نمونه سوالات فصل 1 ریاضی یازدهم همراه با جواب تشریحی

\(\left. \begin{array}{l}{x_M} = \frac{{{x_B} + {x_C}}}{2} = \frac{{2 + ( - 2)}}{2} = 0\\\\{y_M} = \frac{{{y_B} + {y_C}}}{2} = \frac{{0 + ( - 2)}}{2} = - 1\end{array} \right\} \Rightarrow M(0, - 1)\)

\(\begin{array}{l}\left. \begin{array}{l}{x_M} = \frac{{{x_A} + {x_B}}}{2} \Rightarrow 1 = \frac{{2 - b + 1}}{2} \Rightarrow 2 = - b + 3 \Rightarrow b = 1\\\\{y_m} = \frac{{{y_A} + {y_B}}}{2} \Rightarrow 5 = \frac{{a + 7}}{2} \Rightarrow 10 = a + 7 \Rightarrow a = 3\end{array} \right|\\\\ \Rightarrow (b,a) = (1,3)\end{array}\)

\(\begin{array}{l}AB = \sqrt {{{({x_A} - {x_B})}^2} + {{({y_A} - {y_B})}^2}} = \sqrt {{{(2 - 0)}^2} + {{(3 - 7)}^2}} \\\\ = \sqrt {4 + 16} = \sqrt {20} \end{array}\)

\(\begin{array}{l}AM = \sqrt {{{({x_A} - {x_M})}^2} + {{({y_A} - {y_M})}^2}} \\\\XM = \frac{{{x_B} + {x_c}}}{2} = \frac{{3 + 0}}{2} = \frac{3}{2}\\\\YM = \frac{{{y_B} + {y_C}}}{2} = \frac{{0 + 2}}{2} = 1\\\\AM = \sqrt {{{(2 - \frac{3}{2})}^2} + {{(3 - 1)}^2}} = \sqrt {\frac{1}{4} + 4} = \frac{{\sqrt {17} }}{2}\end{array}\)

\(\begin{array}{l}{m_{CH}} \times {m_{AB}} = - 1\\\\{m_{AB}} = \frac{{{y_A} - {y_B}}}{{{x_A} - {x_B}}} = \frac{{3 - 0}}{{2 - 3}} = - 3 \Rightarrow {m_{CH}} = \frac{1}{3}\end{array}\)

\(\begin{array}{l}\left. \begin{array}{l}{x_M} = \frac{{{x_A} + {x_B}}}{2} = \frac{{\frac{1}{3} + \frac{2}{3}}}{2} = \frac{1}{2}\\\\{y_M} = \frac{{{y_A} + {y_{|B}}}}{2} = \frac{{ - 2 + 3}}{2} = \frac{1}{2}\end{array} \right\} \Rightarrow M(\frac{1}{2},\frac{1}{2})\\\\{m_D} = {m_{{D^`}}} = - \frac{a}{b} = \frac{1}{2}\\\\{D^`}:y - {y_M} = m(x - {x_M}) \Rightarrow y - \frac{1}{2} = \frac{1}{2}(x - \frac{1}{2})\\\\ \Rightarrow y = \frac{1}{2}x + \frac{1}{4}\end{array}\)

نقطه M روی ضلعی که معادله ان را داریم قرار ندارد. پس فاصله نقطه M از خط مورد نظر برابر ضلع مربع می باشد.

\(\begin{array}{l}{x_A} + {x_C} = {x_B} + {x_D}\\\\{y_A} + {y_C} = {y_B} + {y_D}\\\\1 + 3 = - 2 + {x_D}\\\\ - 1 + 5 = 3 + {y_D}\\\\{x_D} = 6\\\\{y_B} = 1 \Rightarrow D(6,1)\end{array}\)

الف)

\(\begin{array}{l}{m_{Ab}} \times {m_{CH}} = - 1 \Rightarrow {m_{Ab}} \times \frac{2}{3} = - 1 \Rightarrow {m_{AB}} = - \frac{3}{2}\\\\AB:y - {y_A} = m(x - {x_A}) \Rightarrow y + 1 = - \frac{3}{2}(x - 4)\\\\ \Rightarrow y = - \frac{3}{2}x + 5\end{array}\)

ب)

\(\begin{array}{l}\left\{ \begin{array}{l}CM:2x = - 3y \Rightarrow 2x = - 3( - \frac{3}{2}x + 5)\\\\ \Rightarrow 2x = \frac{9}{2}x - 15 \Rightarrow x = 6\\\\AB:y = - \frac{3}{2}x + 5 \Rightarrow y = - \frac{3}{2}(6) + 5 \Rightarrow y = - 4\end{array} \right.\\\\ \Rightarrow M(6, - 4)\\\\\\\left\{ \begin{array}{l}{x_B} = 2{x_M} - {x_A} \Rightarrow {x_B} = 2(6) - 4 = 8\\\\{y_B} = 2{y_M} - {y_A} \Rightarrow {y_B} = 2( - 4) + 1 = - 7\end{array} \right. \Rightarrow B(8, - 7)\end{array}\)

\(\begin{array}{l}(2m - n + 1)( - 3) + 6m + 9 = 0 \Rightarrow 3n - 3 + 9 = 0 \Rightarrow n = - 2\\\\m + 2n - 3 = 0 \Rightarrow m + 2( - 2) - 3 = 0 \Rightarrow m = 7\end{array}\)

(الف

\(\begin{array}{l}O = \frac{{A + B}}{2} = (1,3)\\\\OA = R = \sqrt {{{( - 1 - 1)}^2} + {{(4 - 3)}^2}} = \sqrt {4 + 1} = \sqrt 5 \end{array}\)

(ب

\(\begin{array}{l}2\pi R = 2\pi (\sqrt 5 ) = 2\sqrt 5 \pi \\\\\pi {R^2} = \pi {(\sqrt 5 )^2} = 5\pi \end{array}\)

\(\begin{array}{l}AB = BC = CD = DA\\\\AB = \sqrt {{{( - 2 - 6)}^2} + {{(2 + 13)}^2}} = \sqrt {289} = 17\\\\CD = \sqrt {{{(21 - 13)}^2} + {{( - 5 - 10)}^2}} = 17\\\\DA = \sqrt {{{(6 - 21)}^2} + {{( - 13 + 5)}^2}} = 17\\\\AC = \sqrt {{{({x_C} - {x_A})}^2} + {{({y_C} - {y_A})}^2}} \\\\ = \sqrt {{{(13 - 6)}^2} + {{(10 + 13)}^2}} = \sqrt {578} \\\\:A{C^2} = A{B^2} + B{C^2} \Rightarrow 578 = 289 + 289 = 578\end{array}\)

\(\begin{array}{l}{x_M} = \frac{{{x_A} + {x_B}}}{2} \Rightarrow - 1 = \frac{{1 + {x_B}}}{2} \Rightarrow {x_B} = - 3\\\\{y_M} = \frac{{{y_A} + {y_B}}}{2} \Rightarrow 4 = \frac{{2 + {y_B}}}{2} \Rightarrow {y_B} = 6\end{array}\)

\(\begin{array}{l}\left\{ \begin{array}{l}3x + 2y - 1 = 0 \Rightarrow 3x + 2(\frac{2}{3}x + \frac{5}{3}) - 1 = 0 \Rightarrow x = \frac{{ - 7}}{{13}}\\\\y = \frac{2}{3}x + \frac{5}{3} \Rightarrow y = \frac{2}{3}( - \frac{7}{{13}}) + \frac{5}{{13}} \Rightarrow y = \frac{{17}}{{13}}\end{array} \right.\\\\ \Rightarrow H( - \frac{7}{{13}},\frac{{17}}{{13}})\end{array}\)

\(\begin{array}{l}\left. \begin{array}{l}AB = \sqrt {{{(2 - 1)}^2} + {{(5 - 2)}^2}} = \sqrt {1 + 9} = \sqrt {10} \\\\AC = \sqrt {{{(4 - 1)}^2} + {{(1 - 2)}^2}} = \sqrt {9 + 1} = \sqrt {10} \end{array} \right\}\\\\AB = AC\\\\{m_{AB}} = \frac{{5 - 2}}{{2 - 1}} = 3\,\,.\,\,{m_{AC}} = \frac{{1 - 2}}{{4 - 1}} = - \frac{1}{3}\\\\{m_{AB}} \times {m_{AC}} = 3 \times - \frac{1}{3} = - 1\end{array}\)

\({x_1}^2 + {x_2}^2 = {S^2} - 2P = {(\frac{7}{2})^2} - 2(\frac{4}{2}) = \frac{{33}}{4}\)

\(\frac{{{x_1}}}{{{x_2}}} + \frac{{{x_2}}}{{{x_1}}} = \frac{{23}}{8}\)

\((\frac{{{x_1}}}{{{x_2}}} + \frac{{{x_2}}}{{{x_1}}}) - 2({x_2}{x_1}) = \frac{{33}}{8} - 4 = \frac{1}{8}\)

\(\begin{array}{l}\alpha + \beta = - \frac{b}{a},\alpha \beta = - \frac{c}{a} = 1\\\\{\alpha ^2} + {\beta ^2} + \frac{1}{\alpha } + \frac{1}{\beta } = {(\alpha + \beta )^2} - 2\alpha \beta + \frac{{\alpha + \beta }}{{\alpha \beta }}\end{array}\)

\(\begin{array}{l}\alpha = 3\beta \Rightarrow \alpha + \beta = 4\beta \Rightarrow \beta = \frac{1}{m}\\\\m{(\frac{1}{m})^2} - 4(\frac{1}{m}) + 1 = 0 \Rightarrow \frac{1}{m} - \frac{4}{m} = - 1 \Rightarrow m = 3\end{array}\)

\(\left\{ \begin{array}{l}\alpha = \beta + 2\\\\\alpha + \phi = 4\\\\\alpha \times \beta = \frac{m}{4}\end{array} \right. \Rightarrow \alpha = 3,\beta = 1,3 \times 1 = \frac{m}{4} \to m = 12\)

\(\begin{array}{l}f(x) = - 2{x^2} + 8x - 5\\\\a = - 2 < 0\\\\x = - \frac{b}{{2a}} \Rightarrow x = 2\\\\f(2) = - 2 \times 4 + 8 \times 2 - 5 = 3 \Rightarrow f(2) = 3\end{array}\)

\(\begin{array}{l}g(x) = 3{x^2} + 6x + 5\\\\a = 3 > 0\\\\x = - \frac{b}{{2a}} \Rightarrow x = - 1\\\\g( - 1) = 3 \times 1 + 6( - 1) + 5 = 2 \Rightarrow f( - 1) = 2\end{array}\)

\(\begin{array}{l}\frac{1}{4}{x^2} - \frac{9}{4}x + 3 < 1{x^2} - = 9x + 12 < 4\\\\ \Rightarrow {x^2} - 9x + < 0 \Rightarrow 1 < x < 8 \Rightarrow \left\{ \begin{array}{l}a = 1\\\\b = 8\end{array} \right.\end{array}\)

\(\begin{array}{l}2{x^2} - 4x + 1 < {x^2} + 4x - 6 \Rightarrow {x^2} - 8x + 7 < 0\\\\ \Rightarrow (x - 7)(x - 1) < 0 \Rightarrow 1 < x < 7\end{array}\)

\(\begin{array}{l}{x_0} = \frac{{ - b}}{{2a}} = - 1 \Rightarrow b = 2a\\\\S( - 1,4)4 = a{( - 1)^2} + b( - 1) + 5\\\\ \Rightarrow a - b + 5 = 4 \Rightarrow a - b = - 1\\\\a - b = - 1a - 2a = - 1\\\\ \Rightarrow - a = - 1 \Rightarrow a = 1 \Rightarrow b = 2\\\\y = {x^2} + 2x + 5\end{array}\)

\(\begin{array}{l}A(0,1) \Rightarrow 1 = a{(0)^2} + b(0) + c \Rightarrow c = 1\\\\B( - 1,0) \Rightarrow 0 = a{( - 1)^2} + b( - 1) + 1 \Rightarrow a - b = - 1\\\\C(4,0) \Rightarrow 0 = a{(4)^2} + b(4) + 1 \Rightarrow 16a + 4b = - 1\\\\4 \times \left\{ \begin{array}{l}a - b = - 1\\\\16a + 4b = - 1\end{array} \right. \Rightarrow \left\{ \begin{array}{l}4a - 4b = - 4\\\\16a + 4b = - 1\end{array} \right. \Rightarrow 20a = - 5\\\\ \Rightarrow a = - \frac{1}{4},b = \frac{3}{4}\\\\y = - \frac{1}{4}{x^2} + \frac{3}{4}x + 1\end{array}\)

\(\begin{array}{l}a < 0\\\\\frac{{ - b}}{a} < 0 \Rightarrow - b > 0 \Rightarrow b < 0\\\\\frac{c}{a} < 0 \Rightarrow c > 0\end{array}\)

تعداد ریشه ها برابر 2 است.

(الف

\(a > 0,b < 0\)

(ب

\(c = 0\)

\(\begin{array}{l}a < 0\\\\b < 0\\\\c < 0\end{array}\)

معادله 2 جواب دارد.

\(\begin{array}{l}3{x^2} - 12x + 2y = - 20 \Rightarrow 2y = 3{x^2} + 12x - 20\\\\ \Rightarrow y = - \frac{3}{2}{x^2} + 6x - 10\\\\{x^2} < 0\\\\{x_{\max }} = \frac{{ - b}}{{2a}} = \frac{{ - 6}}{{2( - \frac{3}{2})}} = 2 \Rightarrow y = \frac{{ - 3}}{2}{(2)^2} + 6(2) - 10 = - 4\end{array}\)

\(\begin{array}{l}\{ (x,y)|{x^2} + 6x + 2y + 5 = 0\} \\\\x = \frac{3}{{2( - \frac{1}{2})}} = - 3 \Rightarrow y = \frac{{ - 9}}{2} + 9 - \frac{5}{2} = 2 \Rightarrow {y_{\max }} = 2\end{array}\)

\(\left\{ \begin{array}{l}2\alpha + \beta = 4 \Rightarrow \alpha + \alpha + \beta = 4\\\\\alpha + \beta = - m\end{array} \right. \Rightarrow \alpha = 4 + m\)

\(\frac{c}{a} < 0 \Rightarrow \frac{{{a^2} - 4a}}{2} < 0 \Rightarrow a \in (0,4)\)

\(\begin{array}{l}|{x_1} - {x_2}| = \frac{{\sqrt \Delta }}{{|a|}} \Rightarrow \frac{{\sqrt \Delta }}{1} = 1 \Rightarrow \Delta = 1 \Rightarrow {P^2} - 48 = 1\\\\ \Rightarrow {P^2} = 49 \Rightarrow \left\{ \begin{array}{l}P = 7\\\\P = - 7\end{array} \right.\end{array}\)