نمونه سوالات فصل 2 ریاضی دوازدهم تجربی همراه با جواب تشریحی

\(\begin{array}{l}f(x) = - 3\pi \sin \frac{1}{2}(x + 5) - 2 = a\sin (bx + d) + c\\\\ \Rightarrow a = - 3\pi \,,\,b = + \frac{1}{2}\,,\,c = - 2 \Rightarrow T = \frac{{2\pi }}{{|b|}} = \frac{{2\pi }}{{\frac{1}{2}}} = 4\pi \\\\\max = |a| + c = 3\pi - 2\\\\\min = - |a| + c = - 3\pi - 2\end{array}\)

مثبت یا منفی بودن b در مقدار cos b تاثیری ندارد.

\(\begin{array}{l}T = \frac{\pi }{6} = \frac{{2\pi }}{{|b|}} \Rightarrow \frac{1}{6} = \frac{2}{{|b|}} \Rightarrow b = \pm 12\\\\\left. \begin{array}{l}\min = - 2 = - |a| + c\\\\\max = + 4 = |a| + c\end{array} \right\} \Rightarrow 2c = 2 \Rightarrow c = 1\\\\ \Rightarrow |a| = 3 \Rightarrow a = \pm 3\\\\ \Rightarrow f(x) = \pm 3\cos 12x + 1\end{array}\)

مثبت و منفی بودن a میتواند تابع \(a\cos bx\) را نسبت به محور X ها قرینه کند.

نمودار شبیه به تابع sin است. پس فرض می کنیم که \(f(x) = a\sin (bx + d) + c\) باشد. نمودار تابع به ازای هر 2 واحد، دوباره تکرار می شود، پس :

\(T = 2 = \frac{{2\pi }}{{|b|}}b = \pi \)

از طرفی نمودار sin در راستای محور x ها انتقالی نداشته ولی در راستای محور yها، 3 واحد بالا رفته است. پس :

\(d = 0\,,\,c = 3\)

نمودار انبساط یا انقباض عمودی ندارد. پس :

\(a = 1\)

در نتیجه معادله این نمودار برابر است با :

\(f(x) = \sin (\pi x) + 3\)

نمودار توابع sin و tan را در بازه \([ - \frac{{3\pi }}{2}\,,\,\pi ]\) رسم می کنیم و یکنوایی این توابع را با توجه به نمودارشان مشخص می کنیم :

\([ - \frac{\pi }{2}\,,\,\frac{\pi }{2}]\)اکیدا صعودی :

\([ - \frac{{3\pi }}{2}\,,\, - \frac{\pi }{2}]\)اکیدا نزولی :

کل دامنه : نه صعودی نه نزولی

\(( - \frac{{3\pi }}{2}\,,\,\frac{\pi }{2})\,\,,\,\,(\frac{\pi }{2}\,,\,\frac{\pi }{2})\,\,,\,\,(\frac{\pi }{2}\,,\,\pi )\)اکیدا صعودی :

کل دامنه : نه صعودی نه نزولی

\(\begin{array}{l}\sin \alpha = - \frac{3}{5} \to {\sin ^2}\alpha + \cos 2\alpha = 1 \Rightarrow {\cos ^2}\alpha \\\\ = 1 - \frac{9}{{25}} = \frac{{16}}{{25}}\cos \alpha = - \frac{4}{5}\\\\\sin 2\alpha = 2\sin \alpha \cos \alpha = 2 \times \frac{{ - 3}}{5} \times \frac{{ - 4}}{5} = \frac{{24}}{{25}}\\\\\cos 2\alpha = {\cos ^2}\alpha - {\sin ^2}\alpha = {( - \frac{4}{5})^2} - {( - \frac{3}{5})^2}\\\\ = \frac{{16}}{{25}} - \frac{9}{{25}} = \frac{7}{{25}}\end{array}\)

\(\begin{array}{l}\sin 2\alpha = 2\sin \alpha \cos \alpha \\\\\cos 2\alpha = {\cos ^2}\alpha - {\sin ^2}\alpha \\\\\tan 2\alpha = \frac{{\sin 2\alpha }}{{\cos 2\alpha }} = \frac{{2\sin \alpha \cos \alpha }}{{2\sin \alpha \cos \alpha }}\\\\\frac{{\frac{{2\sin \alpha \cos \alpha }}{{{{\cos }^2}\alpha }}}}{{\frac{{{{\cos }^2}\alpha }}{{{{\cos }^2}\alpha }} - \frac{{{{\sin }^2}\alpha }}{{{{\cos }^2}\alpha }}}} = \frac{{2\tan \alpha }}{{1 - {{\tan }^2}\alpha }}\\\\\cot 2\alpha = \frac{{\cos 2\alpha }}{{\sin 2\alpha }} = \frac{{{{\cos }^2}\alpha - {{\sin }^2}\alpha }}{{2\sin \alpha \cos \alpha }}\\\\\frac{{\frac{{{{\cos }^2}\alpha }}{{{{\sin }^2}\alpha }} - \frac{{{{\sin }^2}\alpha }}{{{{\sin }^2}\alpha }}}}{{\frac{{2\sin \alpha \cos \alpha }}{{{{\sin }^2}\alpha }}}} = \frac{{{{\cot }^2}\alpha - 1}}{{2\cot \alpha }}\end{array}\)

\(\begin{array}{l}\sin x\cos x = \frac{{ - \sqrt 2 }}{4} \Rightarrow \frac{1}{2}\sin 2x = \frac{{ - \sqrt 2 }}{4}\\\\ \Rightarrow \sin 2x = \frac{{ - \sqrt 2 }}{2} = - \sin \frac{\pi }{{4 = \sin ( - \frac{\pi }{4})}}\\\\ \Rightarrow \left\{ \begin{array}{l}2x = 2k\pi + ( - \frac{\pi }{4})\\\\2x = 2k\pi + \pi - ( - \frac{\pi }{4})\end{array} \right.\\\\ \Rightarrow \left\{ \begin{array}{l}x = k\pi - \frac{\pi }{8}\\\\x = k\pi + \frac{\pi }{2} + \frac{\pi }{8} = k\pi + \frac{{5\pi }}{8}\end{array} \right.\end{array}\)

\(\begin{array}{l}2\cos 3x + \sqrt 3 = 0 \Rightarrow \cos 3x = \frac{{ - \sqrt 3 }}{2} = - \cos \frac{\pi }{6}\\\\ = \cos (\pi - \frac{\pi }{6}) = \cos \frac{{5\pi }}{6} \Rightarrow 3x = 2k\pi \pm \frac{{5\pi }}{6}\\\\ \Rightarrow x = \frac{{2k\pi }}{3} \pm \frac{{5\pi }}{{18}}\end{array}\)

\(\begin{array}{l}\sin x - \cos 2x = 0 \Rightarrow \sin x - ({\cos ^2}x - {\sin ^2}x) = 0\\\\ \Rightarrow \sin x - 1 + 2{\sin ^2}x = 0\\\\\sin x = t \Rightarrow 2{t^2} + t - 1 = 0t = \frac{{ - 1 \pm 3}}{4} = - 1\,,\,\frac{1}{2}\\\\ \Rightarrow \left\{ \begin{array}{l}\sin x = - 1 \Rightarrow x = 2k\pi + \frac{{3\pi }}{2}\\\\\sin x = \frac{1}{2} = \sin (\frac{\pi }{6}) \Rightarrow \left\{ \begin{array}{l}x = 2k\pi + \frac{\pi }{6}\\\\x = 2k\pi + \pi - \frac{\pi }{6} = 2k\pi + \frac{{5\pi }}{6}\end{array} \right.\end{array} \right.\end{array}\)

\(\begin{array}{l}2T = \frac{{7\pi }}{2} - ( - \frac{\pi }{2}) = 4\pi \to T = 4\pi \to \frac{{2\pi }}{{|b|}} = 2\pi \to b = \pm 1\\\\c = \frac{{4 + ( - 2)}}{2} = 1\\\\|a| = \frac{{4 - ( - 2)}}{2} = 3 \to a = - 3\end{array}\)

\(\sin x = \frac{1}{2} \Rightarrow \sin x = \sin \frac{\pi }{6} \to \left\{ \begin{array}{l}x = 2k\pi + \frac{\pi }{6}\\\\x = 2k\pi + \frac{{5\pi }}{6}\end{array} \right.(k \in Z)\)

\(\begin{array}{l}T = \frac{{2\pi }}{{|b|}} = \frac{{2\pi }}{{|\pi |}} = 2\\\\\max = |a| + c = 5\\\\\min = - |a| + c = - 1\end{array}\)

\(\begin{array}{l}1 - 2{\sin ^2}x - 3\sin x + 4 = 0 \Rightarrow - 2{\sin ^2}x - 3\sin x + 5 = 0\\\\ \Rightarrow \left\{ \begin{array}{l}\sin x = - \frac{5}{2} \otimes \\\\\sin x = 1 \Rightarrow 2k\pi + \frac{\pi }{2}\end{array} \right.\end{array}\)

\(\begin{array}{l}\left. \begin{array}{l}|b| = \frac{{2\pi }}{2} = \pi \to b = \pm \pi \\\\|a| = \frac{{4 - ( - 4)}}{2} = 4 \to a = \pm 4\end{array} \right\} \Rightarrow y = \pm 4\sin ( \pm \pi x)\\\\c = \frac{{4 + ( - 4)}}{2} = 0\end{array}\)

\(\begin{array}{l}\sin 2x = \sin x\\\\ \Rightarrow \left\{ \begin{array}{l}2x = 2k\pi + x \to x = 2k\pi \,,\,k \in Z\\\\2x = 2k\pi + \pi - x \to x = \frac{{2k\pi }}{3} + \frac{\pi }{3}\,,\,k \in Z\end{array} \right.\end{array}\)

\(\left\{ \begin{array}{l}|a| = \frac{{2 - ( - 2)}}{2} = 2 \to a = - 2\\\\|b| = \frac{{2\pi }}{{2\pi }} = 1 \to b = 1\\\\c = \frac{{2 + ( - 2)}}{2} = 0\end{array} \right.f(x) = - 2\cos x\)

\(\begin{array}{l}2{\sin ^2}x + \sin x - 1 = 0 \Rightarrow \left\{ \begin{array}{l}\sin x + - 1\\\\\sin x = \frac{1}{2}\end{array} \right.\\\\ \Rightarrow x = 2k\pi - \frac{\pi }{2}\\\\\left\{ \begin{array}{l}x = 2k\pi + \frac{\pi }{6}\\\\x = 2k\pi + \pi - \frac{\pi }{6}\end{array} \right.\end{array}\)

\(\begin{array}{l}\left\{ \begin{array}{l}c = \frac{{5 + 1}}{2} = 3\\\\|a| = \frac{{5 - 1}}{2} = 2\,\,a > 0\,,\,a = 2\end{array} \right.b = \frac{{2\pi }}{{4\pi }} = \frac{1}{2} \to y = 2\cos (\frac{x}{2}) + 3\\\\ \to y = 2\cos ( - \frac{x}{2}) + 3\end{array}\)

\(\begin{array}{l}\sin 2x = \frac{{\sqrt 2 }}{2} = \sin \frac{\pi }{2}\\\\ \Rightarrow \left\{ \begin{array}{l}2x = 2k\pi + \frac{\pi }{3} \to x = k\pi + \frac{\pi }{6}\,,\,k \in Z\\\\2x = 2k\pi + \pi - \frac{\pi }{2} \to x = k\pi + \frac{\pi }{2}\,,\,k \in Z\end{array} \right.\end{array}\)

بیشتر

\(T = \pi \)

\(\begin{array}{l}\min = - |a| + c = - 3 + 5 = 2\\\\\max = |a| + c = 3 + 5 = 8\\\\{R_y} = [2,8]\end{array}\)

\(\begin{array}{l}f(x) = 12\cos (5x - 3) + 4 \Rightarrow T = \frac{{2\pi }}{{|b|}} = \frac{{2\pi }}{5}\\\\\min = - |a| + c = - 12 + 4 = - 8\\\\\max = |a| + c = 12 + 4 = 16\end{array}\)

\(\begin{array}{l}g(x) = \frac{1}{2}\tan \frac{1}{2}x \Rightarrow T = 2 \times \pi = 2\pi \\\\\min = - \infty ,\,(0/25)\\\\\max = + \infty \end{array}\)

\(\begin{array}{l}1 + {\tan ^2}\alpha = \frac{1}{{{{\cos }^2}\alpha }} \Rightarrow 1 + {( - \frac{2}{5})^2}\\\\ = 1 + \frac{4}{{25}} = \frac{{29}}{{25}} = \frac{1}{{{{\cos }^2}\alpha }} \Rightarrow {\cos ^2}x = \frac{{25}}{{29}}\\\\\cos 2\alpha + \sin 2\alpha = {\cos ^2}\alpha - {\sin ^2}\alpha + 2\sin \alpha \cos \alpha \\\\ = {\cos ^2}\alpha (1 - {\tan ^2}\alpha + 2\tan \alpha ) = \frac{{25}}{{29}}(1 - {( - \frac{2}{5})^2} + 2( - \frac{2}{5}))\\\\ = \frac{{25}}{{29}}(1 - \frac{4}{{25}} - \frac{4}{5}) = \frac{{25}}{{29}} \times \frac{1}{{25}} = \frac{1}{{29}}\end{array}\)

\(\begin{array}{l}\left. \begin{array}{l}\max = |a| + c = 5\\\\\min = - |a| + c = 1\end{array} \right\} \Rightarrow 2c = 6\\\\ \Rightarrow c = 3 \Rightarrow |a| = 2a = 2\\\\d = 0 \otimes \\\\y = 2\cos (\frac{1}{2}x) + 3\\\\ \Rightarrow f(x)\sin 2x\cos 2x = (\frac{1}{2}\sin 4x)\end{array}\)

\(\begin{array}{l}\sin 2x - \cos \frac{x}{2} = 0\sin 2x = \sin (\frac{\pi }{2} - \frac{\pi }{2})\\\\ \Rightarrow \left\{ \begin{array}{l}2x = 2k\pi + \frac{\pi }{2} - \frac{\pi }{2} \Rightarrow \frac{{5x}}{2} = 2kV + \frac{\pi }{2}\\\\ \Rightarrow x = \frac{{4k\pi }}{5} + \frac{\pi }{5}\\\\2x = 2k\pi + \pi - \frac{\pi }{2} + \frac{\pi }{2} \Rightarrow \frac{{3x}}{2} = 2k\pi + \frac{\pi }{2}\\\\ \Rightarrow x = \frac{{4k\pi }}{3} + \frac{\pi }{3}\end{array} \right.\end{array}\)

\(\begin{array}{l}2{\sin ^3}x - \sin x = 0 \Rightarrow \sin x(2{\sin ^3}x - 1) = 0\\\\ \Rightarrow \left\{ \begin{array}{l}\sin x = 0 \Rightarrow x = k\pi \\\\{\sin ^2}x = \frac{1}{2} \Rightarrow \left\{ \begin{array}{l}\sin x = \frac{{\sqrt 2 }}{2} \Rightarrow \sin x = \sin (\frac{\pi }{4})\\\\ \Rightarrow \left\{ \begin{array}{l}x = 2k\pi + \frac{\pi }{4}\\\\x = 2k\pi + \frac{{3\pi }}{4}\end{array} \right.\\\\\sin x = - \frac{{\sqrt 2 }}{2} \Rightarrow \left\{ \begin{array}{l}x = 2k\pi - \frac{\pi }{4}\\\\x = 2k\pi + \frac{{5\pi }}{4}\end{array} \right.\end{array} \right.\end{array} \right.\end{array}\)

\(\begin{array}{l}\sin x + \cos x = \frac{{\sqrt 3 }}{5}\\\\{\sin ^2}x + {\cos ^2}x + 2\sin x\cos x = \frac{3}{{25}}\\\\2\sin x\cos x = \sin 2x = \frac{3}{{25}} - 1 = - \frac{{22}}{{25}}\\\\{\sin ^2}2x + {\cos ^2}2x = 1 \Rightarrow {\cos ^2}2x = 1 - \frac{{486}}{{625}} = \frac{{141}}{{625}}\\\\ \Rightarrow \cos 2x = \pm \frac{{\sqrt {141} }}{{25}}\end{array}\)

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

\(\begin{array}{l}x - 1 \ge 0 \Rightarrow x \ge 1 \Rightarrow {D_f} = [1, + \infty )\\\\\sqrt {x - 1} \ge 0 \Rightarrow \sqrt {x - 1} + 1 \ge 1 \Rightarrow {R_f} = [1, + \infty )\end{array}\)

\(x \in {D_f} - 1:fo{f^{ - 1}}(x) = x \Rightarrow fo{f^{ - 1}}(\sqrt 3 ) = \sqrt 3 \)

\(f(g(x)) = \sqrt {x - 1 + 1} = \sqrt x \)

\({f^{ - 1}}(15) = \alpha \Rightarrow f(\alpha ) = 15 \Rightarrow 2{\alpha ^3} - 1 = 15 \Rightarrow \alpha = 2\)

برای مثال \(0,{5^2} > 0,{5^3}\) است.

\(\begin{array}{l}{D_{fog}} = \{ x \in {D_f}|f(x) \in {D_g}\} = \{ x \in R|4{x^2} - 1 \in [ - 1,1]\} \\\\ - 1 \le 4{x^2} - 1 \le 1 \Rightarrow 0 \le {x^2} \le \frac{1}{2} \Rightarrow - \frac{{\sqrt 2 }}{2} \le x \le \frac{{\sqrt 2 }}{2}\\\\ \Rightarrow {D_{fog}} = [ - \frac{{\sqrt 2 }}{2},\frac{{\sqrt 2 }}{2}]\end{array}\)