گام به گام تمرین صفحه 18 درس 1 ریاضی (2) (هندسۀ تحلیلی و جبر)

1)

(الف \(2{x^4} - 7{x^2} - 4 = 0\)

\(\begin{array}{l} \Rightarrow t = {x^2} \Rightarrow 2{t^2} - 7t - 4 = 0 \Rightarrow \left. \begin{array}{l}a = 2\\b = - 7\\c = - 4\end{array} \right\}\\ \Rightarrow \Delta = {b^2} - 4ac = {\left( { - 7} \right)^2} - 4\left( 2 \right)\left( { - 4} \right) = 49 + 32 = 81\\t = \frac{{ - b \pm \sqrt \Delta }}{{2a}} = \frac{{7 \pm \sqrt {81} }}{4}\\ \Rightarrow \left\{ \begin{array}{l}{t_1} = {x^2} = \frac{{7 + 9}}{4} = 4 \Rightarrow x = \pm 2 \Rightarrow \left\{ \begin{array}{l}{x_1} = 2\\{x_2} = - 2\end{array} \right.\\{t_2} = {x^2} = \frac{{7 - 9}}{4} = - \frac{1}{2}\;\; \otimes \end{array} \right.\end{array} \)

(ب \({x^4} + 3{x^2} + 2 = 0\)

\(\begin{array}{l} \Rightarrow t = {x^2} \Rightarrow {t^2} + 3t + 2 = 0\\ \Rightarrow \left( {t + 2} \right)\left( {t + 1} \right) = 0\\ \Rightarrow \left\{ \begin{array}{l}t + 1 = 0 \Rightarrow {t_1} = {x^2} =  - 1\;\; \otimes \\t + 2 = 0 \Rightarrow {t_2} = {x^2} =  - 2\;\; \otimes \end{array} \right.\end{array}\)

معادله جواب حقیقی ندارد.

2)

\(\begin{array}{l}\left. \begin{array}{l}\alpha  = 1 - \sqrt 2 \\\beta  = 1 + \sqrt 2 \end{array} \right\}\\ \Rightarrow \left\{ \begin{array}{l}S = \alpha  + \beta  = 1 - \sqrt 2  + 1 + \sqrt 2  = 2\\P = \alpha  \cdot \beta  = \left( {1 - \sqrt 2 } \right) \times \left( {1 + \sqrt 2 } \right) = {1^2} - {\left( {\sqrt 2 } \right)^2} =  - 1\end{array} \right.\\ \Rightarrow {x^2} - Sx + P = 0 \Rightarrow {x^2} - 2x - 1 = 0\end{array}\)

3)

(الف \(f\left( x \right) =  - 2{x^2} + 8x - 5\)

\(\Rightarrow a =  - 2 < 0 \Rightarrow\)

تابع دارای ماکزیمم است.

\(\begin{array}{l} \Rightarrow \left\{ \begin{array}{l}{x_S} =  - \frac{b}{{2a}} =  - \frac{8}{{2\left( { - 2} \right)}} = 2\\ \Rightarrow {y_S} = f\left( 2 \right) =  - 2{\left( 2 \right)^2} + 8\left( 2 \right) - 5 = 3\end{array} \right.\\ \Rightarrow S\left( {2\;,\;3} \right)\end{array}\)

(ب \(g\left( x \right) = 3{x^2} + 6x + 5\)

\(\Rightarrow a = 3 > 0 \Rightarrow\)

تابع دارای مینیمم است

\(\begin{array}{l} \Rightarrow \left\{ \begin{array}{l}{x_S} =  - \frac{b}{{2a}} =  - \frac{6}{{2\left( 3 \right)}} =  - 1\\ \Rightarrow {y_S} = g\left( { - 1} \right) = 3{\left( { - 1} \right)^2} + 6\left( { - 1} \right) + 5 = 2\end{array} \right.\\ \Rightarrow S\left( { - 1\;,\;2} \right)\end{array}\)

4)

الف)

\(\begin{array}{l}h\left( t \right) = 100t - 5{t^2}\quad \left( {t \ge 0} \right)\quad \\ \Rightarrow {t_S} =  - \frac{b}{{2a}} =  - \;\frac{{100}}{{2\left( { - \;5} \right)}} = 10\end{array}\)

ب)

\({h_S} = h\left( {{t_S}} \right) = 100\left( {10} \right) - 5{\left( {10} \right)^2} = 1000 - 500 = 500\)

پ)

\(\begin{array}{l}h\left( t \right) = 100t - 5{t^2} = 0 \Rightarrow 5t\left( {20 - t} \right) = 0\\ \Rightarrow \left\{ \begin{array}{l}5t = 0 \Rightarrow t = 0\\20 - t = 0 \Rightarrow t = 20\end{array} \right.\end{array}\)

t = 0 غ ق ق   ,   t = 20  ق ق

5)

الف)

\(\begin{array}{l}P = {P_c} + 2a \Rightarrow P = 2\pi  \times \frac{b}{2} + 2a\\ \Rightarrow \pi b + 2a = 1500 \Rightarrow a = 750 - \frac{\pi }{2}b\\{S_r} = a \cdot b \Rightarrow {S_r} = \left( {750 - \frac{\pi }{2}b} \right)b =  - \frac{\pi }{2}{b^2} + 750b \\\Rightarrow  - \frac{\pi }{2} < 0 \Rightarrow {b_S} =  - \frac{{750}}{{2\left( { - \frac{\pi }{2}} \right)}} = \frac{{750}}{\pi } \\\Rightarrow {a_S} = 375 \Rightarrow {S_{rS}} = {a_S} \cdot {b_S} = \frac{{281,250}}{\pi }\end{array} \)

\({P_c}\) محیط دایره  ,  \({S_r}\) مساحت مستطیل  ,  \({S_c}\) مساحت دایره

ب)

\(\begin{array}{l}a = 750 - \frac{\pi }{2}b\\S = {S_r} + {S_c} \Rightarrow S =  - \frac{\pi }{2}{b^2} + 750b + {\left( {\frac{b}{2}} \right)^2}\pi \\ \Rightarrow S =  - \frac{\pi }{4}{b^2} + 750b\\ \Rightarrow {b_S} =  - \frac{{750}}{{2\left( { - \frac{\pi }{4}} \right)}} = \frac{{1500}}{\pi } \Rightarrow {a_S} = 0\\ \Rightarrow S =  - \frac{\pi }{4}{\left( {\frac{{1500}}{\pi }} \right)^2} + 750\left( {\frac{{1500}}{\pi }} \right) = \frac{{562,500}}{\pi }\end{array}\)

6)

الف)

\(\begin{array}{*{20}{l}}{f\left( x \right) = a{x^2} + bx + c}\\{f\left( 0 \right) = 0 \Rightarrow c = 0}\\{f\left( 2 \right) = 0 \Rightarrow 4a + 2b = 0}\\{f\left( { - \frac{b}{{2a}}} \right) = {\rm{\;}} - 4 \Rightarrow a{{\left( { - \frac{b}{{2a}}} \right)}^2} + b\left( { - \frac{b}{{2a}}} \right) = {\rm{\;}} - 4}\\{ \Rightarrow \frac{{{b^2}}}{{4a}} - \frac{{{b^2}}}{{2a}} = {\rm{\;}} - 4 \Rightarrow {b^2} = 16a}\\{\left\{ {\begin{array}{*{20}{l}}{{b^2} = 16a}\\{4a + 2b = 0}\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{l}}{{b^2} = 16a}\\{16a + 8b = 0}\end{array}} \right. \Rightarrow {b^2} + 8b = 0}\\{ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{b = 0 \Rightarrow a = 0\; \otimes }\\{b = {\rm{\;}} - 8 \Rightarrow a = 4}\end{array}} \right. \Rightarrow f\left( x \right) = 4{x^2} - 8x}\end{array}\)

ب)

\(\begin{array}{l}f\left( x \right) = a{x^2} + bx + c\\f\left( 0 \right) = 4 \Rightarrow c = 4\\f\left( 2 \right) = 0 \Rightarrow 4a + 2b + 4 = 0\\\left\{ \begin{array}{l}\frac{{ - \;b}}{{2a}} = 3\\4a + 2b + 4 = 0\end{array} \right. \Rightarrow \left\{ \begin{array}{l}6a + b = 0\\2a + b =  - 2\end{array} \right. \Rightarrow \left\{ \begin{array}{l}a = \frac{1}{2}\\b =  - 3\end{array} \right.\\f\left( x \right) = \frac{1}{2}{x^2} - 3x + 4\end{array}\)

پ)

\(\begin{array}{l}f\left( x \right) = a{x^2} + bx + c\\f\left( 0 \right) = 4 \Rightarrow c = 4\\f\left( 2 \right) = 0 \Rightarrow 4a + 2b + 4 = 0\\\left\{ \begin{array}{l}\frac{{ - \;b}}{{2a}} = 2\\4a + 2b + 4 = 0\end{array} \right. \Rightarrow \left\{ \begin{array}{l}4a + b = 0\\2a + b =  - 2\end{array} \right. \Rightarrow \left\{ \begin{array}{l}a = 1\\b =  - 4\end{array} \right.\\f\left( x \right) = {x^2} - 4x + 4\end{array}\)

ت)

\(\begin{array}{l}f\left( x \right) = a{x^2} + bx + c\\f\left( 0 \right) = 3 \Rightarrow c = 3\\f\left( { - 1} \right) = 0 \Rightarrow a - b + 3 = 0\\\left\{ \begin{array}{l}\frac{{ - \;b}}{{2a}} = 0\\a - b + 3 = 0\end{array} \right. \Rightarrow \left\{ \begin{array}{l}b = 0\\a - b =  - 3\end{array} \right. \Rightarrow \left\{ \begin{array}{l}a =  - 3\\b = 0\end{array} \right.\\f\left( x \right) =  - 3{x^2} + 3\end{array}\)

ث)

\(\begin{array}{l}f\left( x \right) = a{x^2} + bx + c\\f\left( 1 \right) = 0 \Rightarrow a + b + c = 0\\f\left( 2 \right) = 0 \Rightarrow 4a + 2b + c = 1\\\frac{{ - \;b}}{{2a}} = 2 \Rightarrow \left\{ \begin{array}{l}4a + b = 0\\a + b + c = 0\\4a + 2b + c = 1\end{array} \right. \Rightarrow \left\{ \begin{array}{l}a =  - 3\\b = 4\\c =  - 3\end{array} \right.\\f\left( x \right) =  - 3{x^2} + 4x - 3\end{array} \)

ج)

\(\begin{array}{l}f\left( x \right) = a{x^2} + bx + c\\f\left( 0 \right) =  - 2 \Rightarrow c =  - 2\\f\left( 1 \right) =  - 1 \Rightarrow a + b - 2 =  - 1\\\left\{ \begin{array}{l}\frac{{ - \;b}}{{2a}} = 1\\a + b - 2 =  - 1\end{array} \right. \Rightarrow \left\{ \begin{array}{l}2a + b = 0\\a + b = 1\end{array} \right. \Rightarrow \left\{ \begin{array}{l}a =  - 1\\b = 2\\c =  - 2\end{array} \right.\\f\left( x \right) =  - \;{x^2} + 2x - 2\end{array}\)