نمونه سوالات فصل 1 حسابان دوازدهم همراه با جواب تشریحی

5

کوچکتر

ابتدا مولفه های اول را از کوچک به بزرگ مرتب می کنیم :

\(f = \left\{ {( - 3, - 3a + 1),( - 1,2a),(1,{a^2} + 1)} \right\}\)

می دانیم که در تابع صعودی اگر \({x_2} > {x_1}\) آنگاه باید \({y_2} \ge {y_1}\) باشد، پس داریم :

\(\begin{array}{l}{a^2} + 1 \ge 2a \Rightarrow {a^2} - 2a + 1 \ge 0 \Rightarrow {(a - 1)^2} \ge 0\\\\ \Rightarrow 2a \ge - 3a + 1 \Rightarrow 5a \ge 1 \Rightarrow a \ge \frac{1}{5}\end{array}\)

همواره به ازای جمیع مقادیر a برقرار می باشد.

برد – دامنه

\(\begin{array}{l}\left\{ \begin{array}{l}x = - 3 \Rightarrow f( - 3) = 2 \Rightarrow g(2) = - 2 \Rightarrow ( - 3, - 2)\\\\x = 4 \Rightarrow f(4) = 5 \Rightarrow g(5) = 1 \Rightarrow (4,1)\\\\x = - 1 \Rightarrow f( - 1) = - 2 \Rightarrow g( - 2) = \otimes \end{array} \right.\\\\ \Rightarrow gof = \left\{ {( - 3, - 2),(4,1)} \right\}\\\\\left\{ \begin{array}{l}x = 5 \Rightarrow g(5) = 1 \Rightarrow f(1) = \otimes \\\\x = 2 \Rightarrow g(2) = - 2 \Rightarrow f( - 2) = \otimes \\\\x = 7 \Rightarrow g(7) = - 1 \Rightarrow f( - 1) = - 2 \Rightarrow (7, - 2)\end{array} \right.\\\\ \Rightarrow fog = \left\{ {(7, - 2)} \right\}\end{array}\)

\(\begin{array}{l}{D_f} = [\frac{2}{3}, + \infty )\,,\,{D_g} = R - \left\{ {\frac{1}{2}} \right\}\\\\{D_{gof}} = \left\{ {x \in {D_f}|f(x) \in {D_g}} \right\} = \left\{ {x \ge \frac{2}{3} \cap \sqrt {3x - 2} \ne \frac{1}{2}} \right\}\\\\ \Rightarrow \sqrt {3x - 2} \ne \frac{1}{2} \Rightarrow 3x - 2 \ne \frac{1}{4} \Rightarrow 3x \ne \frac{1}{4} + 2 \Rightarrow x \ne \frac{3}{4}\\\\ \Rightarrow {D_{gof}} = \left\{ {x \ge \frac{2}{3} \cap \sqrt {3x - 2} \ne \frac{1}{2}} \right\} = [\frac{2}{3}, + \infty ) - \left\{ {\frac{3}{4}} \right\}\\\\gof(x) = \frac{2}{{2(\sqrt {3x - 2} ) - 1}}\end{array}\)

\(\begin{array}{l}g(f(x)) = \frac{{x + 1}}{{x - 1}} \Rightarrow g(3x + 1) = \frac{{x + 1}}{{x - 1}}\\\\ \Rightarrow 3x + 1 = t \Rightarrow x = \frac{{t - 1}}{3}\\\\ \Rightarrow g(t) = \frac{{\frac{{t - 1}}{3} + 1}}{{\frac{{t - 1}}{3} - 1}} = \frac{{\frac{{t + 2}}{3}}}{{\frac{{t - 4}}{3}}} = \frac{{t + 2}}{{t - 4}}\\\\ \Rightarrow g(x) = \frac{{x + 2}}{{x - 4}}\end{array}\)

نمودار را رسم می کنیم :

تابع یک به یک است.\( \Rightarrow \)  هیچ خط افقی نمودار تابع را در بیشتر از یک نقطه قطع نمی کند.

 \(y = {(x + 1)^3} - 3 \Rightarrow {(x + 1)^3} = y + 3 \Rightarrow x{ = ^3}\sqrt {y + 3} - 1\)

\( \Rightarrow {f^{ - 1}}(x){ = ^3}\sqrt {x + 3} - 1\)

می دانیم توابع سهمی در \(( - \infty ,\frac{{ - b}}{{2a}}]\) و \([\frac{{ - b}}{{2a}}, + \infty )\) یک به یک هستند، پس :

 \(( - \infty ,2],[2, + \infty )\,,\,y = {(x - 2)^2}\) \( \Rightarrow \) : محدوده یک به یک بودن تابع

\(f = \{ ( - 1,10),(0,2),(3,2)\} \)

\(\begin{array}{l}g = \{ (3, - 1),(0,2),( - 5,1)\} \Rightarrow {g^{ - 1}} = \{ ( - 1,3),(2,0),(1, - 5)\} \\\\\left\{ \begin{array}{l}x = - 1 \Rightarrow {g^{ - 1}}( - 1) = 3 \Rightarrow g(3) = - 1 \Rightarrow ( - 1, - 1)\\\\x = 2 \Rightarrow {g^{ - 1}}(2) = 0 \Rightarrow g(0) = 2 \Rightarrow (2,2)\\\\x = 1 \Rightarrow {g^{ - 1}}(1) = - 5 \Rightarrow g( - 5) = 1 \Rightarrow (1,1)\end{array} \right.\\\\ \Rightarrow go{g^{ - 1}} = \{ ( - 1, - 1),(2,2),(1,1)\} \\\\x = - 1 \Rightarrow {g^{ - 1}}( - 1) = 3 \Rightarrow f(3) = 2 \Rightarrow ( - 1,2)\\\\x = 2 \Rightarrow {g^{ - 1}}(2) = 0 \Rightarrow f(0) = - 2 \Rightarrow (2, - 2)\\\\x = 1 \Rightarrow {g^{ - 1}}(1) = - 5 \Rightarrow f( - 5) = \otimes \end{array}\)

\(\begin{array}{l}{g^{ - 1}}o{f^{ - 1}}(3) = {(fog)^{ - 1}}(3)\\\\fog(x) = 5({x^3} - 3) + 3 = 5{x^3} - 12\\\\ \Rightarrow {(fog)^{ - 1}}(x) = \sqrt[3]{{\frac{{x + 12}}{5}}} \Rightarrow {g^{ - 1}}o{f^{ - 1}}(3) = \sqrt[3]{{\frac{{15}}{5}}} = \sqrt[3]{3}\end{array}\)

2

\(^3\sqrt x \)

الف)

\(\begin{array}{l}{D_{fog}} = \{ x \in {D_g}|g(x) \in {D_f}\} = \\\\\{ x \in [ - 3, + \infty )|\sqrt {x + 3} \in R\} = [ - 3, + \infty )\end{array}\)

ب)

\((gof)(1) = g(3) = \sqrt 6 \)

\(gof = \{ (0,4),(3,7),(5,0)\} \)

2

\(x = 3 \to 2(3) - 2 = 4 \to \frac{4}{{\sqrt {(4)} + 1}} = \frac{4}{3}\)

پایین تر

الف)

\(\begin{array}{l}{D_f} = [1, + \infty )\,,\,{D_g} = R\\\\{D_{fog}} = \{ x \in {D_g}|g(x) \in {D_f}\} = \{ x \in R|2{x^2} - 1 \in {D_f}\} = \\\\( - \infty , - 1] \cup [1, + \infty )\end{array}\)

ب)

\((gof)(2) = 1\)

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

اکیدا نزولی

\( \Rightarrow D = ( - \infty ,1] \Rightarrow \) اکیدا نزولی

\(h(x) = {x^2}\)

\(\begin{array}{l}{D_f} = R - \{ \pm 1\} \\\\{D_{{g}}} = ( - \infty ,1]\\\\{D_{fog}} = \{ x \in {D_g}|g(x) \in {D_f}\} \\\\ = \{ x \le 1 \cap \sqrt {1 - x} \ne 1\,,\,\sqrt {1 - x} \ne - 1\} \\\\\sqrt {1 - x} \ne - 1 \Rightarrow \,\sqrt {1 - x} \ne 1 \Rightarrow 1 - x \ne 1\\\\ \Rightarrow x \ne 0 \Rightarrow {D_{fog}} = ( - \infty ,1] - \{ 0\} \\\\f(g(x)) = \frac{{1 + {{(\sqrt {1 - x} )}^2}}}{{1 - {{(\sqrt {1 - x} )}^2}}} = \frac{{1 + 1 - x}}{{1 - 1 + x}} = \frac{{2 - x}}{x}\end{array}\)

\(\begin{array}{l}f(g(x)) = \frac{{1 + {{(\sqrt {1 - x} )}^2}}}{{1 - {{(\sqrt {1 - x} )}^2}}} = \frac{{1 + 1 - x}}{{1 - 1 + x}} = \frac{{2 - x}}{x}\\\\f(g(x)) = {x^2} + 3x + 2\\\\ \Rightarrow \left\{ \begin{array}{l}f(g(x)) = g{(x)^2} + 2g(x) + 2\,\,\,\,\,\,\,(1)\\\\f(g(x)) = {x^2} - 4x + 5\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(2)\end{array} \right.\\\\(1) = (2) \Rightarrow g{(x)^2} + 2g(x) + 2 = {x^2} - 4x + 5\\\\ \Rightarrow (g{(x)^2} + 2g(x) + 1) = ({x^2} - 4x + 4)\\\\ \Rightarrow {(g(x) + 1)^2} = {(x - 2)^2}\\\\ \Rightarrow g(x) + 1 = \pm (x - 2) \Rightarrow \left\{ \begin{array}{l}g(x) = + (x - 2) - 1 = x - 3\\\\g(x) = - (x - 2) - 1 = x + 1\end{array} \right.\end{array}\)

\(( - \infty ,0]\)اکیدا نزولی (نزولی)

\([0, + \infty )\)اکیدا صعودی (صعودی)

\(\begin{array}{l}\\M = 0\end{array}\)

\(\begin{array}{l}f(x) = 2 - \sqrt x \Rightarrow {f^{ - 1}}(x) = {(2 - x)^2}\\\\g(x) = 1 - {x^2} \to {g^{ - 1}}(x) = \sqrt {1 - x} \\\\{g^{ - 1}}o{f^{ - 1}}(x) = \sqrt {1 - {{(2 - x)}^2}} = \sqrt {1 - 4 - {x^2} + 4x} = \\\\ = \sqrt { - {x^2} + 4x - 3} \end{array}\)

باید ثابت کنیم که \(gof(x) = x\,,\,fog(x) = x\) پس داریم :

\(\begin{array}{l}fog(x) = \frac{1}{{\frac{1}{{x - 3}}}} + 3 = x - 3 + 3 = x\\\\gof(x) = \frac{1}{{\frac{1}{x} + 3 - 3}} = \frac{1}{{\frac{1}{x}}} = x\end{array}\)

پس \(g(x)\,,\,f(x)\) وارون یکدیگرند.