نمونه سوالات تشریحی فصل 3 ریاضی یازدهم همراه با جواب تشریحی

\(\begin{array}{l}f(x) = - 3 + 2 = - 1\,\,\,\,\,\, - 3 \le x < - 2\\\\f(x) = - 2 + 2 = 0\,\,\,\,\,\,\,\,\,\,\, - 2 \le x < - 1\\\\f(x) = - 1 + 2 = 1\,\,\,\,\,\,\,\,\,\,\,\,\, - 1 \le x < 0\\\\f(x) = 0 + 2 = 2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 \le x < 1\\\\f(x) = 1 + 2 = 3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 \le x < 2\\\\f(x) = 2 + 2 = 4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2 \le x < 3\end{array}\)

بله دو تابع برابرند.

خیر دو تابع برابر نیستند.

\(\begin{array}{l}{D_f}:x \ge 0 \cap x \ge 1 \Rightarrow {D_f} = [1, + \infty )\\\\{D_g} = {x^2} - x \ge 0 \Rightarrow {D_g} = ( - \infty ,0] \cup [1, + \infty )\\\\{D_f} \ne {D_g}\end{array}\)

\(\begin{array}{l}{D_f}:{x^3} - 25x \ge 0 \Rightarrow \left\{ \begin{array}{l}x = 0\\\\x = 5\\\\x = - 5\end{array} \right.\\\\p = {x^3} - 25x\\\\{D_f} = [ - 5,0] \cup [5, + \infty )\end{array}\)

\(\begin{array}{l}{D_g}:\left\{ \begin{array}{l}{x^2} - 1 \ne 0 \Rightarrow x \ne \pm 1\\\\{x^2} - 3x \ne 0 \Rightarrow x \ne 0,3\end{array} \right. \Rightarrow {D_g} = R - \{ 0,3, \pm 1\} \\\\x = 2 \Rightarrow x - 2 = 0{x^2} - 4x + 4 = 0\\\\ \Rightarrow \left\{ \begin{array}{l}a = - 4\\\\b - 1 = 4 \Rightarrow b = 5\end{array} \right.\end{array}\)

\(\begin{array}{l}f(x) = x(x \ne - 2) \Rightarrow \frac{{{x^2} + ax + b - 7}}{{x + 2}} = x\\\\ \Rightarrow ax + b - 7 = + 2x \Rightarrow \left\{ \begin{array}{l}a = 2\\\\b - 7 = 0 \Rightarrow b = 7\end{array} \right.\\\\D = R - \{ - 2\} \end{array}\)

\(\left\{ \begin{array}{l}4 - {x^2} \ge 0 \Rightarrow - 2 \le x \le 2\\ - x > 0 \Rightarrow x < 0\end{array} \right. \to {D_f} = [ - 2,0)\)

\(\left\{ \begin{array}{l} - x \ge 0 \Rightarrow x \le 0\\\\3 - |x| > 0 \Rightarrow |x| < 3 \Rightarrow - 3 < x < 3\end{array} \right. \to {D_g} = ( - 3,0]\)

\(\begin{array}{l}f(2x - 4) = \frac{{2(2x - 4)}}{{4(2x - 4)}} = \frac{{4x - 7}}{{8x - 17}} \Rightarrow 8x - 17 \ne 0\\\\ \Rightarrow x \ne \frac{{17}}{8}\\\\{D_{f(2x - 4)}} - R - \{ \frac{{17}}{8}\} \end{array}\)

\(\begin{array}{l}h(x) = \frac{{f(x)}}{{g(x)}} = \frac{{{x^2} - 1}}{{x + 1}} = \frac{{(x + 1)(x + 1)}}{{x + 1}} = x - 1\\\\{D_h} = R - \{ 1\} \\\\h(x) = x - 1y \ne - 2\end{array}\)

\(\begin{array}{l}{x^2} + |x| - 90 \ge 0 \Rightarrow |x{|^2} + |x| - 90 \ge 0\\\\ \Rightarrow (|x| + 10) \ge 0 \Rightarrow |x| - 9 \ge 0 \Rightarrow |x| \ge 9\\\\ \Rightarrow \left\{ \begin{array}{l}x \ge 9\\\\x \le - 9\end{array} \right.\\\\{D_f} = ( - \infty , - 9] \cup [9, + \infty )\end{array}\)

\(\begin{array}{l}{x^2} - 7|x| + 10 = 0|x{|^2} - 7|x| + 10 = 0\\\\ \Rightarrow (|x| - 2)(|x| - 5) = 0 \Rightarrow \left\{ \begin{array}{l}|x| = 2 \Rightarrow x = \pm 2\\\\|x| = 5 \Rightarrow x = \pm 5\end{array} \right.\\\\{D_g} = R - \{ \pm 2, \pm 5\} \end{array}\)

\(\left\{ \begin{array}{l}{D_f} = R - \{ 0\} \\\\{R_f} = R - \{ 2\} \end{array} \right.\)

\(\left\{ \begin{array}{l}{D_g} = R - \{ - 2\} \\\\{R_g} = R - \{ 0\} \end{array} \right.\)

\(\begin{array}{l}\Delta = 0 \Rightarrow 16 - 4a = 0 \Rightarrow a = 4\\\\{x^2} + 4x + 4 = 0 \Rightarrow {(x + 2)^2} = 0\\\\ \Rightarrow x = - 2 \Rightarrow b = - 2\end{array}\)

(الف

\(f(x) = \sqrt {x + 2} + 2 \Rightarrow \left\{ \begin{array}{l} - a = 2 \Rightarrow a = - 2\\\\b - 1 = 2 \Rightarrow b = 3\end{array} \right.\)

(ب

\(f(\frac{1}{4}) = \sqrt {\frac{1}{4} + 2} + 2 = \sqrt {\frac{9}{4}} + 2 = \frac{3}{2} + 2 = \frac{7}{2}\)

\(\begin{array}{l}{D_f}:2 - x \ge 0 \Rightarrow - x \ge - 2 \Rightarrow x \le 2 \Rightarrow {D_f} = ( - \infty ,2]\\\\{R_f}:\sqrt {2 - x} \ge 0 - \sqrt {2 - x} \le 0\\\\2 - \sqrt {2 - x} \le 2 \Rightarrow y \le 2 \Rightarrow {R_f} = ( - \infty ,2]\end{array}\)

\(\begin{array}{l}{D_g}:5 - x \ge 0 \Rightarrow - x \ge - 5 \Rightarrow x \le 5 \Rightarrow {D_g} = ( - \infty ,5]\\\\{R_g}:\sqrt {5 - x} \ge 02\sqrt {5 - x} \ge 02\sqrt {5 - x} + 2 \ge 7\\\\ \Rightarrow y \ge 7 \Rightarrow {R_f} = [7, + \infty )\end{array}\)

\(\begin{array}{l}y = 5x - 2 \Rightarrow 5x - 2 = y \Rightarrow 5x = y + 2\\\\ \Rightarrow x = \frac{{y + 2}}{5} \Rightarrow {f^{ - 1}}(x) = \frac{{x + 2}}{5}\end{array}\)

\(\begin{array}{l}y = \frac{3}{4}x + 4 \Rightarrow 5y = 3x + 20 \Rightarrow 3x + 20 = 5y\\\\ \Rightarrow 3x = 5y - 20 \Rightarrow x = \frac{{5y - 20}}{3}\\\\{f^{ - 1}}(x) = \frac{{5x - 20}}{3}\end{array}\)

\(\begin{array}{l}f(x) = \frac{{ - 7x + 3}}{5} \Rightarrow 5y = - 7x + 3 \Rightarrow 7x = - 5y + 3\\\\ \Rightarrow x = \frac{{ - 5y + 3}}{7} \Rightarrow {f^{ - 1}}(x) = \frac{{ - 5x + 3}}{7}\end{array}\)

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

\(\begin{array}{l}f = \{ ({m^4} + 2,5),({n^3} + 1,4)\} \\\\{f^{ - 1}} = f = \{ (5,{m^4} + 2),(4,{n^3} + 1)\} \\\\ \Rightarrow {R_{{f^{ - 1}}}} = \{ {m^4} + 2,{n^3} + 1\} \\\\\left\{ \begin{array}{l}{m^4} + 2 = 18 \Rightarrow {m^4} = 16 \Rightarrow m = \pm 2\\\\{n^3} + 1 = - 7 \Rightarrow {n^3} = - 8 \Rightarrow {n^3} = {( - 2)^3} \Rightarrow n = - 2\end{array} \right.\end{array}\)

\(\begin{array}{l}(\frac{f}{g})(1) = \frac{{f(1)}}{{g(1)}} = \frac{{\sqrt 6 }}{2}\\\\2 < \sqrt 6 < 31 < \frac{{\sqrt 6 }}{2} < \frac{3}{2} \Rightarrow [\frac{{\sqrt 6 }}{2}] = 1\end{array}\)

\(\begin{array}{l}f(x) = \frac{{x(x - 2)}}{x} \Rightarrow y = x - 2 \Rightarrow y + 2 = x\\\\y = x + 2 \Rightarrow {f^{ - 1}}(x) = x + 2\end{array}\)

\(\left| \begin{array}{l}{D_f} = R - \{ 0\} \\\\{D_f} = R - \{ - 2\} \end{array} \right. \Rightarrow {D_{f - {f^{ - 1}}}} = {D_f} \cap {D_{{f^{ - 1}}}} = R - \{ 0, - 2\} \)

\(\left| \begin{array}{l}{D_f} = R\\\\{D_g} = R - \{ 2\} \end{array} \right. \Rightarrow {D_{f \times g}} - {D_f} \cap {D_g} = R - \{ 2\} \)

\(\begin{array}{l}(f \times g)(x) = f(x) \times g(x) = ({x^2} - 4) \times (\frac{{14}}{{x - 2}})\\\\ = \frac{{{x^2} - 4}}{{x - 2}} = \frac{{(x - 2)(x + 2)}}{{x - 2}} = x + 2\end{array}\)

\(\begin{array}{l}{D_f} = R,{D_g} = x \ge 3\,,\,{D_{\frac{f}{g}}}:{D_f} \cap {D_g} - \{ x|g(x) = 0\} = x > 3\\\\\sqrt {x - 3} = 0\,,\,x = 3\\\\(\frac{f}{g})(x) = \frac{{x - 1}}{{\sqrt {x - 3} }}\end{array}\)

\((3f + 4g)(7) = 3f(7) + 4g(7) = 3(6) + 4(2) = 26\)

\(\begin{array}{l}{D_f}:36 - {x^2} \ge 0 \Rightarrow - 6 \le x \le 6 \Rightarrow {D_f} = [ - 6,6]\\\\{D_g}:b - x \ge 0 \Rightarrow - x \ge - b \Rightarrow x \le b \Rightarrow {D_g} = ( - \infty ,b)\\\\{D_{f - g}} = {D_f} \cap {D_g} = [ - 6,6] \cap ( - \infty ,b] = [ - 6,b] \Rightarrow b = 5\end{array}\)

\((f + g)(2) = f(2) + g(2) = 4 + n + 8 + 1 = 8 \Rightarrow n = - 5\)

\(\begin{array}{l}f(1) = 4 \Rightarrow a + b = 4\\\\(f + g)(2) = f(2) + g(2) = 2a + b + 4 + 7 = 17\\\\ \Rightarrow 2a + b = 6\\\\\left\{ \begin{array}{l}a + b = 4\\\\2a + b = 6\end{array} \right. \Rightarrow a = 2 \Rightarrow b = 2\end{array}\)

\(\begin{array}{l}(2f + g)(3) = 2f(3) + g(3) = 2 \times 5 + \sqrt {3 + 6} \\\\ = 10 + \sqrt 9 = 13\end{array}\)