جواب تمرین صفحه 81 درس 4 ریاضی دهم (معادله ها و نامعادله ها)

الف

\(\begin{array}{*{20}{l}}{y = - {{\left( {x + 1} \right)}^2} - 3}\\{x + 1 = 0 \Rightarrow {x_0} = - 1 \Rightarrow {y_0} = - {{\left( { - 1 + 1} \right)}^2} - 3 = 0 - 3 = - 3}\\{ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{{x_0} = - 1}\\{{y_0} = - 3}\end{array}} \right.}\\{ \Rightarrow {x_1} = - 2 \Rightarrow {y_1} = - {{\left( { - 2 + 1} \right)}^2} - 3 = - 1 - 3 = - 4}\\{ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{{x_1} = - 2}\\{{y_1} = - 4}\end{array}} \right.}\\{ \Rightarrow {x_2} = 0 \Rightarrow {y_2} = - {{\left( {0 + 1} \right)}^2} - 3 = - 1 - 3 = - 4}\\{ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{{x_2} = 0}\\{{y_2} = - 4}\end{array}} \right.}\end{array}\)

ب

\(\begin{array}{*{20}{l}}{y = 3{x^2} - 2}\\{{x_0} = 0 \Rightarrow {y_0} = 3{{\left( 0 \right)}^2} - 2 = 0 - 2 = - 2}\\{ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{{x_0} = 0}\\{{y_0} = - 2}\end{array}} \right.}\\{{x_1} = - 1 \Rightarrow {y_1} = 3{{\left( { - 1} \right)}^2} - 2 = 3 - 2 = 1}\\{ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{{x_1} = - 1}\\{{y_1} = 1}\end{array}} \right.}\\{{x_2} = 1 \Rightarrow {y_2} = 3{{\left( 1 \right)}^2} - 2 = 3 - 2 = 1}\\{ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{{x_2} = 1}\\{{y_2} = 1}\end{array}} \right.}\end{array}\)

پ

\(\begin{array}{*{20}{l}}{y = x - {x^2} = - {{\left( {x - \frac{1}{2}} \right)}^2} + \frac{1}{4}}\\{x - \frac{1}{2} = 0 \Rightarrow {x_0} = \frac{1}{2} \Rightarrow {y_0} = - {{\left( {\frac{1}{2} - \frac{1}{2}} \right)}^2} + \frac{1}{4} = 0 + \frac{1}{4} = \frac{1}{4}}\\{ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{{x_0} = \frac{1}{2}}\\{{y_0} = \frac{1}{4}}\end{array}} \right.}\\{{x_1} = - \frac{1}{2} \Rightarrow {y_1} = - {{\left( { - \frac{1}{2} - \frac{1}{2}} \right)}^2} + \frac{1}{4} = - 1 + \frac{1}{4} = - \frac{3}{4}}\\{ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{{x_1} = - \frac{1}{2}}\\{{y_1} = - \frac{3}{4}}\end{array}} \right.}\\{{x_2} = \frac{3}{2} \Rightarrow {y_2} = - {{\left( {\frac{3}{2} - \frac{1}{2}} \right)}^2} + \frac{1}{4} = - 1 + \frac{1}{4} = - \frac{3}{4}}\\{ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{{x_2} = \frac{3}{2}}\\{{y_2} = - \frac{3}{4}}\end{array}} \right.}\end{array}\)

ت

\(\begin{array}{*{20}{l}}{y = \frac{{{x^2}}}{2} + x - 4 = \frac{1}{2}{{\left( {x + 1} \right)}^2} - \frac{9}{2}}\\{x + 1 = 0 \Rightarrow {x_0} = - 1 \Rightarrow {y_0} = \frac{1}{2}{{\left( { - 1 + 1} \right)}^2} - \frac{9}{2} = 0 - \frac{9}{2} = - \frac{9}{2}}\\{ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{{x_0} = - 1}\\{{y_0} = - \frac{9}{2}}\end{array}} \right.}\\{{x_1} = - 2 \Rightarrow {y_1} = \frac{1}{2}{{\left( { - 2 + 1} \right)}^2} - \frac{9}{2} = \frac{1}{2} - \frac{9}{2} = - 4}\\{ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{{x_1} = - 2}\\{{y_1} = - 4}\end{array}} \right.}\\{{x_2} = 0 \Rightarrow {y_2} = \frac{1}{2}{{\left( {0 + 1} \right)}^2} - \frac{9}{2} = \frac{1}{2} - \frac{9}{2} = - 4}\\{ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{{x_2} = 0}\\{{y_2} = - 4}\end{array}} \right.}\end{array}\)

با توجه به اینکه عرض نقاط یکسان است، این دو نقطه نسبت به محور تقارن قرینه یکدیگرند؛ به عبارت دیگر محور تقارن از وسط طول های این دو نقطه می گذرد ؛ یعنی:

\({x_0} = \frac{{0 + ( - 2)}}{2} = - 1\)

در نتیجه خط 1-=x محور تقارن سهمی می باشد.

\(\begin{array}{l}\begin{array}{*{20}{l}}\begin{array}{l}\left( {0{\kern 1pt} ,{\kern 1pt} 2} \right) \Rightarrow 2 = a{\left( 0 \right)^2} + b\left( 0 \right) + c \Rightarrow c = 2\\\end{array}\\\begin{array}{l}\left( { - 1{\kern 1pt} ,{\kern 1pt} 0} \right) \Rightarrow 0 = a{\left( { - 1} \right)^2} + b\left( { - 1} \right) + 2 \Rightarrow a - b = - 2\\\end{array}\\\begin{array}{l}\left( {2{\kern 1pt} ,{\kern 1pt} 0} \right) \Rightarrow 0 = a{\left( 2 \right)^2} + b\left( 2 \right) + 2 \Rightarrow 4a + 2b = - 2\\\end{array}\\{\left\{ {\begin{array}{*{20}{l}}\begin{array}{l}a - b = - 2\\\end{array}\\{4a + 2b = - 2}\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{l}}\begin{array}{l}a = - 1\\\end{array}\\{b = - 1}\end{array}} \right. \Rightarrow y = - {x^2} + x + 2}\end{array}\\\\\begin{array}{*{20}{l}}{y = - {x^2} + x + 2 = - {{\left( {x - \frac{1}{2}} \right)}^2} + \frac{9}{4}}\\\begin{array}{l}\\x - \frac{1}{2} = 0 \Rightarrow {x_0} = \frac{1}{2} \Rightarrow {y_0} = - {\left( {\frac{1}{2} - \frac{1}{2}} \right)^2} + \frac{9}{4} = \frac{9}{4}\end{array}\\\begin{array}{l}\\ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{{x_0} = \frac{1}{2}}\\{{y_0} = \frac{9}{4}}\end{array}} \right.\;\;{\mkern 1mu} {\kern 1pt} ,\;\;{\mkern 1mu} {\kern 1pt} \left\{ {\begin{array}{*{20}{l}}{{x_1} = - 1}\\{{y_1} = 0}\end{array}} \right.\;\;{\mkern 1mu} {\kern 1pt} ,\;\;{\mkern 1mu} {\kern 1pt} \left\{ {\begin{array}{*{20}{l}}{{x_2} = 2}\\{{y_2} = 0}\end{array}} \right.\end{array}\end{array}\end{array}\)

الف

مسیر پرتابه A:

\(\begin{array}{*{20}{l}}{y = - \frac{{{x^2}}}{2} + \frac{3}{2}x + 2 = - \frac{1}{2}{{\left( {x - \frac{3}{2}} \right)}^2} + \frac{{25}}{8}}\\{ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{{x_0} = \frac{3}{2}}\\{{y_0} = \frac{{25}}{8}}\end{array}} \right.\;\;{\mkern 1mu} {\kern 1pt} ,\;\;{\mkern 1mu} {\kern 1pt} \left\{ {\begin{array}{*{20}{l}}{{x_1} = \frac{1}{2}}\\{{y_1} = \frac{{21}}{8}}\end{array}} \right.\;\;{\mkern 1mu} {\kern 1pt} ,\;\;{\mkern 1mu} {\kern 1pt} \left\{ {\begin{array}{*{20}{l}}{{x_2} = \frac{3}{2}}\\{{y_2} = \frac{{21}}{8}}\end{array}} \right.}\end{array}\)

مسیر پرتابه B:

\(\begin{array}{*{20}{l}}{y = - 2{x^2} + 3x + 2 = - 2{{\left( {x - \frac{3}{4}} \right)}^2} + \frac{{25}}{8}}\\{ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{{x_0} = \frac{3}{4}}\\{{y_0} = \frac{{25}}{8}}\end{array}} \right.\;\;{\mkern 1mu} {\kern 1pt} ,\;\;{\mkern 1mu} {\kern 1pt} \left\{ {\begin{array}{*{20}{l}}{{x_1} = - \frac{1}{4}}\\{{y_1} = \frac{9}{8}}\end{array}} \right.\;\;{\mkern 1mu} {\kern 1pt} ,\;\;{\mkern 1mu} {\kern 1pt} \left\{ {\begin{array}{*{20}{l}}{{x_2} = \frac{7}{4}}\\{{y_2} = \frac{9}{8}}\end{array}} \right.}\end{array}\)

ب

طول های برخورد پرتابه A با زمین:

\(\begin{array}{*{20}{l}}{ - \frac{{{x^2}}}{2} + \frac{3}{2}x + 2 = - \frac{1}{2}{{\left( {x - \frac{3}{2}} \right)}^2} + \frac{{25}}{8} = 0}\\{ \Rightarrow \frac{1}{2}{{\left( {x - \frac{3}{2}} \right)}^2} = \frac{{25}}{8} \Rightarrow {{\left( {x - \frac{3}{2}} \right)}^2} = \frac{{25}}{4}}\\{ \Rightarrow x - \frac{3}{2} = \pm \frac{5}{2} \Rightarrow x = \frac{3}{2} \pm \frac{5}{2} \Rightarrow \left\{ {\begin{array}{*{20}{l}}{{x_1} = - 1}\\{{x_2} = 4}\end{array}} \right.}\\{}\\{ \Rightarrow \Delta {x_A} = {x_2} - {x_1} = 4 + 1 = 5}\end{array}\)

طول های برخورد پرتابه B با زمین:

\(\begin{array}{*{20}{l}}{ - 2{x^2} + 3x + 2 = - 2{{\left( {x - \frac{3}{4}} \right)}^2} + \frac{{25}}{8} = 0}\\{ \Rightarrow 2{{\left( {x - \frac{3}{4}} \right)}^2} = \frac{{25}}{8} \Rightarrow {{\left( {x - \frac{3}{4}} \right)}^2} = \frac{{25}}{{16}}}\\{ \Rightarrow x - \frac{3}{4} = \pm \frac{5}{4} \Rightarrow x = \frac{3}{4} \pm \frac{5}{4} \Rightarrow \left\{ {\begin{array}{*{20}{l}}{{x_1} = - \frac{1}{4}}\\{{x_2} = 2}\end{array}} \right.}\\{ \Rightarrow \Delta {x_B} = {x_2} - {x_1} = 2 + \frac{1}{4} = \frac{9}{4}}\end{array}\)

پرتابه A مسیر افقی بیشتری رفته است:

\(\Delta {x_B} < \Delta {x_A}\)

پ

میزان ارتفاع همان عرض نقطه سهمی است که در این مورد هر دو عرض یکسان است، پس ارتفاع یکسانی پیموده اند؛ یعنی ارتفاع پیموده شده آنها \(\frac{{25}}{8}\)  است.