نمونه سوالات تشریحی فصل 3 حسابان یازدهم همراه با جواب تشریحی

\({2^{3n - 2}} = \frac{1}{{{{32}^2}}} = {2^{ - 10}} \Rightarrow \left| \begin{array}{l}3n - 2 = - 10\\\\n = - \frac{8}{3}\end{array} \right.\)

\({9^{3y - 3}} = {27^{y + 1}} \Rightarrow {3^{6y - 6}} = {3^{3y + 3}}\)

\(\begin{array}{l}{4^{3x + 2}} = \frac{1}{{{{64}^3}}} \Rightarrow {2^{6x + 4}} - 18 = 2\\\\6x = - 22 \Rightarrow x = - \frac{{11}}{3}\end{array}\)

\(\begin{array}{l}{9^x} = {3^{{x^2} - 4x}} = {3^{2x}} = {3^{{x^2} - 4x}} \Rightarrow {x^2} - 4x - 2x = 0\\\\ \Rightarrow x = 0\,,\,x = 6\end{array}\)

\({(\frac{3}{5})^{x + 1}} = \frac{{25}}{9} \Rightarrow {(\frac{3}{5})^{x + 1}} = {(\frac{3}{5})^{ - 2}} \Rightarrow x + 1 = - 2 \Rightarrow x = - 3\)

(الف

\(f(3) = {3^3} = 27\)

(ب

\(g( - 1) = {(\frac{1}{{16}})^{ - 1}} = 16\)

(پ

\(h( - 2) = 10 = \frac{1}{{100}}\)

خیر

\({2^{2x + 1}} = 32 \Rightarrow {2^{2x + 1}} = {2^5} \Rightarrow 2x + 1 = 5 \Rightarrow x = 2\)

\(\begin{array}{l}{9^{{x^2} + 2}} = {(\frac{1}{{27}})^{2x}} \Rightarrow {({3^2})^{{x^2} + 2}} = {({3^{ - 3}})^{2x}} \Rightarrow {3^{2{x^2} + 4}} = {3^{ - 6x}}\\\\ \Rightarrow 2{x^2} + 4 = - 6x\\\\2{x^2} + 6x + 4 = 0 \Rightarrow \left| \begin{array}{l}x = - 1\\\\x = - 2\end{array} \right.\end{array}\)

\(\begin{array}{l}\left. \begin{array}{l}{2^{x + 2y}} = {2^{ - 5}}\\\\{2^{x - 2y}} = {2^3}\end{array} \right| \Rightarrow \left\{ \begin{array}{l}x + 2y = - 5\\\\x - 2y = 3\end{array} \right. \Rightarrow 2x = - 2 \Rightarrow x = - 1\\\\ \Rightarrow y = - 2 \Rightarrow x + y = - 3\end{array}\)

\({2^{2n - 6}} > {2^4} \Rightarrow 2n - 6 > 4 \Rightarrow 2n > 10 \Rightarrow n > 5\)

\({(\frac{1}{7})^{5 - x}} < {(\frac{1}{7})^{7 + x}} \Rightarrow 5 - x > 7 + x \Rightarrow - 2x > 2 \Rightarrow x < - 1\)

\(\begin{array}{l}{27^{4 + 4n}} = {81^{2n + 8}} \Rightarrow {({3^3})^{4 + 4n}} = {({3^4})^{2n + 8}}\\\\ \Rightarrow {3^{12 + 12n}} = {3^{8n + 32}} \Rightarrow 12 + 12n = 8n + 32\\\\ \Rightarrow 4n = 20 \Rightarrow n = 5\end{array}\)

\(\begin{array}{l}{2^{{x^2} - x + 2}} = {({2^7})^{x - 2}} \Rightarrow {2^{{x^2} - x + 2}} = {2^{7x - 14}} \Rightarrow {x^2} - x + 2\\\\ = 7x - 14 \Rightarrow {x^2} - 8x + 16 = 0 \Rightarrow {(x - 4)^2} = 0 \Rightarrow x = 4\end{array}\)

\(\begin{array}{l}E = 2 \times {10^{19}}Erg\\\\M = ?\\\\Log2 \times {10^{19}} = 11/8 + 1/5M \Rightarrow Log2 + Log{10^{19}}\\\\ = 11/8 + 1/5M \Rightarrow 0/3 + 19 = 11/8 + 1/5M\\\\ \Rightarrow 1/5M = 7/5 \Rightarrow M = 5\end{array}\)

\(\begin{array}{l}LogE = 11/8 + 1/5M\\\\LogE = 11/8 + 1/5(8/8)\\\\ \Rightarrow LogE = 11/8 + 13/2 \Rightarrow LogE = 25\\\\ \Rightarrow E = {10^{25}}Erg\end{array}\)

ابتدا نقطه ی (2,5) را درون ضابطه تابع قرار می دهیم تا a بدست آید.

\(\begin{array}{l}\left\{ \begin{array}{l}x = 2\\\\y = 5\end{array} \right. \Rightarrow {a^2} \Rightarrow a = \sqrt 5 \Rightarrow y = {(\sqrt 5 )^x}\\\\x = 1 \Rightarrow y = {(\sqrt 5 )^1} \Rightarrow y = \sqrt 5 \Rightarrow A(1,\sqrt 5 )\\\\x = - 1 \Rightarrow y = {(\sqrt 5 )^{ - 1}} = \frac{1}{{\sqrt 5 }} \Rightarrow y = \frac{{\sqrt 5 }}{5} \Rightarrow B( - 1,\frac{{\sqrt 5 }}{5})\\\\x = - 4 \Rightarrow y = {(\sqrt 5 )^{ - 4}} = \frac{1}{{25}} \Rightarrow C( - 4,\frac{1}{{25}})\end{array}\)

(الف

\(\left\{ \begin{array}{l}{x^6} > 0 \Rightarrow x \ne 0\\\\{x^2} > 0 \Rightarrow x \ne 0\\\\{x^2} \ne 1 \Rightarrow x \ne \pm 1\end{array} \right. \Rightarrow D = R\{ 0, \pm \} \)

(ب

\(f(x) = \frac{6}{2}Log_x^x = 3\)

\(\begin{array}{l}{2^{4x - 2}} > \frac{1}{{{2^{10}}}} \Rightarrow {2^{4x - 2}} > {2^{ - 10}} \Rightarrow 4x - 2 > - 10\\\\4x > - 8 \Rightarrow x > - 2\end{array}\)

\(\begin{array}{l}f(x) = {2^x}\\\\{D_t} = R\\\\{R_t} = (0, + \infty )\end{array}\)

\({10^{ - 3}} = 0/001\)

\(\log {x^3} + \log \sqrt[3]{{{y^2}}} - \log a = \log \frac{{{x^3}\sqrt[3]{{{y^2}}}}}{a}\)

\(\begin{array}{l}\log (2x + 7) - \log 9 = \log x \Rightarrow \log \frac{{2x + 7}}{9} = \log x\\\\ \Rightarrow \frac{{2x + 7}}{9} = x\\\\2x + 7 = 9x \Rightarrow x = 1\end{array}\)

\(\begin{array}{l}A = \frac{1}{{\log _3^7 + \log _3^3}} + \frac{1}{{\log _7^3 + \log _7^7}} = \frac{1}{{\log _3^{21}}} + \frac{1}{{\log _7^{21}}}\\\\ = \log _{21}^3 + \log _{21}^7 = \log _{21}^{21} = 1\end{array}\)