جواب تمرین صفحه 15 درس 1 حسابان یازدهم

\(\begin{array}{l}\left\{ \begin{array}{l}\alpha = \frac{1}{3}\\\\\beta = \frac{2}{3}\end{array} \right. \Rightarrow \left\{ \begin{array}{l}\alpha + \beta = 1\\\\\alpha \beta = \frac{2}{9}\end{array} \right.\\\\ \Rightarrow {x^2} - (\alpha + \beta )x + \alpha \beta = 0\\\\ \Rightarrow {x^2} - x + \frac{2}{9} = 0\end{array}\)

\(\begin{array}{l}\left\{ \begin{array}{l}\alpha = \alpha \\\\\beta = 2\alpha \end{array} \right. \Rightarrow \left\{ \begin{array}{l}\alpha + \beta = 3\alpha \\\\\alpha \beta = 2{\alpha ^2}\end{array} \right.\\\\ \Rightarrow {x^2} - (\alpha + \beta )x + \alpha \beta = 0\\\\ \Rightarrow {x^2} - 3\alpha x + 2{\alpha ^2} = 0\end{array}\)

مسئله بیشمار جواب دارد.

\(\begin{array}{l}\left\{ \begin{array}{l}x = 2 \Rightarrow x - 2 = 0\\\\x = 2 \Rightarrow x - 2 = 0\end{array} \right. \Rightarrow P\left( x \right) = a{\left( {x - 2} \right)^2}\\\\ \Rightarrow P(0) = - 2 \Rightarrow a{(0 - 2)^2} = 4a = - 2\\\\ \Rightarrow a = - \frac{1}{2}\\\\P\left( x \right) = - \frac{1}{2}{\left( {x - 2} \right)^2}\end{array}\)

\(\begin{array}{l}\left\{ \begin{array}{l}x = 1 \Rightarrow x - 1 = 0\\\\x = - 3 \Rightarrow x + 3 = 0\end{array} \right. \Rightarrow P(x) = a(x - 1)(x + 3)\\\\ \Rightarrow P( - 1) = - 2 \Rightarrow a( - 1 - 1)( - 1 + 3) = - 4a = - 2\\\\ \Rightarrow a = \frac{1}{2}\\\\ \Rightarrow P(x) = \frac{1}{2}(x - 1)(x + 3)\end{array}\)

\(\begin{array}{l}h(x) = - 0/03x(x - 36) \Rightarrow h(x) = 0\\\\ \Rightarrow \left\{ \begin{array}{l} - 0/03x = 0 \Rightarrow x = 0\\\\x - 36 = 0 \Rightarrow x = 36\end{array} \right.\end{array}\)

بیشترین ارتفاع توپ از زمین همان عرض نقطه رأس می باشد؛ بنابراین:

\(\begin{array}{l}{x_ \circ } = \frac{{ - b}}{{2a}} = \frac{{ - \left( { - 0/03 \times 36} \right)}}{{2 \times \left( { - 0/03} \right)}} = 18\\\\ \Rightarrow {y_ \circ } = h\left( {{x_ \circ }} \right) = h\left( {18} \right) = 9/72\end{array}\)

9/72 متر

\(\begin{array}{l}f\left( x \right) = {x^3} - 4x = x\left( {{x^2} - 4} \right) = x\left( {x - 2} \right)\left( {x + 2} \right)\\\\ \Rightarrow \left\{ \begin{array}{l}{x_1} = 0\\\\x - 2 = 0 \Rightarrow {x_2} = 2\\\\x + 2 = 0 \Rightarrow {x_3} = - 2\end{array} \right.\end{array}\)

\(\begin{array}{l}g(x) = 2{x^3} + {x^2} + 3x = x(2{x^2} + x + 3)\\\\ \Rightarrow \left\{ \begin{array}{l}{x_1} = 0\\\\2{x^2} + x + 3 = 0 \Rightarrow \Delta = {1^2} - 4(2)(3) = - 23 < 0\end{array} \right.\end{array}\)

\(\begin{array}{l}\left\{ \begin{array}{l}h(x) = {x^4} + 3{x^2} + 5\\\\t = {x^2}\end{array} \right. \Rightarrow {x^4} + 3{x^2} + 5 = {t^2} + 3x + 5\\\\ \Rightarrow \Delta = {3^2} - 4(1)(5) = - 11 < 0\end{array}\)

این تابع، صفر ندارد.

\(\begin{array}{l}t = {x^2} \Rightarrow {x^4} - 3{x^2} - 4 = 0 \Rightarrow {t^2} - 3t - 4 = 0\\\\ \Rightarrow (t + 1)(t - 4) = 0\\\\ \Rightarrow \left\{ \begin{array}{l}t + 1 = 0 \Rightarrow t = - 1 \Rightarrow {x^2} = - 1 \otimes \\\\t - 4 = 0 \Rightarrow t = 4 \Rightarrow {x^2} = 4 \Rightarrow x = \pm 2\end{array} \right.\\\\ \Rightarrow \left\{ \begin{array}{l}x = - 2\\\\x = 2\end{array} \right.\end{array}\)

\(\begin{array}{l}t = \frac{{{x^2}}}{3} - 2 \Rightarrow {(\frac{{{x^2}}}{3} - 2)^2} - 7(\frac{{{x^2}}}{3} - 2) + 6 = 0\\\\ \Rightarrow {t^2} - 7t + 6 = 0 \Rightarrow (t - 1)(t - 6) = 0\\\\ \Rightarrow \left\{ \begin{array}{l}t - 1 = 0 \Rightarrow t = 1\\\\t - 6 = 0 \Rightarrow t = 6\end{array} \right.\\\\t = 1 \Rightarrow \frac{{{x^2}}}{3} - 2 = 1 \Rightarrow \frac{{{x^2}}}{3} = 3 \Rightarrow {x^2} = 9 \Rightarrow x = \pm 3\\\\t = 6 \Rightarrow \frac{{{x^2}}}{3} - 2 = 6 \Rightarrow \frac{{{x^2}}}{3} = 8 \Rightarrow {x^2} = 24 \Rightarrow x = \pm 2\sqrt 6 \\\\ \Rightarrow \left\{ \begin{array}{l}x = - 3\\\\x = 3\\\\x = - 2\sqrt 6 \\\\x = 2\sqrt 6 \end{array} \right.\end{array}\)

\(\begin{array}{l}t = 4 - {x^2} \Rightarrow {(4 - {x^2})^2} - (4 - {x^2}) = 12\\\\ \Rightarrow {t^2} - t = 12 \Rightarrow {t^2} - t - 12 = 0 \Rightarrow (t - 4)(t + 3) = 0\\\\ \Rightarrow \left\{ \begin{array}{l}t - 4 = 0 \Rightarrow t = 4\\\\t + 3 = 0 \Rightarrow t = - 3\end{array} \right.\\\\t = 4 \Rightarrow 4 - {x^2} = 4 \Rightarrow {x^2} = 0 \Rightarrow x = 0\\\\t = - 3 \Rightarrow 4 - {x^2} = - 3 \Rightarrow {x^2} = 7 \Rightarrow x = \pm \sqrt 7 \\\\ \Rightarrow \left\{ \begin{array}{l}x = 0\\\\x = - \sqrt 7 \\\\x = \sqrt 7 \end{array} \right.\end{array}\)

\(\begin{array}{l}|x - 1| = {x^2} - x - 1\\\\ \Rightarrow \left\{ \begin{array}{l}f(x) = |x - 1|\\\\g(x) = {x^2} - x - 1\end{array} \right.\end{array}\)

2 ریشه دارد و این ریشه ها عبارتند از:

\(\left\{ \begin{array}{l}x \simeq - 2\\\\x \simeq 2\end{array} \right.\)

\(\begin{array}{l}a = 1\\\\f\left( x \right) = {x^2} + bx + c\\\\ \Rightarrow {x_0} = \frac{{ - b}}{{2a}} = \frac{{ - b}}{2} = - 3 \Rightarrow b = 6\\\\f\left( { - 3} \right) = 5 = {\left( { - 3} \right)^2} + 6\left( { - 3} \right) + c \Rightarrow c = 14\\\\ \Rightarrow f\left( x \right) = {x^2} + 6x + 14\end{array}\)

\(\begin{array}{l}a = - 1\\\\g(x) = - {x^2} + bx + c\\\\ \Rightarrow {x_0} = \frac{{ - b}}{{2a}} = \frac{{ - b}}{{ - 2}} = - 2 \Rightarrow b = - 4\\\\g( - 2) = 2 \Rightarrow - {( - 2)^2} - 4( - 2) + c = 2 \Rightarrow c = - 2\\\\g(x) = - {x^2} - 4x - 2\end{array}\)

\(\begin{array}{l}a = 1\\\\h\left( x \right) = {x^2} + bx + c\\\\ \Rightarrow {x_0} = \frac{{ - b}}{{2a}} = \frac{{ - b}}{2} = 3 \Rightarrow b = - 6\\\\h(3) = - 3 \Rightarrow {(3)^2} - 6(3) + c = - 3 \Rightarrow c = 6\\\\h\left( x \right) = {x^2} - 6x + 6\end{array}\)

\(\begin{array}{l}a = - 1\\\\i(x) = - {x^2} + bx + c\\\\ \Rightarrow {x_0} = \frac{{ - b}}{{2a}} = \frac{{ - b}}{{ - 2}} = - 2 \Rightarrow b = - 4\\\\i( - 2) = - 1 \Rightarrow - {( - 2)^2} - 4( - 2) + c = - 1 \Rightarrow c = - 5\\\\ \Rightarrow i(x) = - {x^2} - 4x - 5\end{array}\)

پهنای آبراه = x

\(\begin{array}{l}S = 2 \times x\left( {3 + 2x} \right) + 2 \times x\; \times \;10 = 14\\\\ \Rightarrow 4{x^2} + 26x = 14 \Rightarrow 4{x^2} + 26x - 14 = 0\\\\ \Rightarrow 2{x^2} + 13x - 7 = 0 \Rightarrow \Delta = {(13)^2} - 4(2)( - 7) = 225\\\\ \Rightarrow x = \frac{{ - b \pm \sqrt \Delta }}{{2a}} = \frac{{ - 13 \pm 15}}{4} \Rightarrow \left\{ \begin{array}{l}{x_1} = - 7 \otimes \\\\{x_2} = \frac{1}{2}m = 50cm\end{array} \right.\end{array}\)

طول کاشی = x

عرض کاشی = y

مساحت هر کاشی = S

\(\begin{array}{l}x = 4y + 1\\\\ \Rightarrow S = \frac{{52/8{\;^{{m^2}}}}}{{2000}} \times \frac{{10000{\;^{c{m^2}}}}}{{1{\;^{{m^2}}}}} = 264{\;^{c{m^2}}}\\\\ \Rightarrow S = x \times y = 4{y^2} + y = 264 \Rightarrow 4{y^2} + y - 264 = 0\\\\ \Rightarrow \Delta = {b^2} - 4ac = {(1)^{}} - 4(4)( - 246) = 4225\\\\ \Rightarrow y = \frac{{ - b \pm \sqrt \Delta }}{{2a}} = \frac{{ - 1 \pm 65}}{8} \Rightarrow \left\{ \begin{array}{l}{y_1} = - \frac{{33}}{4} \otimes \\\\{y_2} = 8\end{array} \right.\\\\y = 8 \Rightarrow x = 4(8) + 1 = 33cm\end{array}\)