جواب تمرین صفحه 22 درس 1 ریاضی دوازدهم تجربی (تابع)

\(\begin{array}{l}fog = \left\{ {(5\;,\;8)\;,\;(3\;,\;3)\;,\;(7\;,\;8)\;,\;(9\;,\;4)} \right\}\\gof = \left\{ {(5\;,\;5)} \right\}\end{array}\)

الف

\(\begin{array}{l}\left\{ \begin{array}{l}f(x) = {x^2} - 5\quad \; \to \quad {D_f} = \mathbb{R}\\g(x) = \sqrt {x + 6} \quad \to \quad {D_g} = \left[ { - 6\;,\; + \infty } \right)\end{array} \right.\\\\ \Rightarrow {D_{fog}} = \left\{ {x \in {D_g}|g(x) \in {D_f}} \right\}\\\\ = \left\{ {\left[ { - 6\;,\; + \infty } \right)|\sqrt {x + 6} \in \mathbb{R}} \right\} = \left\{ {\left[ { - 6\;,\; + \infty } \right)|\sqrt {x + 6} \ge 0} \right\}\\\\ \Rightarrow {D_{fog}} = \left[ { - 6\;,\; + \infty } \right) \cap \left[ { - 6\;,\; + \infty } \right) = \left[ { - 6\;,\; + \infty } \right)\\\\fog(x) = f(g(x)) = {(g(x))^2} - 5 = {(\sqrt {x + 6} )^2} - 5 = x + 1\end{array}\)

ب

\(\begin{array}{l}\left\{ \begin{array}{l}f(x) = \sqrt {3 - 2x} \quad \; \to \quad {D_f} = \left( { - \;\infty \;,\;\frac{3}{2}} \right]\\g(x) = \frac{6}{{3x - 5}}\quad \;\;\;\, \to \quad {D_g} = \mathbb{R} - \left\{ {\frac{5}{3}} \right\}\end{array} \right.\\\\ \Rightarrow {D_{fog}} = \left\{ {x \in {D_g}|g(x) \in {D_f}} \right\}\\\\ = \left\{ {\mathbb{R} - \left\{ {\frac{5}{3}} \right\}|\frac{6}{{3x - 5}} \le \frac{3}{2}} \right\}\\\\\frac{6}{{3x - 5}} - \frac{3}{2} \le 0 \Rightarrow \frac{{ - 9x + 27}}{{6x - 10}} \le 0\\\\ \Rightarrow x \in \mathbb{R} - \left[ {\frac{5}{3}\;,\;3} \right)\\\\ \Rightarrow {D_{fog}} = \left\{ {\mathbb{R} - \left\{ {\frac{5}{3}} \right\}|\mathbb{R} - \left[ {\frac{5}{3}\;,\;3} \right)} \right\}\\\\ = \mathbb{R} - \left\{ {\frac{5}{3}} \right\} \cap \mathbb{R} - \left[ {\frac{5}{3}\;,\;3} \right) = \mathbb{R} - \left[ {\frac{5}{3}\;,\;3} \right)\\\\fog(x) = f(g(x)) = \sqrt {3 - 2g(x)} \\\\ = \sqrt {3 - 2\left( {\frac{6}{{3x - 5}}} \right)} = \sqrt {\frac{{9x - 27}}{{3x - 5}}} \end{array}\)

تعیین علامت عبارت \(\frac{{ - 9x + 27}}{{6x - 10}} \le 0\) :

پ

\(\begin{array}{l}\left\{ \begin{array}{l}f(x) = \sqrt {x + 2} \quad \;\,\;\, \to \quad {D_f} = \left[ { - 2\;,\; + \infty } \right)\\g(x) = \sqrt {{x^2} - 16} \quad \to \quad {D_g} = \mathbb{R} - \left( { - 4\;,\;4} \right)\end{array} \right.\\\\ \Rightarrow {D_{gof}} = \left\{ {x \in {D_f}|f(x) \in {D_g}} \right\}\\\\ = \left\{ {\left[ { - 2\;,\; + \infty } \right)|\sqrt {x + 2} \in \mathbb{R} - \left( { - 4\;,\;4} \right)} \right\}\\\\ = \left\{ {\left[ { - 2\;,\; + \infty } \right)|\sqrt {x + 2} \ge 4} \right\}\\\\ = \left\{ {\left[ { - 2\;,\; + \infty } \right)|x \ge 14} \right\} = \left\{ {\left[ { - 2\;,\; + \infty } \right)|} \right\} = \left[ { - 2\;,\; + \infty } \right) \cap \left[ {14\;,\; + \infty } \right)\\\\ \Rightarrow {D_{gof}} = \left[ {14\;,\; + \infty } \right)\\\\gof(x) = g(f(x)) = \sqrt {{{\left( {\sqrt {x + 2} } \right)}^2} - 16} = \sqrt {x - 14} \end{array}\)

ت

\(\begin{array}{l}\left\{ \begin{array}{l}f(x) = \sin x\quad \; \to \quad {D_f} = \mathbb{R}\\g(x) = \sqrt x \quad \;\;\,\, \to \quad {D_g} = \left[ {0\;,\; + \infty } \right)\end{array} \right.\\\\ \Rightarrow {D_{gof}} = \left\{ {x \in {D_f}|f(x) \in {D_g}} \right\}\\\\ = \left\{ {\mathbb{R}|\sin x \in \left[ {0\;,\; + \infty } \right)} \right\}\\\\ = \left\{ {\mathbb{R}|\sin x \ge 0} \right\} = \left\{ {\mathbb{R}|\left[ {2k\pi \;,\;2k\pi + \pi } \right]\quad ,\quad k \in \mathbb{Z}} \right\}\\\\ = \mathbb{R} \cap \left[ {2k\pi \;,\;2k\pi + \pi } \right]\quad ,\quad k \in \mathbb{Z}\\\\ \Rightarrow {D_{gof}} = \left[ {2k\pi \;,\;2k\pi + \pi } \right]\quad ,\quad k \in \mathbb{Z}\\\\gof(x) = g(f(x)) = \sqrt {\sin x} \end{array}\)

\(\begin{array}{l}\left\{ \begin{array}{l}f(g(x)) = 3{x^2} - 6x + 14\\f(x) = 3x - 4\\g(x) = ?\end{array} \right.\\\\ \Rightarrow \left\{ \begin{array}{l}f(g(x)) = 3g(x) - 4 = 3{x^2} - 6x + 14\\ \Rightarrow 3g(x) = 3{x^2} - 6x + 18\end{array} \right.\\\\ \Rightarrow g(x) = {x^2} - 2x + 6\end{array}\)

الف

نادرست

\(\begin{array}{l}fog(5) = f(g(5)) = {(g(5))^2} - 4\\ = {(\sqrt {{5^2} - 4} )^2} - 4 = 21 - 4 = 17\end{array}\)

ب

نادرست

\(\begin{array}{l}\left\{ \begin{array}{l}f(x) = 3x\\g(x) = 2x\end{array} \right. \to \left\{ \begin{array}{l}fog(x) = f(g(x)) = 3(2x) = 6x\\gof(x) = g(f(x)) = 2(3x) = 6x\end{array} \right.\\\\ \to fog(x) = gof(x)\end{array}\)

پ

درست

\(fog(4) = f(g(4)) = f(7) = 5\)

ت

درست

\(\left\{ \begin{array}{l}fog(5) = f(g(5)) = \sqrt {2 \times 5 - 1} = \sqrt 9 = 3\\g(2) = 2 \times 2 - 1 = 3\end{array} \right.\)

الف

حالت الف مقرون به صرفه است؛ چرا که :

\(\begin{array}{l}\left\{ \begin{array}{l}f(x) = x - \frac{2}{{10}}x = \frac{8}{{10}}x = \frac{4}{5}x\quad (x > 0)\\g(x) = x - 200,000\quad (x > 1,500,000)\end{array} \right.\\\\ \to g(f(x)) = f(x) - 200,000 = \frac{4}{5}x - 200,000\\\\\frac{{80}}{{100}} \times 2,000,000 - 200,000 = 1,400,000\end{array}\)

ب

\(\begin{array}{l}f(g(x)) = f(x - 200,000) = \frac{4}{5}x - 160,000\\\frac{{80}}{{100}} \times (2,000,000 - 200,000) = 1,440,000\end{array}\)

الف

\(\left\{ \begin{array}{l}fog(x) = f(g(x)) = \sqrt[5]{{3{x^2} - 4x + 1}}\quad \quad \to \quad \times \\gof(x) = g(f(x)) = 3\sqrt[5]{{{x^2}}} - 4\sqrt[5]{x} + 1\quad \to \quad \times \end{array} \right.\)

ب

\(\begin{array}{l}kol(x) = k(l(x)) = {(3{x^2} - 4x + 1)^5} = h(x)\\lok(x) = l(k(x)) = 3{({x^5})^2} - 4({x^5}) + 1\quad \quad \to \quad \times \end{array}\)

الف

\(h(x) = \sqrt[3]{{{x^2} + 1}} \to \left\{ \begin{array}{l}f(x) = {x^2} + 1\\g(x) = \sqrt[3]{x}\end{array} \right.\)

ب

\(l(x) = \sqrt {{x^2} + 5} \to \left\{ \begin{array}{l}f(x) = {x^2} + 5\\g(x) = \sqrt x \end{array} \right.\)

خیر؛ منحصر بفرد نیست.

الف

\(\left( {fog} \right)\left( { - 1} \right) = f(g( - 1)) = f( - 3) = 1\)

ب

\(\left( {gof} \right)\left( 0 \right) = g(f(0)) = g(2) = - 6\)

پ

\(\left( {fog} \right)\left( 1 \right) = f(g(1)) = f( - 5) = 3\)

ت

\(\left( {gof} \right)\left( { - 1} \right) = g(f( - 1)) = g(1) = - 5\)

الف

\(\begin{array}{l}f(g(x)) = 2({x^2} - 3x + 8) - 5 = 7\\\\ \Rightarrow 2{x^2} - 6x + 11 = 7 \Rightarrow 2{x^2} - 6x + 4 = 0\\\\ \to \Delta = {b^2} - 4ac = {( - 6)^2} - 4(2)(4) = 4\\\\ \Rightarrow x = \frac{{ - b \pm \sqrt \Delta }}{{2a}} = \frac{{6 \pm \sqrt 4 }}{4} \Rightarrow \left\{ \begin{array}{l}x = 1\\x = 2\end{array} \right.\end{array}\)

ب

\(\begin{array}{l}g(f(x)) = 1 - 2(3{x^2} + x - 1) = - 5 \Rightarrow - 6{x^2} - 2x + 3 = - 5\\\\ \Rightarrow - 6{x^2} - 2x + 8 = 0\\\\ \to \Delta = {b^2} - 4ac = {( - 2)^2} - 4( - 6)(8) = 196\\\\ \Rightarrow x = \frac{{ - b \pm \sqrt \Delta }}{{2a}} = \frac{{2 \pm \sqrt {196} }}{{ - 12}} \Rightarrow \left\{ \begin{array}{l}x = 1\\x = - \frac{4}{3}\end{array} \right.\end{array}\)

الف

4

ب

1

پ

2

ت

3

الف

ب

پ

ت