publish","type":"post","link":"https:\/\/y-page.y.kumamoto-nct.ac.jp\/wp\/Math_23\/?p=2222","title":{"rendered":"\u6570\u5b66II_2Q\u6a21\u64ec"},"content":{"rendered":"\n\n\n\n
\u518d\u751f\u30ea\u30b9\u30c8<\/span><\/a><\/td>\n<\/td>\n<\/td>\n<\/td>\n<\/td>\n<\/td>\n<\/tr>\n
4H1_\u524d\u534a<\/a><\/td>\n4H1_\u5f8c\u534a<\/a><\/td>\n2H1<\/a><\/td>\n4H2_\u524d\u534a<\/a><\/td>\n4H2_\u5f8c\u534a<\/a><\/td>\n2H2<\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
4H1_\u524d\u534a<\/span>4H1_\u5f8c\u534a<\/span>2H1<\/span>4H2_\u524d\u534a<\/span>4H2_\u5f8c\u534a<\/span>2H2<\/span><\/div>
\n

\u95a2\u6570\u306e\u5fae\u5206I<\/span><\/p>\n

1\uff0e \u6b21\u306e\u95a2\u6570\u3092\u5fae\u5206\u305b\u3088\u3002<\/p>\n

(1) $y=3x^2-5x+4$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u516c\u5f0f\u3000$(c)’=0$\u3000\u3001$(x)’=1$\u3001$(x^n)’=nx^{n-1}$\u304a\u3088\u3073\u5fae\u5206\u306e\u7dda\u5f62\u6027\u3092\u7528\u3044\u308b
\n$y’=6x-5$ (\u7b54)<\/span><\/div><\/div>\n

(2) $y=\\dfrac{1}{3x^6}$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$y=\\dfrac{1}{3}\\cdot\\dfrac{1}{x^6}$\u3000\u306b\u6ce8\u610f\u3057\u3066\u3001\u516c\u5f0f\u3000$\\dfrac{1}{x^n}=\\dfrac{-n}{x^{n+1}}$\u3000\u3092\u7528\u3044\u308b
\n$y’=\\dfrac{1}{3}\\cdot\\dfrac{-6}{x^7}=\\dfrac{-2}{x^7}$ (\u7b54)<\/span><\/div><\/div>\n

(3) $y=\\sqrt[3]{x^4}$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$y=x^{\\scriptsize\\dfrac{4}{3}}$\u3000\u306b\u6ce8\u610f\u3057\u3066\u3001\u516c\u5f0f\u3000$(x^a)=ax^{a-1}$\u3000\u3092\u7528\u3044\u308b
\n$y’=\\dfrac{4}{3}x^{\\scriptsize\\dfrac{1}{3}}=\\dfrac{4}{3}\\sqrt[3]{x}$ (\u7b54)<\/span><\/div><\/div>\n

(4) $y=\\dfrac{1}{\\sqrt[6]{x}}$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$y=x^{\\scriptsize\\dfrac{-1}{6}}$\u3000\u306b\u6ce8\u610f\u3057\u3066\u3001\u516c\u5f0f\u3000$(x^a)=ax^{a-1}$\u3000\u3092\u7528\u3044\u308b
\n$y’=\\dfrac{-1}{6}x^{\\scriptsize{\\dfrac{-7}{6}}}=\\dfrac{-1}{6\\sqrt[6]{x^7}}=\\dfrac{-1}{6x\\sqrt[6]{x}}$(\u7b54)<\/span><\/div><\/div>\n

(5) $y=(x^2-3x+5)^3$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u516c\u5f0f\u3000$(f^n)’=n\\cdot f’\\cdot f^{n-1}$\u3000\u3092\u7528\u3044\u308b
\n$y’=3(x^2-3x+5)'(x^2-3x+5)^2=3(2x-3)(x^2-3x+5)^2$ (\u7b54)<\/span><\/div><\/div>\n

(6) $y=\\dfrac{1}{(4x-7)^6}$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u516c\u5f0f\u3000$\\left( \\dfrac{1}{f^n} \\right)^{\\prime}= \\dfrac{-nf’}{f^{n+1}}$\u3000\u3092\u7528\u3044\u308b
\n$y’=\\dfrac{-6\\cdot(4x-7)’}{(4x-7)^7}=\\dfrac{-6\\cdot4}{(4x-7)^7}=\\dfrac{-24}{(4x-7)^7}$ (\u7b54)<\/span><\/div><\/div>\n

(7) $y=\\sqrt[3]{3x+5}$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$y=(3x+5)^{\\scriptsize\\dfrac{1}{3}}$\u3000\u306b\u6ce8\u610f\u3057\u3066\u3001\u516c\u5f0f\u3000$(f^a)’=a\\cdot f’\\cdot f^{a-1}$\u3000\u3092\u7528\u3044\u308b
\n$y’=\\dfrac{1}{3}(3x+5)'(3x+5)^{\\scriptsize{-\\dfrac{2}{3}}}=\\dfrac{1}{3}\\cdot 3 \\dfrac{1}{\\sqrt[3]{(3x+5)^2}}=\\dfrac{1}{\\sqrt[3]{(3x+5)^2}}$ (\u7b54)<\/span><\/div><\/div>\n

(8) $y=\\sqrt{1-x^2}$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u516c\u5f0f\u3000$(\\sqrt{f})’=\\dfrac{f’}{2\\sqrt{f}}$\u3000\u3092\u7528\u3044\u308b
\n$y’=\\dfrac{(1-x^2)’}{2\\sqrt{1-x^2}}=\\dfrac{-2x}{2\\sqrt{1-x^2}}=\\dfrac{-x}{\\sqrt{1-x^2}}$ (\u7b54)<\/span><\/div><\/div>\n

(9) $y=\\dfrac{3x-1}{x^2+1}$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u516c\u5f0f\u3000$\\left( \\dfrac{f}{g} \\right)^{\\prime}=\\dfrac{f’g-fg’}{g^2}$\u3000\u3092\u7528\u3044\u308b
\n$(\u5206\u5b50)=(3x-1)'(x^2+1)-(3x-1)(x^2+1)’=3(x^2+1)-2x(3x-1)=-3x^2+2x+3$
\n$y’=\\dfrac{-3x^2+2x+3}{(x^2+1)^2}$ (\u7b54)<\/span><\/div><\/div>\n

(10) $y=(2x+1)^4(3x-1)^5$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u516c\u5f0f\u3000$(f^ng^m)’=(nf’g+mfg’)f^{n-1}g^{m-1}$\u3000\u3092\u7528\u3044\u308b
\n$4(2x+1)'(3x-1)+5(2x+1)(3x-1)’=4\\cdot 2(3x-1)+5\\cdot3(2x+1)=54x+7$
\n$y’=(54x+7)(2x+1)^3(3x-1)^4$ (\u7b54)<\/span><\/div><\/div>\n

(11) $y=\\dfrac{(2x+1)^6}{(3x-1)^3}$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u516c\u5f0f\u3000$\\left(\\dfrac{f^n}{g^m}\\right)^{\\prime}=\\dfrac{(nf’g-mfg’)f^{n-1}}{g^{m+1}}$\u3000\u3092\u7528\u3044\u308b
\n$6(2x+1)'(3x-1)-3(2x+1)(3x-1)’=6\\cdot 2(3x-1)-3\\cdot3(2x+1)=18x-21$
\n$y’=\\dfrac{(18x-21)(2x+1)^5}{(3x-1)^4}=\\dfrac{3(6x-7)(2x+1)^5}{(3x-1)^4}$ (\u7b54)<\/span><\/div><\/div>\n

(12) $y=x^x\\;(x>0)$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u5bfe\u6570\u5fae\u5206\u306e\u516c\u5f0f\u3000$y’=y({\\mathrm{log}}y)’$\u3000\u3092\u7528\u3044\u308b
\n$({\\mathrm{log}}y)’=({\\mathrm{log}}x^x)’=(x{\\mathrm{log}}x)’=(x)'{\\mathrm{log}}x+x({\\mathrm{log}}x)’=1\\cdot{\\mathrm{log}}x+x\\cdot \\dfrac{1}{x}=1+{\\mathrm{log}}x$
\n$y’=x^x(1+{\\mathrm{log}}x)$ (\u7b54)<\/span><\/div><\/div>\n<\/div>\n
\n

\u95a2\u6570\u306e\u5fae\u5206II<\/span><\/p>\n

1\uff0e \u6b21\u306e\u95a2\u6570\u3092\u5fae\u5206\u305b\u3088\u3002<\/p>\n

(1) $y={\\mathrm{log}}(x^2+x-1)$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u516c\u5f0f\u3000$({\\mathrm{log}}f)’=({\\mathrm{log}}|f|)’=\\dfrac{f’}{f}$\u3092\u7528\u3044\u308b
\n$y’=\\dfrac{(x^2+x-1)’}{x^2+x-1}=\\dfrac{2x+1}{x^2+x-1}$ (\u7b54)<\/span><\/div><\/div>\n

(2) $y={\\mathrm{log}}|4x-3|$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u516c\u5f0f\u3000$({\\mathrm{log}}f)’=({\\mathrm{log}}|f|)’=\\dfrac{f’}{f}$ \u3092\u7528\u3044\u308b
\n$y’=\\dfrac{(4x-3)’}{4x-3}=\\dfrac{4}{4x-3}$ (\u7b54)<\/span><\/div><\/div>\n

(3) $y={\\mathrm{log}}_3|7x+4|$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u516c\u5f0f\u3000$({\\mathrm{log}}_af)’=({\\mathrm{log}}_a|f|)’=\\dfrac{f’}{f{\\mathrm{log}}a}$ \u3092\u7528\u3044\u308b
\n$y’=\\dfrac{(7x+4)’}{(7x+4){\\mathrm{log}}3}=\\dfrac{7}{(7x+4){\\mathrm{log}}3}$ (\u7b54)<\/span><\/div><\/div>\n

(4) $y=3^{2x+1}$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u516c\u5f0f\u3000$(a^f)’=f’a^f{\\mathrm{log}}a$ \u3092\u7528\u3044\u308b
\n$y’=(2x+1)’3^{2x+1}{\\mathrm{log}}3=2\\cdot3^{2x+1}{\\mathrm{log}}3$ (\u7b54)<\/span><\/div><\/div>\n

(5) $y=e^{x^2-3x-2}$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u516c\u5f0f\u3000$(e^f)’=f’e^f$ \u3092\u7528\u3044\u308b
\n$y’=(x^2-3x-2)’e^{x^2-3x-2}=(2x-3)e^{x^2-3x-2}$ (\u7b54)<\/span><\/div><\/div>\n

(6) $y=x{\\mathrm{sin}}x$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u516c\u5f0f\u3000$(fg)’=f’g+fg’$ \u304a\u3088\u3073 $({\\mathrm{sin}}x)’={\\mathrm{cos}}x$ \u3092\u7528\u3044\u308b
\n$y’=(x)'{\\mathrm{sin}}x+x({\\mathrm{sin}}x)’={\\mathrm{sin}}x+x{\\mathrm{cos}}x$ (\u7b54)<\/span><\/div><\/div>\n

(7) $y=\\dfrac{{\\mathrm{sin}}x}{1+{\\mathrm{cos}}x}$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u516c\u5f0f\u3000$\\left(\\dfrac{f}{g}\\right)^{\\prime}=\\dfrac{f’g-fg’}{g^2}$\u3000\u3001$({\\mathrm{sin}}x)’={\\mathrm{cos}}x$\u3000\u304a\u3088\u3073 $({\\mathrm{cos}}x)’=-{\\mathrm{sin}}x$ \u3092\u7528\u3044\u308b
\n(\u5206\u5b50)$=({\\mathrm{sin}}x)'(1+{\\mathrm{cos}}x)-{\\mathrm{sin}}x(1+{\\mathrm{cos}}x)’={\\mathrm{cos}}x(1+{\\mathrm{cos}}x)-(-{\\mathrm{sin}}x)$
\n$\\quad \\quad \\;={\\mathrm{cos}}x+{\\mathrm{cos}}^2x+{\\mathrm{sin}}^2x={\\mathrm{cos}}x+1$ $({\\mathrm{cos}}^2x+{\\mathrm{sin}}^2x=1)$ \u306b\u6ce8\u610f\u3059\u308b
\n$y’=\\dfrac{1+{\\mathrm{cos}}x}{(1+{\\mathrm{cos}}x)^2}=\\dfrac{1}{1+{\\mathrm{cos}}x}$ (\u7b54)<\/span><\/div><\/div>\n

(8) $y={\\mathrm{tan}}(2x+1)$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u516c\u5f0f\u3000$({\\mathrm{tan}}f)’=\\dfrac{f’}{{\\mathrm{cos}}^2f}$\u3000\u3092\u7528\u3044\u308b
\n$y’=\\dfrac{(2x+1)’}{{\\mathrm{cos}}^2(2x+1)}=\\dfrac{2}{{\\mathrm{cos}}^2(2x+1)}$ (\u7b54)<\/span><\/div><\/div>\n

(9) $y={\\mathrm{Sin}}^{-1}3x$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$({\\mathrm{Sin}}^{-1}x)’=\\dfrac{1}{\\sqrt{1-x^2}}\u3001 ({\\mathrm{Sin}}^{-1}f)’=\\dfrac{f’}{\\sqrt{1-f^2}}\u3001 \\left({\\mathrm{Sin}}^{-1}\\dfrac{x}{a}\\right)^{\\prime}=\\dfrac{1}{\\sqrt{a^2-x^2}}$ \u306b\u6ce8\u610f\u3059\u308b
\n$y’=\\dfrac{(3x)’}{\\sqrt{1-(3x)^2}}=\\dfrac{3}{\\sqrt{1-9x^2}}$ (\u7b54)<\/span><\/div><\/div>\n

(10) $y={\\mathrm{Tan}}^{-1}\\dfrac{x}{3}$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$({\\mathrm{Tan}}^{-1}x)’=\\dfrac{1}{x^2+1}\u3001({\\mathrm{Tan}}^{-1}f)’=\\dfrac{f’}{f^2+1}\u3001\\left({\\mathrm{Tan}}^{-1}\\dfrac{x}{a}\\right)^{\\prime}=\\dfrac{a}{x^2+a^2}$ \u306b\u6ce8\u610f\u3059\u308b
\n$y’=\\dfrac{3}{x^2+3^2}=\\dfrac{3}{x^2+9}$ (\u7b54)<\/span><\/div><\/div>\n

(11) $y=e^{2x}{\\mathrm{sin}}3x$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u516c\u5f0f\u3000$(fg)’=f’g+fg’$\u3001$(e^f)’=f’e^f$\u3001$({\\mathrm{sin}}f)’=f'{\\mathrm{cos}}f$\u3000\u3092\u7528\u3044\u308b
\n$y’=(e^{2x})'({\\mathrm{sin}}3x)+(e^{2x})({\\mathrm{sin}}3x)’=2e^{2x}{\\mathrm{sin}}3x+e^{2x}(3{\\mathrm{cos}}3x)=e^{2x}(2{\\mathrm{sin}}3x+3{\\mathrm{cos}}3x)$ (\u7b54)<\/span><\/div><\/div>\n

(12) $y={\\mathrm{cos}}^{5}2x$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\u516c\u5f0f\u3000$(f^n)’=nf’f^{n-1}$\u3001$({\\mathrm{cos}}f)’=-f'{\\mathrm{sin}}f$\u3000\u3092\u7528\u3044\u308b
\n$y={\\mathrm{cos}}^{5}2x=({\\mathrm{cos}}2x)^5$ \u306b\u6ce8\u610f\u3059\u308b
\n$y’=5({\\mathrm{cos}}2x)'({\\mathrm{cos}}2x)^4=5(-2{\\mathrm{sin}}2x)({\\mathrm{cos}}2x)^4=-10{\\mathrm{sin}}2x{\\mathrm{cos}}^42x$ (\u7b54)<\/span><\/div><\/div>\n<\/div>\n
\n

\u30d9\u30af\u30c8\u30eb\u306e\u7dda\u5f62\u548c<\/span><\/p>\n

1\uff0e \u4e0b\u56f3\u306e\u30d9\u30af\u30c8\u30eb $\\vec{c}$\u3001$\\vec{d}$ \u3092\u30d9\u30af\u30c8\u30eb $\\vec{a}$\u3001$\\vec{b}$ \u3092\u7528\u3044\u3066\u8868\u305b<\/p>\n

\"\"<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$\\vec{c}=2\\vec{a}+\\vec{b}$\u3001$\\vec{d}=\\vec{a}-2\\vec{b}$ (\u7b54) (\u4e0b\u56f3\u53c2\u7167)<\/span><\/p>\n

\"\"<\/div><\/div>\n

\u4ea4\u70b9\u306e\u4f4d\u7f6e\u30d9\u30af\u30c8\u30eb<\/span><\/p>\n

2. $\\triangle {\\mathrm{ABC}}$ \u306b\u304a\u3044\u3066\u3001$\\overrightarrow{\\mathrm{OA}}=\\vec{a}$\u3001$\\overrightarrow{\\mathrm{OB}}=\\vec{b}$ \u3068\u3057\u3001\u70b9${\\mathrm{C}}$\u306f\u7dda\u5206${\\mathrm{OA}}$\u30921:2\u306b\u5185\u5206\u3059\u308b\u70b9\u3001\u70b9${\\mathrm{D}}$\u306f\u7dda\u5206${\\mathrm{OB}}$\u30922:1\u306b\u5185\u5206\u3059\u308b\u70b9\u3068\u3059\u308b \u7dda\u5206${\\mathrm{BC}}$\u3068\u7dda\u5206${\\mathrm{AD}}$\u306e\u4ea4\u70b9\u3092${\\mathrm{P}}$\u3068\u3059\u308b\u3068\u304d\u3001\u30d9\u30af\u30c8\u30eb$\\overrightarrow{\\mathrm{OP}}$\u3092$\\vec{a}$\u3001$\\vec{b}$\u3067\u8868\u305b<\/p>\n

\"\"<\/p>\n

<\/span>\u89e3\u7b54<\/div>
(\u516c\u5f0f) \u76f4\u7dda${\\mathrm{AB}}$\u306e$t$\u3092\u5a92\u4ecb\u5909\u6570\u3068\u3059\u308b\u30d9\u30af\u30c8\u30eb\u65b9\u7a0b\u5f0f $\\;\\vec{p}=\\overrightarrow{\\mathrm{OP}}=(1-t)\\vec{a}+t\\vec{b}$\u3000\u3092\u7528\u3044\u308b
\n$\\vec{c}=\\overrightarrow{\\mathrm{OC}}=\\dfrac{1}{3}\\vec{a}$\u3001$\\vec{d}=\\overrightarrow{\\mathrm{OD}}=\\dfrac{2}{3}\\vec{b}$ \u3088\u308a
\n\u76f4\u7dda${\\mathrm{CB}}:\\vec{p}=(1-t)\\vec{c}+t\\vec{b}=\\dfrac{1-t}{3}\\vec{a}+t\\vec{b}$\u3001\u76f4\u7dda${\\mathrm{AD}}:\\vec{p}=(1-s)\\vec{a}+s\\vec{d}=(1-s)\\vec{a}+\\dfrac{2s}{3}\\vec{b}$
\n\u3060\u304b\u3089$\\;\\begin{cases}\\dfrac{1-t}{3}=1-s\\\\t=\\dfrac{2s}{3}\\end{cases} \\Rightarrow \\begin{cases}t+3s=1\\\\3t+2s=2\\end{cases} \\Rightarrow \\begin{cases}t=\\dfrac{4}{7}\\\\s=\\dfrac{1}{7}\\end{cases}$ \u3088\u3063\u3066\u3000 $\\vec{p}=\\dfrac{1}{7}\\vec{a}+\\dfrac{4}{7}\\vec{b}$ (\u7b54)<\/span><\/p>\n

(\u5225\u89e3\uff11\uff1a\u30e1\u30cd\u30e9\u30a6\u30b9\u306e\u5b9a\u7406\u3092\u7528\u3044\u308b\uff09
\n$\\overrightarrow{\\mathrm{CP}}=t\\overrightarrow{\\mathrm{CB}}$ \u3068\u304a\u304f\u3068 $\\vec{p}=\\vec{c}+t\\overrightarrow{\\mathrm{CB}}=\\vec{c}+t(\\vec{b}-\\vec{c})=(1-t)\\vec{c}+t\\vec{b}$
\n\u3053\u3053\u3067\u3001\u30e1\u30cd\u30e9\u30a6\u30b9\u306e\u5b9a\u7406\u3088\u308a $\\dfrac{3}{2}\\cdot\\dfrac{t}{1-t}\\cdot\\dfrac{1}{2}=1 \\Rightarrow 3t=4(1-t) \\Rightarrow t=\\dfrac{4}{7}$
\n\u3088\u3063\u3066 $\\vec{p}=\\dfrac{3}{7}\\vec{c}+\\dfrac{4}{7}\\vec{b}$ \u3053\u3053\u3067\u3001$\\vec{c}=\\dfrac{1}{3}\\vec{a}$ \u3088\u308a$\\vec{p}=\\dfrac{1}{3}\\cdot\\dfrac{3}{7}\\vec{a}+\\dfrac{4}{7}\\vec{b}=\\dfrac{1}{7}\\vec{a}+\\dfrac{4}{7}\\vec{b}$\u3000(\u7b54)<\/span><\/p>\n

(\u5225\u89e32\uff1a\u7c21\u6613\u516c\u5f0f\u3092\u7528\u3044\u308b\uff09
\n\u4e00\u822c\u306b\u3001${\\mathrm{OA}}$ \u3092 $a_1:a_2$ \u306b\u5185\u5206\u3059\u308b\u70b9\u3092 ${\\mathrm{C}}$\u3001${\\mathrm{OB}}$ \u3092 $b_1:b_2$ \u306b\u5185\u5206\u3059\u308b\u70b9\u3092 ${\\mathrm{D}}$ \u3068\u3059\u308b\u3068 $\\alpha=a_1b_2, \\beta=b_1a_2, \\gamma=a_2b_2 \\Rightarrow \\vec{p}=\\dfrac{\\alpha}{\\alpha+\\beta+\\gamma}\\vec{a}+\\dfrac{\\beta}{\\alpha+\\beta+\\gamma}\\vec{b}\\;$ (\u7c21\u6613\u516c\u5f0f)$\\quad$\u304c\u6210\u308a\u7acb\u3064
\n$\\alpha=1\\cdot, \\beta=2\\cdot 2=4, \\gamma=2\\cdot 1=2$ \u3088\u308a $\\vec{p}=\\dfrac{1}{1+4+2}\\vec{a}+\\dfrac{4}{1+4+2}\\vec{b}=\\dfrac{1}{7}\\vec{a}+\\dfrac{4}{7}\\vec{b}$ (\u7b54)<\/div><\/div><\/span><\/p>\n

\u30d9\u30af\u30c8\u30eb\u306e\u6210\u5206\u8868\u793a<\/span><\/p>\n

3. $\\vec{a}=\\begin{pmatrix}1\\\\-2\\end{pmatrix}$\u3001$\\vec{b}=\\begin{pmatrix}2\\\\1\\end{pmatrix}$\u3001$\\vec{c}=\\begin{pmatrix}-1\\\\3\\end{pmatrix}$ \u3068\u3059\u308b\u3068\u304d\u3001\u6b21\u3092\u6c42\u3081\u3088<\/p>\n

(1) $2\\vec{a}-\\vec{b}+3\\vec{c}\\quad$ (2) $|\\vec{b}-\\vec{a}|\\quad$ (3) $\\vec{a}\\cdot(\\vec{b}-2\\vec{c})$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
(\u516c\u5f0f) $\\vec{a}=\\begin{pmatrix}a_1\\\\a_2\\end{pmatrix}$\u3001$\\vec{b}=\\begin{pmatrix}b_1\\\\b_2\\end{pmatrix}$ \u306e\u3068\u304d\u3001$\\vec{a}\\cdot\\vec{b}=a_1b_1+a_2b_2$\u3001\u3000$|\\vec{a}|=a_1^2+a_2^2$ \u3092\u7528\u3044\u308b<\/span><\/p>\n

(1) $2\\begin{pmatrix}1\\\\-2\\end{pmatrix}-\\begin{pmatrix}2\\\\1\\end{pmatrix}+3\\begin{pmatrix}-1\\\\3\\end{pmatrix}=\\begin{pmatrix}2\\cdot 1-2+3\\cdot(-1)\\\\2\\cdot(-2)-1+3\\cdot3\\end{pmatrix}=\\begin{pmatrix}-3\\\\4\\end{pmatrix}\\quad$ (\u7b54)<\/span><\/p>\n

(2) $\\left |\\begin{pmatrix}2\\\\1\\end{pmatrix}-\\begin{pmatrix}1\\\\-2\\end{pmatrix}\\right |=\\left |\\begin{pmatrix}1\\\\3\\end{pmatrix}\\right |=\\sqrt{1^2+3^2}=\\sqrt{10}\\quad$ (\u7b54)<\/span><\/p>\n

(3) $\\begin{pmatrix}1\\\\-2\\end{pmatrix}\\cdot \\left \\lbrace\\begin{pmatrix}2\\\\1\\end{pmatrix}-2\\begin{pmatrix}-1\\\\3\\end{pmatrix} \\right \\rbrace =\\begin{pmatrix}1\\\\-2\\end{pmatrix}\\cdot\\begin{pmatrix}4\\\\-5\\end{pmatrix}=1\\cdot 4 +(-2)\\cdot(-5)=14\\quad$ (\u7b54)<\/div><\/div><\/span><\/p>\n

\u30d9\u30af\u30c8\u30eb\u306e\u5185\u7a4d<\/span><\/p>\n

4. \u6b21\u306e\u554f\u3044\u306b\u7b54\u3048\u3088<\/p>\n

(1) $|\\vec{a}|=2$\u3001$|\\vec{b}|=3$\u3001$\\theta=\\dfrac{2\\pi}{3}$ \u306e\u3068\u304d\u5185\u7a4d $\\vec{a}\\cdot\\vec{b}$ \u3092\u6c42\u3081\u3088 ($\\theta$ \u306f\u30d9\u30af\u30c8\u30eb $\\vec{a}$ \u3068 $\\vec{b}$ \u306e\u306a\u3059\u89d2)<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$\\begin{matrix} \\theta & 0^{\\circ} & 30^{\\circ} & 45^{\\circ} & 60^{\\circ} & 90^{\\circ} & 120^{\\circ} & 135^{\\circ} & 150^{\\circ} & 180^{\\circ} \\\\ {\\mathrm{cos}}\\theta & 1&\u3000\\dfrac{\\sqrt{3}}{2} & \\dfrac{\\sqrt{2}}{2} & \\dfrac{1}{2} & 0 & \\dfrac{-1}{2} & \\dfrac{-\\sqrt{2}}{2} & \\dfrac{-\\sqrt{3}}{2} & -1 \\end{matrix}$<\/span><\/p>\n

(\u5b9a\u7fa9) $\\vec{a}\\cdot\\vec{b}=|\\vec{a}||\\vec{b}|{\\mathrm{cos}}\\theta$\u3000\u3092\u7528\u3044\u308b<\/span><\/p>\n

$\\vec{a}\\cdot\\vec{b}=2\\cdot3{\\mathrm{cos}}\\dfrac{2\\pi}{3}=6{\\mathrm{cos}}120^{\\circ}=6\\cdot\\dfrac{-1}{2}=-3$ (\u7b54)<\/div><\/div><\/span><\/p>\n

(2) $\\vec{a}=\\begin{pmatrix}2\\sqrt{3}\\\\1\\end{pmatrix}$\u3001$\\vec{b}=\\begin{pmatrix}-7\\\\\\sqrt{3}\\end{pmatrix}$ \u306e\u3068\u304d\u3001$\\vec{a}$ \u3068 $\\vec{b}$ \u306e\u306a\u3059\u89d2 $\\theta$ \u3092\u6c42\u3081\u3088<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$\\begin{matrix}\\theta & 0^{\\circ} & 30^{\\circ} & 45^{\\circ} & 60^{\\circ} & 90^{\\circ} & 120^{\\circ} & 135^{\\circ} & 150^{\\circ} & 180^{\\circ} \\\\ {\\mathrm{cos}}\\theta & 1&\u3000\\dfrac{\\sqrt{3}}{2} & \\dfrac{\\sqrt{2}}{2} & \\dfrac{1}{2} & 0 & \\dfrac{-1}{2} & \\dfrac{-\\sqrt{2}}{2} & \\dfrac{-\\sqrt{3}}{2} & -1 \\end{matrix}$<\/span><\/p>\n

(\u516c\u5f0f) ${\\mathrm{cos}}\\theta=\\dfrac{\\vec{a}\\cdot\\vec{b}}{|\\vec{a}||\\vec{b}|}$\u3000\u3088\u308a $\\theta={\\mathrm{Cos}}^{-1}\\left(\\dfrac{\\vec{a}\\cdot\\vec{b}}{|\\vec{a}||\\vec{b}|}\\right)$ \u3092\u7528\u3044\u308b<\/span><\/p>\n

$\\vec{a}\\cdot\\vec{b}=-14\\sqrt{3}+\\sqrt{3}=-13\\sqrt{3}$\u3001$|\\vec{a}|=\\sqrt{12+1}=\\sqrt{13}$\u3001$|\\vec{b}|=\\sqrt{49+3}=\\sqrt{52}=2\\sqrt{13}$
\n\u3088\u3063\u3066\u3001$\\theta={\\mathrm{Cos}}^{-1}\\left( \\dfrac{-13\\sqrt{3}}{\\sqrt{13}\\cdot\u30002\\sqrt{13}} \\right)={\\mathrm{Cos}}^{-1}\\left(\\dfrac{-\\sqrt{3}}{2}\\right)=120^{\\circ}=\\dfrac{5\\pi}{6}$ (\u7b54)<\/div><\/div><\/span><\/p>\n

(3) $|\\vec{a}|=3$\u3001$|\\vec{b}|=2$\u3001$|2\\vec{a}-3\\vec{b}|=6$ \u306e\u3068\u304d\u3001\u5185\u7a4d $\\vec{a}\\cdot\\vec{b}$ \u3092\u6c42\u3081\u3088<\/p>\n

<\/span>\u89e3\u7b54<\/div>
(\u516c\u5f0f) $|\\vec{a}|^2=\\vec{a}\\cdot\\vec{a}$\u3000\u3092\u7528\u3044\u308b
\n$|2\\vec{a}-3\\vec{b}|^2=4|\\vec{a}|^2-12\\vec{a}\\cdot\\vec{b}+9|\\vec{b}|^2$ \u3088\u308a\u3000 $\\vec{a}\\cdot\\vec{b}=\\dfrac{4|\\vec{a}|^2+9|\\vec{b}|^2-|2\\vec{a}-3\\vec{b}|^2}{12}=\\dfrac{4\\cdot3^2+9\\cdot2^2-6^2}{12}=3$ (\u7b54)<\/div><\/div><\/span><\/p>\n

(4) $|\\vec{a}|=4$\u3001$|\\vec{b}|=3$\u3001$\\vec{a}\\cdot\\vec{b}=-6$ \u306e\u3068\u304d $|\\vec{a}-2\\vec{b}|$ \u3092\u6c42\u3081\u3088<\/p>\n

<\/span>\u89e3\u7b54<\/div>
(\u516c\u5f0f) $|\\vec{a}|^2=\\vec{a}\\cdot\\vec{a}$\u3000\u3092\u7528\u3044\u308b
\n$|\\vec{a}-2\\vec{b}|^2=|\\vec{a}|^2-4\\vec{a}\\cdot\\vec{b}+4|\\vec{b}|^2=4^2-4\\cdot(-6)+4\\cdot3^2=76$ \u3088\u308a\u3000 $|\\vec{a}-2\\vec{b}|=\\sqrt{76}=2\\sqrt{19}$ (\u7b54)<\/div><\/div><\/span><\/p>\n<\/div>\n
\n

\u63a5\u7dda\u306e\u65b9\u7a0b\u5f0f<\/span><\/p>\n

1\uff0e \u6b21\u306e\u66f2\u7dda\u306e\uff08\u3000\u3000\uff09\u5185\u306e $x$ \u306e\u5024\u306b\u5bfe\u5fdc\u3059\u308b\u70b9\u306b\u304a\u3051\u308b\u63a5\u7dda\u306e\u65b9\u7a0b\u5f0f\u3092\u6c42\u3081\u3088<\/p>\n

(1) $y=\\sqrt{x}\\quad(x=1)$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\uff08$x=a$\u306b\u5bfe\u5fdc\u3059\u308b\u70b9\u306b\u304a\u3051\u308b\u63a5\u7dda\u306e\u516c\u5f0f\uff09$y=f'(a)(x-a)+f(a)$ \u3092\u7528\u3044\u308b
\n$f(x)=\\sqrt{x}$ \u3068\u304a\u304f\u3068\u3001$f(1)=\\sqrt{1}=1$\u3001$f'(x)=\\dfrac{1}{2\\sqrt{x}}$ \u3088\u308a $f'(1)=\\dfrac{1}{2\\sqrt{1}}=\\dfrac{1}{2}$
\n$y=f'(1)(x-1)+f(1)=\\dfrac{1}{2}(x-1)+1=\\dfrac{1}{2}x+\\dfrac{1}{2}$<\/span><\/div><\/div>\n

(2) $y={\\mathrm{sin}}x\\quad(x=\\dfrac{\\pi}{3})$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
\uff08$x=a$\u306b\u5bfe\u5fdc\u3059\u308b\u70b9\u306b\u304a\u3051\u308b\u63a5\u7dda\u306e\u516c\u5f0f\uff09$y=f'(a)(x-a)+f(a)$ \u3092\u7528\u3044\u308b
\n$f(x)={\\mathrm{sin}}x$ \u3068\u304a\u304f\u3068\u3001$f\\left(\\dfrac{\\pi}{3}\\right)=\\dfrac{\\sqrt{3}}{2}$\u3001$f'(x)={\\mathrm{cos}}x$ \u3088\u308a $f’\\left(\\dfrac{\\pi}{3}\\right)={\\mathrm{cos}}\\left(\\dfrac{\\pi}{3}\\right)=\\dfrac{1}{2}$
\n$y=f’\\left(\\dfrac{\\pi}{3}\\right)\\left(x-\\dfrac{\\pi}{3}\\right)+f\\left(\\dfrac{\\pi}{3}\\right)=\\dfrac{1}{2}\\left(x-\\dfrac{\\pi}{3}\\right)+\\dfrac{\\sqrt{3}}{2}=\\dfrac{1}{2}x-\\dfrac{\\pi}{6}+\\dfrac{\\sqrt{3}}{2}$ (\u7b54)<\/span><\/div><\/div>\n

\u95a2\u6570\u306e\u5897\u6e1b\u30fb\u6975\u5024<\/span><\/p>\n

2\uff0e \u6b21\u306e\u95a2\u6570\u306e\u5897\u6e1b\u30fb\u6975\u5024\u3092\u8abf\u3079\u3066\u30b0\u30e9\u30d5\u3092\u304b\u3051<\/p>\n

(1) $y=x^3+3x^2+1$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$y’=3x^2+6x=3x(x+2)=0 \\Rightarrow x=-2,0$\u3088\u308a\u5897\u6e1b\u8868\u3092\u304b\u304f
\n$\\begin{matrix}
\nx & \\cdots & -2 & \\cdots & 0 &\\cdots \\\\
\ny’ & + & 0 & – & 0 & + \\\\
\ny & \\nearrow & \\underset{\u6975\u5927}{5} & \\searrow & \\underset{\u6975\u5c0f}{1} & \\nearrow \\\\
\n\\end{matrix}$<\/span><\/p>\n

\"\"<\/p>\n<\/div><\/div>\n

(2) $y=x^4-2x^2$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$y’=4x^3-4x=4x(x^2-1)=4x(x+1)(x-1)=0 \\Rightarrow x=-1,0,1$ \u3088\u308a\u5897\u6e1b\u8868\u3092\u304b\u304f
\n$\\begin{matrix}
\nx & \\cdots & -1 & \\cdots & 0 &\\cdots & 1 & \\cdots\\\\y’ & – & 0 & + & 0 & – & 0 & +\\\\y & \\searrow & \\underset{\u6975\u5c0f}{-1} & \\nearrow & \\underset{\u6975\u5927}{0} & \\searrow &\\underset{\u6975\u5c0f}{-1} & \\nearrow \\\\
\n\\end{matrix}$<\/span><\/p>\n

\"\"<\/p>\n<\/div><\/div>\n

(3) $y=3x^4-4x^3$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$y’=12x^3-12x^2=12x^2(x-1)=0 \\Rightarrow x=0,1$\u3088\u308a\u5897\u6e1b\u8868\u3092\u304b\u304f
\n$\\begin{matrix}
\nx & \\cdots & 0 & \\cdots & 1 &\\cdots \\\\y’ & – & 0 & – & 0 & + \\\\y & \\searrow & 0 & \\searrow & \\underset{\u6975\u5c0f}{-1} & \\nearrow \\\\\\end{matrix}$<\/span><\/p>\n

\"\"<\/p>\n<\/div><\/div>\n<\/div>\n

\n

\u6975\u9650\u5024\uff08\u30ed\u30d4\u30bf\u30eb\u306e\u5b9a\u7406\uff09<\/span><\/p>\n

\u6ce8\u610f\uff1a$\\dfrac{0}{0}, \\dfrac{\\infty}{\\infty}$ \u306e\u4e0d\u5b9a\u5f62\u3067\u3042\u3063\u3066\u3082\u3001\u6b21\u306e\u5834\u5408\u306f\u30ed\u30d4\u30bf\u30eb\u306e\u5b9a\u7406\u3092\u9069\u7528\u3059\u308b\u3053\u3068\u306f\u3067\u304d\u306a\u3044<\/span><\/p>\n

(1) $\\displaystyle{\\lim_{x\\to a}}\\dfrac{f'(x)}{g'(x)}$ \u304c\u632f\u52d5\u3059\u308b\u5834\u5408
\n(2) $a$ \u306e\u3044\u304f\u3089\u3067\u3082\u201d\u8fd1\u304f\u201d\u306b $a$ \u3068\u7570\u306a\u308b $g'(x)=0$ \u306e\u89e3\u304c\u5b58\u5728\u3059\u308b\u5834\u5408<\/span><\/p>\n

<\/span>\u4f8b<\/div>
(1) $\\displaystyle{\\lim_{x\\to \\infty}}\\dfrac{x+{\\mathrm{sin}}x}{x}=1$ \u3057\u304b\u3057\u3001$\\displaystyle{\\lim_{x\\to \\infty}}\\dfrac{(x+{\\mathrm{sin}}x)’}{(x)’}=\\displaystyle{\\lim_{x\\to \\infty}}(1+{\\mathrm{cos}}x)$ \u306f\u632f\u52d5\u3059\u308b
\n(\u30b3\u30fc\u30b7\u30fc\u306e\u5e73\u5747\u5024\u306e\u5b9a\u7406\u306e$’c=c_x’$ \u306f\u3001$a$ \u306b\u8fd1\u3065\u304f\u304c\u3001$x$ \u306b\u4f9d\u5b58\u3057\u3066\u6c7a\u307e\u308b\u306e\u3067\u3001\u7279\u5225\u306a\u8fd1\u3065\u304d\u65b9\u3092\u3057\u3066\u3044\u308b\u3053\u3068\u304c\u539f\u56e0)
\n(2) $\\displaystyle{\\lim_{x\\to \\infty}}\\dfrac{x+\\dfrac{1}{2}{\\mathrm{sin}}2x}{\\left(x+\\dfrac{1}{2}{\\mathrm{sin}}2x\\right)e^{{\\mathrm{sin}}x}}=\\displaystyle{\\lim_{x\\to \\infty}}\\dfrac{1}{e^{{\\mathrm{sin}}x}}$ \u306f\u632f\u52d5\u3059\u308b\u304c\u3001
\n$\\displaystyle{\\lim_{x\\to \\infty}}\\dfrac{\\left(x+\\dfrac{1}{2}{\\mathrm{sin}}2x \\right)^{\\prime}}{\\left \\lbrace \\left(x+\\dfrac{1}{2}{\\mathrm{sin}}2x \\right)e^{{\\mathrm{sin}}x} \\right\\rbrace ^{\\prime}}=\\displaystyle{\\lim_{x\\to \\infty}}\\dfrac{2{\\mathrm{cos}}^2x}{{\\mathrm{cos}}x \\left( 2{\\mathrm{cos}}x +x+\\dfrac{1}{2}{\\mathrm{sin}}2x\\right)e^{{\\mathrm{sin}}x}}=0$
\n($a$ \u306e\u201d\u5341\u5206\u8fd1\u304f\u201d\u3067 $x\\ne a$ \u3067\u3042\u308c\u3070\u3001$g'(x) \\ne 0$ \u304c\u6210\u308a\u7acb\u305f\u306a\u3051\u308c\u3070\u3001\u30b3\u30fc\u30b7\u30fc\u306e\u5e73\u5747\u5024\u306e\u5b9a\u7406\u3092\u9069\u7528\u3067\u304d\u306a\u3044\u3053\u3068\u304c\u539f\u56e0)<\/span><\/div><\/div>\n

1\uff0e \u6b21\u306e\u6975\u9650\u5024\u3092\u6c42\u3081\u3088<\/p>\n

(1) $\\displaystyle{\\lim_{x\\to 1}}\\dfrac{x^3-1}{x-1}$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$\\displaystyle{\\lim_{x\\to 1}}\\dfrac{x^3-1}{x-1}\\underset{\\scriptsize{\\dfrac{0}{0}}}{=}\\displaystyle{\\lim_{x\\to 1}}\\dfrac{3x^2}{1}=\\dfrac{3\\cdot1^2}{1}=3$ \uff08\u7b54\uff09<\/span><\/div><\/div>\n

(2) $\\displaystyle{\\lim_{x\\to 0}}\\dfrac{{\\mathrm{sin}}3x}{{\\mathrm{tan}}2x}$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$\\displaystyle{\\lim_{x\\to 0}}\\dfrac{{\\mathrm{sin}}3x}{{\\mathrm{tan}}2x}\\underset{\\scriptsize{\\dfrac{0}{0}}}{=}\\displaystyle{\\lim_{x\\to 0}}\\dfrac{3{\\mathrm{cos}}3x}{\\dfrac{2}{{\\mathrm{cos}}^{2}2x}}\\underset{{\\mathrm{cos}}0=1}{=}\\dfrac{3\\cdot 1}{\\dfrac{2}{1^2}}=\\dfrac{3}{2}$ \uff08\u7b54\uff09<\/span><\/div><\/div>\n

(3) $\\displaystyle{\\lim_{x\\to 0}}\\dfrac{1-{\\mathrm{cos}}^2x}{x^2}$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$\\displaystyle{\\lim_{x\\to 0}}\\dfrac{1-{\\mathrm{cos}}^2x}{x^2}\\underset{\\scriptsize{\\dfrac{0}{0}}}{=}\\displaystyle{\\lim_{x\\to 0}}\\dfrac{2{\\mathrm{sin}}x{\\mathrm{cos}}x}{2x}\\underset{\\scriptsize{\\dfrac{0}{0}}}{=}\\displaystyle{\\lim_{x\\to 0}}\\dfrac{2{\\mathrm{cos}}^2x-2{\\mathrm{sin}}^2x}{2}=\\dfrac{2\\cdot 1^2-2\\cdot 0^2}{2}=1$ \uff08\u7b54\uff09<\/span><\/div><\/div>\n

(4) $\\displaystyle{\\lim_{x\\to \\infty}}\\dfrac{e^{2x}}{x^2}$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$\\displaystyle{\\lim_{x\\to \\infty}}\\dfrac{e^{2x}}{x^2}\\underset{\\scriptsize{\\dfrac{\\infty}{\\infty}}}{=}\\displaystyle{\\lim_{x\\to \\infty}}\\dfrac{2e^{2x}}{2x}\\underset{\\scriptsize{\\dfrac{\\infty}{\\infty}}}{=}\\displaystyle{\\lim_{x\\to \\infty}}\\dfrac{4e^{2x}}{2}=\\infty$ \uff08\u7b54\uff09<\/span><\/div><\/div>\n

(5) $\\displaystyle{\\lim_{x\\to +0}}x{\\mathrm{log}}x$<\/p>\n

\u300c$+0\\cdot\\infty=\\dfrac{\\infty}{\\dfrac{1}{+0}}=\\dfrac{\\infty}{\\infty}$ \u4e0d\u5b9a\u5f62\u300d\u306b\u6ce8\u610f\u3059\u308b<\/span><\/p>\n

<\/span>\u89e3\u7b54<\/div>
\n$\\displaystyle{\\lim_{x\\to +0}}x{\\mathrm{log}}x=\\displaystyle{\\lim_{x\\to +0}}\\dfrac{{\\mathrm{log}}x}{\\dfrac{1}{x}}\\underset{\\scriptsize{\\dfrac{\\infty}{\\infty}}}{=}\\displaystyle{\\lim_{x\\to +0}}\\dfrac{\\dfrac{1}{x}}{\\dfrac{-1}{x^2}}\\underset{\\scriptsize{\\dfrac{1}{x}}\\div\\scriptsize{\\dfrac{-1}{x^2}}=-x}{=}\\displaystyle{\\lim_{x\\to +0}}(-x)=0$ \uff08\u7b54\uff09<\/span><\/div><\/div>\n

\u95a2\u6570\u306e\u6700\u5927\u30fb\u6700\u5c0f<\/span><\/p>\n

2\uff0e \u6b21\u306e\u95a2\u6570\u306e\u6307\u5b9a\u3055\u308c\u305f\u533a\u9593\u306b\u304a\u3051\u308b\u6700\u5927\u5024\u3068\u6700\u5c0f\u5024\u3092\u6c42\u3081\u3088<\/p>\n

(1) $y=x\\sqrt{1-x^2}\\quad(-1\\leq x \\leq1)$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$y’=\\sqrt{1-x^2}+x\\cdot\\dfrac{-\\cancel{2}x}{\\cancel{2}\\sqrt{1-x^2}}=\\dfrac{1-x^2}{\\sqrt{1-x^2}}+\\dfrac{-x^2}{\\sqrt{1-x^2}}=\\dfrac{(1+\\sqrt{2}x)(1-\\sqrt{2}x)}{\\sqrt{1-x^2}}=0$
\n$ \\Rightarrow x=-\\dfrac{1}{\\sqrt{2}},\\dfrac{1}{\\sqrt{2}}=-\\dfrac{\\sqrt{2}}{2},\\dfrac{\\sqrt{2}}{2}$ \u3088\u308a\u533a\u9593 $\\lbrack -1, 1 \\rbrack$ \u3067\u5897\u6e1b\u8868\u3092\u304b\u304f <\/span><\/p>\n

$\\begin{matrix}
\nx & -1\u3000&\\cdots & -\\dfrac{\\sqrt{2}}{2} & \\cdots & \\dfrac{\\sqrt{2}}{2} &\\cdots & 1\\\\
\ny’ & \\cancel{} & – & 0 & + & 0 & – & \\cancel{} \\\\
\ny & 0 & \\searrow & \\underset{\u6700\u5c0f\u5024}{-\\dfrac{1}{2}} & \\nearrow & \\underset{\u6700\u5927\u5024}{\\dfrac{1}{2}} & \\searrow & 0 \\\\
\n\\end{matrix}$<\/span><\/p>\n

\u5897\u6e1b\u8868\u3088\u308a\u3000$x=-\\dfrac{\\sqrt{2}}{2}$ \u306e\u3068\u304d\u3000\u6700\u5c0f\u5024\u3000$-\\dfrac{1}{2}$\u3001$x=\\dfrac{\\sqrt{2}}{2}$ \u306e\u3068\u304d\u3000\u6700\u5927\u5024\u3000$\\dfrac{1}{2}$\u3000\uff08\u7b54\uff09<\/span><\/div><\/div>\n

(2) $y=\\dfrac{2x}{x^2+1}\\quad(-2\\leq x \\leq2)$<\/p>\n

<\/span>\u89e3\u7b54<\/div>
$y’=\\dfrac{2(x^2+1)-2x\\cdot2x}{(x^2+1)^2}=\\dfrac{-2x^2+2}{(x^2+1)^2}=\\dfrac{-2(x+1)(x-1)}{(x^2+1)^2}=0$
\n$ \\Rightarrow x=-1,1$ \u3088\u308a\u533a\u9593 $\\lbrack -2, 2 \\rbrack$ \u3067\u5897\u6e1b\u8868\u3092\u304b\u304f<\/span><\/p>\n

$\\begin{matrix}
\nx & -2\u3000&\\cdots & -1 & \\cdots & 1 &\\cdots & 2\\\\
\ny’ & \\cancel{} & – & 0 & + & 0 & – & \\cancel{} \\\\
\ny & -\\dfrac{4}{5} & \\searrow & \\underset{\u6700\u5c0f\u5024}{-1} & \\nearrow & \\underset{\u6700\u5927\u5024}{1} & \\searrow & \\dfrac{4}{5} \\\\
\n\\end{matrix}$<\/span><\/p>\n

\u5897\u6e1b\u8868\u3088\u308a\u3000$x=-1$ \u306e\u3068\u304d\u3000\u6700\u5c0f\u5024\u3000$-1$\u3001$x=1$ \u306e\u3068\u304d\u3000\u6700\u5927\u5024\u3000$1$\u3000\uff08\u7b54\uff09<\/span><\/div><\/div>\n<\/div>\n

\n

\u30d9\u30af\u30c8\u30eb\u306e\u5e73\u884c\u6761\u4ef6\u30fb\u5782\u76f4\u6761\u4ef6<\/span><\/p>\n

1. 3\u70b9 ${\\mathrm{A}}(1, 3), {\\mathrm{B}}(3, y), {\\mathrm{C}}(7, 5)$ \u306b\u3064\u3044\u3066\u6b21\u306e\u554f\u3044\u306b\u7b54\u3048\u3088<\/p>\n

(1) 3\u70b9 ${\\mathrm{A}}, {\\mathrm{B}}, {\\mathrm{C}}$ \u304c\u4e00\u76f4\u7dda\u4e0a\u306b\u4e26\u3076\u3088\u3046\u306b $y$ \u306e\u5024\u3092\u5b9a\u3081\u3088<\/p>\n

<\/span>\u89e3\u7b54<\/div>
(\u516c\u5f0f) 3\u70b9 ${\\mathrm{A}}, {\\mathrm{B}}, {\\mathrm{C}}$ \u304c\u4e00\u76f4\u7dda\u4e0a\u306b\u4e26\u3076 $\\iff \\overrightarrow{\\mathrm{AB}}\/\\!\/\\overrightarrow{\\mathrm{AC}} \\quad \\left( \\overrightarrow{\\mathrm{AB}}=t\\overrightarrow{\\mathrm{AC}} \\right)$
\n$\\quad \\quad \\quad \\begin{pmatrix}a_1\\\\a_2\\end{pmatrix} \/\\!\/ \\begin{pmatrix}b_1\\\\b_2\\end{pmatrix}$ \u304b\u3064 $ b_1\\cdot b_2 \\ne 0 \\iff \\dfrac{a_1}{b_1}=\\dfrac{a_2}{b_2}$<\/span><\/p>\n

$\\overrightarrow{\\mathrm{AB}}=\\begin{pmatrix}3\\\\y\\end{pmatrix}-\\begin{pmatrix}1\\\\3\\end{pmatrix}=\\begin{pmatrix}2\\\\y-3\\end{pmatrix}, \\; \\overrightarrow{\\mathrm{AC}}=\\begin{pmatrix}7\\\\5\\end{pmatrix}-\\begin{pmatrix}1\\\\3\\end{pmatrix}=\\begin{pmatrix}6\\\\2\\end{pmatrix},\\quad \\overrightarrow{\\mathrm{AB}}\/\\!\/\\overrightarrow{\\mathrm{AC}}\\;$ \u306a\u306e\u3067
\n$\\dfrac{2}{6}=\\dfrac{y-3}{2} \\Rightarrow 2=3(y-3) \\Rightarrow y=\\dfrac{11}{3}$ \uff08\u7b54\uff09<\/span><\/div><\/div>\n

(2) $\\triangle{\\mathrm{ABC}}$ \u304c ${\\mathrm{B}}=90^{\\circ}$ \u3067\u3042\u308b\u76f4\u89d2\u4e09\u89d2\u5f62\u3068\u306a\u308b\u3088\u3046\u306b $y$ \u306e\u5024\u3092\u5b9a\u3081\u3088<\/p>\n

<\/span>\u89e3\u7b54<\/div>
(\u516c\u5f0f) $\\overrightarrow{\\mathrm{AB}}=\\overrightarrow{\\mathrm{OB}}-\\overrightarrow{\\mathrm{OA}}$\u3001$\\vec{a}\\perp\\vec{b} \\iff \\vec{a}\\cdot\\vec{b}=0$<\/span><\/p>\n

$\\overrightarrow{\\mathrm{BA}}=\\begin{pmatrix}1\\\\3\\end{pmatrix}-\\begin{pmatrix}3\\\\y\\end{pmatrix}=\\begin{pmatrix}-2\\\\3-y\\end{pmatrix}, \\overrightarrow{\\mathrm{BC}}=\\begin{pmatrix}7\\\\5\\end{pmatrix}-\\begin{pmatrix}3\\\\y\\end{pmatrix}=\\begin{pmatrix}4\\\\5-y\\end{pmatrix}$ \u306a\u306e\u3067
\n$\\overrightarrow{\\mathrm{BA}}\\cdot\\overrightarrow{\\mathrm{BC}}=\\begin{pmatrix}-2\\\\3-y\\end{pmatrix}\\cdot\\begin{pmatrix}4\\\\5-y\\end{pmatrix}=0 \\Rightarrow (-2)\\cdot4+(3-y)(5-y)=0 \\Rightarrow$
\n$y^2-8y+7=(y-1)(y-7)=0 \\Rightarrow y=1,7$ \uff08\u7b54\uff09<\/span><\/div><\/div>\n

\u30d9\u30af\u30c8\u30eb\u306e\u9577\u3055<\/span><\/p>\n

2. 3\u70b9 ${\\mathrm{A}}(1, -1), {\\mathrm{B}}(3, 1), {\\mathrm{C}}(-2, 2)$ \u3092\u9802\u70b9\u3068\u3059\u308b\u4e09\u89d2\u5f62\u306f\u3069\u3093\u306a\u4e09\u89d2\u5f62\u304b<\/p>\n

<\/span>\u89e3\u7b54<\/div>
(\u516c\u5f0f)\u3000$\\left |\\begin{pmatrix}a_1\\\\a_2\\end{pmatrix}\\right |=\\sqrt{a_1^2+a_2^2}$ \u3001$\\overrightarrow{\\mathrm{AB}}=\\overrightarrow{\\mathrm{OB}}-\\overrightarrow{\\mathrm{OA}}$<\/span><\/p>\n

$\\overrightarrow{\\mathrm{AB}}=\\begin{pmatrix}3\\\\1\\end{pmatrix}-\\begin{pmatrix}1\\\\-1\\end{pmatrix}=\\begin{pmatrix}2\\\\2\\end{pmatrix},\\quad {\\mathrm{AB}}^2=\\left|\\overrightarrow{\\mathrm{AB}}\\right|^2=2^2+2^2=8 $
\n$\\overrightarrow{\\mathrm{AC}}=\\begin{pmatrix}-2\\\\2\\end{pmatrix}-\\begin{pmatrix}1\\\\-1\\end{pmatrix}=\\begin{pmatrix}-3\\\\3\\end{pmatrix},\\quad {\\mathrm{AC}}^2=\\left|\\overrightarrow{\\mathrm{AC}}\\right|^2=(-3)^2+3^2=18$
\n$\\overrightarrow{\\mathrm{BC}}=\\begin{pmatrix}-2\\\\2\\end{pmatrix}-\\begin{pmatrix}3\\\\1\\end{pmatrix}=\\begin{pmatrix}-5\\\\1\\end{pmatrix},\\quad {\\mathrm{BC}}^2=\\left|\\overrightarrow{\\mathrm{BC}}\\right|^2=(-5)^2+1^2=26$
\n\u4ee5\u4e0a\u3088\u308a ${\\mathrm{BC}}^2={\\mathrm{AB}}^2+{\\mathrm{AC}}^2$ \u304c\u6210\u308a\u7acb\u3064\u306e\u3067\u3001${\\mathrm{A}}=90^{\\circ}$ \u306e\u76f4\u89d2\u4e09\u89d2\u5f62 \uff08\u7b54\uff09<\/span><\/div><\/div>\n

\u5206\u70b9\u306e\u4f4d\u7f6e\u30d9\u30af\u30c8\u30eb<\/span><\/p>\n

3. 3\u70b9 ${\\mathrm{A}}(1, 2), {\\mathrm{B}}(5, 2), {\\mathrm{C}}(3, 2)$ \u306b\u3064\u3044\u3066\u6b21\u306e\u554f\u3044\u306b\u7b54\u3048\u3088<\/p>\n

(1) ${\\mathrm{AB}}$ \u3092 1:3 \u306b\u5185\u5206\u3059\u308b\u70b9 ${\\mathrm{P}}$ \u306e\u4f4d\u7f6e\u30d9\u30af\u30c8\u30eb\u3092\u6c42\u3081\u3088<\/p>\n

<\/span>\u89e3\u7b54<\/div>
(\u516c\u5f0f)\u3000${\\mathrm{AB}}$ \u3092 $n:m$ \u306b\u5185\u5206\u3059\u308b\u70b9 ${\\mathrm{P}}$ \u306e\u4f4d\u7f6e\u30d9\u30af\u30c8\u30eb\uff1a $\\overrightarrow{\\mathrm{OP}}=\\dfrac{m}{n+m}\\overrightarrow{\\mathrm{OA}}+\\dfrac{m}{n+m}\\overrightarrow{\\mathrm{OB}}$
\n$\\overrightarrow{\\mathrm{OP}}=\\dfrac{3}{1+3}\\begin{pmatrix}1\\\\2\\end{pmatrix}+\\dfrac{1}{1+3}\\begin{pmatrix}5\\\\2\\end{pmatrix}=\\begin{pmatrix}\\dfrac{3\\cdot1+1\\cdot5}{4}\\\\\\dfrac{3\\cdot2+1\\cdot2}{4}\\end{pmatrix}=\\begin{pmatrix}2\\\\2\\end{pmatrix}$ \uff08\u7b54\uff09<\/span><\/div><\/div>\n

(2) ${\\mathrm{AB}}$ \u3092 1:3 \u306b\u5916\u5206\u3059\u308b\u70b9 ${\\mathrm{Q}}$ \u306e\u4f4d\u7f6e\u30d9\u30af\u30c8\u30eb\u3092\u6c42\u3081\u3088<\/p>\n

<\/span>\u89e3\u7b54<\/div>
(\u516c\u5f0f)\u3000${\\mathrm{AB}}$ \u3092 $n:m$ \u306b\u5916\u5206\u3059\u308b\u5834\u5408\u306f\u3001$n$ \u3068 $m$ \u306e\u5c0f\u3055\u3044\u65b9\u306b\u30de\u30a4\u30ca\u30b9\u3092\u4ed8\u3051\u3066
\n$\\quad \\quad \\quad$(\u5185)\u5206\u70b9\u306e\u516c\u5f0f\u3092\u9069\u7528\u3059\u308b
\n$\\overrightarrow{\\mathrm{OQ}}=\\dfrac{3}{-1+3}\\begin{pmatrix}1\\\\2\\end{pmatrix}+\\dfrac{-1}{-1+3}\\begin{pmatrix}5\\\\2\\end{pmatrix}=\\begin{pmatrix}\\dfrac{3\\cdot1+(-1)\\cdot5}{2}\\\\\\dfrac{3\\cdot2+(-1)\\cdot2}{2}\\end{pmatrix}=\\begin{pmatrix}-1\\\\2\\end{pmatrix}$ \uff08\u7b54\uff09<\/span><\/div><\/div>\n

(3) $\\triangle{\\mathrm{ABC}}$ \u306e\u91cd\u5fc3 ${\\mathrm{G}}$ \u306e\u4f4d\u7f6e\u30d9\u30af\u30c8\u30eb\u3092\u6c42\u3081\u3088<\/p>\n

<\/span>\u89e3\u7b54<\/div>
(\u516c\u5f0f)\u3000$\\triangle{\\mathrm{ABC}}$ \u306e\u91cd\u5fc3 ${\\mathrm{G}}$ \u306e\u4f4d\u7f6e\u30d9\u30af\u30c8\u30eb\uff1a $\\overrightarrow{\\mathrm{OG}}=\\dfrac{1}{3}\\left(\\overrightarrow{\\mathrm{OA}}+\\overrightarrow{\\mathrm{OB}}+\\overrightarrow{\\mathrm{OC}}\\right)$
\n$\\overrightarrow{\\mathrm{OG}}=\\dfrac{1}{3}\\left \\lbrace \\begin{pmatrix}1\\\\2\\end{pmatrix}+\\begin{pmatrix}5\\\\2\\end{pmatrix}+\\begin{pmatrix}3\\\\2\\end{pmatrix}\\right \\rbrace=\\begin{pmatrix}\\dfrac{1+5+3}{3}\\\\\\dfrac{2+2+2}{3}\\end{pmatrix}=\\begin{pmatrix}3\\\\2\\end{pmatrix}$ \uff08\u7b54\uff09<\/span><\/div><\/div>\n

\u76f4\u7dda\u306e\u65b9\u5411\u30d9\u30af\u30c8\u3068\u6cd5\u7dda\u30d9\u30af\u30c8\u30eb<\/span><\/p>\n

4. 3\u70b9 ${\\mathrm{A}}(1, 3), {\\mathrm{B}}(4, 5), {\\mathrm{C}}(2, -1)$ \u306b\u3064\u3044\u3066\u6b21\u306e\u554f\u3044\u306b\u7b54\u3048\u3088<\/p>\n

(1) $\\overrightarrow{\\mathrm{AB}}$ \u3092\u65b9\u5411\u30d9\u30af\u30c8\u30eb\u3068\u3057\u3001\u70b9 ${\\mathrm{C}}$ \u3092\u901a\u308b\u76f4\u7dda\u306e\u65b9\u7a0b\u5f0f\u3092\u6c42\u3081\u3088<\/p>\n

<\/span>\u89e3\u7b54<\/div>
(\u516c\u5f0f) \u70b9 ${\\mathrm{P}}_0(x_0, y_0)$ \u3092\u901a\u308a\u3001\u65b9\u5411\u30d9\u30af\u30c8\u30eb $\\vec{v}=\\begin{pmatrix}a\\\\b\\end{pmatrix}$ \u306e\u76f4\u7dda\uff1a $\\dfrac{x-x_0}{a}=\\dfrac{y-y_0}{b}$
\n$\\vec{v}=\\overrightarrow{\\mathrm{AB}}=\\begin{pmatrix}4\\\\5\\end{pmatrix}-\\begin{pmatrix}1\\\\3\\end{pmatrix}=\\begin{pmatrix}3\\\\2\\end{pmatrix}$ \u3088\u308a
\n$\\dfrac{x-2}{3}=\\dfrac{y+1}{2} \\Rightarrow 2(x-2)=3(y+1) \\Rightarrow 2x-3y-7=0 \\Rightarrow y=\\dfrac{2}{3}x-\\dfrac{7}{3}$ \uff08\u7b54\uff09<\/span><\/div><\/div>\n

(2) $\\overrightarrow{\\mathrm{AB}}$ \u3092\u6cd5\u7dda\u30d9\u30af\u30c8\u30eb\u3068\u3057\u3001\u70b9 ${\\mathrm{C}}$ \u3092\u901a\u308b\u76f4\u7dda\u306e\u65b9\u7a0b\u5f0f\u3092\u6c42\u3081\u3088<\/p>\n

<\/span>\u89e3\u7b54<\/div>
(\u516c\u5f0f) \u70b9 ${\\mathrm{P}}_0(x_0, y_0)$ \u3092\u901a\u308a\u3001\u6cd5\u7dda\u30d9\u30af\u30c8\u30eb $\\vec{n}=\\begin{pmatrix}a\\\\b\\end{pmatrix}$ \u306e\u76f4\u7dda\uff1a $a(x-x_0)+b(y-y_0)=0$
\n$\\vec{n}=\\overrightarrow{\\mathrm{AB}}=\\begin{pmatrix}4\\\\5\\end{pmatrix}-\\begin{pmatrix}1\\\\3\\end{pmatrix}=\\begin{pmatrix}3\\\\2\\end{pmatrix}$ \u3088\u308a
\n$3(x-2)+2(y+1)=0 \\Rightarrow 3x+2y-4=0 \\Rightarrow y=\\dfrac{-3}{2}x+2$ \uff08\u7b54\uff09<\/span><\/div><\/div>\n

2\u70b9\u3092\u76f4\u5f84\u306e\u4e21\u7aef\u3068\u3059\u308b\u5186<\/span><\/p>\n

5. 2\u70b9 ${\\mathrm{A}}(-5, 2), {\\mathrm{B}}(3, -6)$ \u3092\u76f4\u5f84\u306e\u4e21\u7aef\u3068\u3059\u308b\u5186\u306e\u65b9\u7a0b\u5f0f\u3092\u6c42\u3081\u3088<\/p>\n

<\/span>\u89e3\u7b54<\/div>
(\u516c\u5f0f) ${\\mathrm{A}}(a_1, a_2) {\\mathrm{B}}(b_1, b_2)$ \u3092\u76f4\u5f84\u306e\u4e21\u7aef\u3068\u3059\u308b\u5186\uff1a $(x-a_1)(x-b_1)+(y-a_2)(y-b_2)=0$<\/span><\/p>\n

$(x+5)(x-3)+(y-2)(y+6)=0 \\Rightarrow x^2+y^2+2x+4y-27=0$ \uff08\u7b54\uff09
\n$\\left( \\Rightarrow (x+1)^2+(y+2)^2=32 \\Rightarrow \u4e2d\u5fc3\\:(-1,-2) \\; \u534a\u5f84\\:\\sqrt{32}=4\\sqrt{2} \\; \u306e\u5186\\right)$ <\/span><\/div><\/div>\n<\/div><\/div><\/div>\n","protected":false},"excerpt":{"rendered":"

\u518d\u751f\u30ea\u30b9\u30c8 4H1_\u524d\u534a 4H1_\u5f8c\u534a 2H1 4H2_\u524d\u534a 4H2_\u5f8c\u534a 2H2<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[42,14,4,8],"tags":[],"class_list":["post-2222","post","type-post","status-publish","format-standard","hentry","category-si","category-top","category-math2","category-math2-2q"],"_links":{"self":[{"href":"https:\/\/y-page.y.kumamoto-nct.ac.jp\/wp\/Math_23\/index.php?rest_route=\/wp\/v2\/posts\/2222","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/y-page.y.kumamoto-nct.ac.jp\/wp\/Math_23\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/y-page.y.kumamoto-nct.ac.jp\/wp\/Math_23\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/y-page.y.kumamoto-nct.ac.jp\/wp\/Math_23\/index.php?rest_route=\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/y-page.y.kumamoto-nct.ac.jp\/wp\/Math_23\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=2222"}],"version-history":[{"count":458,"href":"https:\/\/y-page.y.kumamoto-nct.ac.jp\/wp\/Math_23\/index.php?rest_route=\/wp\/v2\/posts\/2222\/revisions"}],"predecessor-version":[{"id":2998,"href":"https:\/\/y-page.y.kumamoto-nct.ac.jp\/wp\/Math_23\/index.php?rest_route=\/wp\/v2\/posts\/2222\/revisions\/2998"}],"wp:attachment":[{"href":"https:\/\/y-page.y.kumamoto-nct.ac.jp\/wp\/Math_23\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=2222"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/y-page.y.kumamoto-nct.ac.jp\/wp\/Math_23\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=2222"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/y-page.y.kumamoto-nct.ac.jp\/wp\/Math_23\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=2222"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}