نمونه سوالات فصل 5 حسابان دوازدهم همراه با جواب تشریحی

\(\left| \begin{array}{l}x - y = 10\\\\p = xy = x(x - 10) = {x^2} - 10x\\\\{p^`}(x) = 2x - 10 = 0 \to x = 5\,,\,y = - 5\end{array} \right.\)

\(\begin{array}{l}{f^`}(x) = 3{x^2} - 12x = 2\,,\,x = - 2\\\\(2, + \infty )\,,\,( - \infty , - 2)\end{array}\)

\(\begin{array}{l}16 = 2(x + y) \Rightarrow x + y = 8 \Rightarrow y = 8 - x\\\\s(x) = x \times y = - {x^2} + 8x \Rightarrow s'(x) = - 2x + 8 \Rightarrow s'(x) = 0\\\\ \Rightarrow - 2x + 8 = 0 \Rightarrow x = 4 \Rightarrow y = 4\end{array}\)

\(\begin{array}{l}{f^`}(x) = {x^2} - 2x - 3 = 0\\\\x = 3\,,\,x = - 1\end{array}\)

\(\begin{array}{l}p = xy = 5{x^2} - 10x \to {p^`}(x) = 0 \to 10x - 10 = 0\\\\ \to \left| \begin{array}{l}x = 1\\\\y = - 5\end{array} \right.\end{array}\)

\(\begin{array}{l}{f^`}(x) = 6{x^2} + 6x - 12 = 0 \to \left\{ \begin{array}{l}x = 1\\\\x = - 2 \notin [ - 1,3]\end{array} \right.\\\\f( - 1) - 13\\\\f(1) = - 7 \Rightarrow \min (1, - 7)\\\\f(3) = 45 \Rightarrow \max (3,45)\end{array}\)

\(\begin{array}{l}f(2) = 1 \Rightarrow 4b + d = - 7\\\\{f^`}(2) = 0 \Rightarrow b = - 3\,,\,d = 5\end{array}\)

\([ - 1,1]\,/( - 1,1)\)

\(\begin{array}{l}{f^`}(2) = 0 \to 12 + 4b = 0\, \Rightarrow b = - 3\\\\f(2) = 1 \to 4b + d = - 7\\\\ - 12 + d = - 7 \to d = 5\end{array}\)

\(\begin{array}{l}y = 7 - x \to s = (y)(x) = 7x - {x^2} \to {s^`}(x) = 7 - 2x = 0\\\\ \to x = 3/5\,,\,y = 3/5\end{array}\)

\(\begin{array}{l}f(x) = \frac{1}{{1 + {x^2}}} \Rightarrow {f^`}(x) = \frac{{ - 2x}}{{{{(1 + {x^2})}^2}}}\\\\x > 0 \Rightarrow - 2x < 0 \Rightarrow \uparrow \\\\x < 0 \Rightarrow - 2x > 0 \Rightarrow {f^`}(x) > 0 \Rightarrow \downarrow \end{array}\)

\(\begin{array}{l}f(x) = a{x^2} + bx \Rightarrow {f^`}(x) = 2ax + b = 0 \Rightarrow x = \frac{{ - b}}{{2a}} = 1\\\\ \Rightarrow b = - 2a\\\\ \to a - 2a = 7 \Rightarrow a = - 7\,,\,b = 14\end{array}\)

\(\begin{array}{l}f(x) = {x^3} - 12x - 5 \Rightarrow {f^`}(x) = 3{x^2} - 12 = 0 \Rightarrow x = \pm 2\\\\\left. \begin{array}{l}x = + 5 \Rightarrow f(5) = {5^3} - 12 \times 5 - 5 = 126 - 60 - 5 = 60\\\\x = + 2 \Rightarrow f(2) = {2^3} - 12 \times 2 - 5 = 8 - 24 - 5 = - 21\\\\x = - 2 \Rightarrow f( - 2) = {( - 2)^3} - 12( - 2) - 5 = - 8 + 24 - 5 = 11\end{array} \right\}\\\\ \Rightarrow \left| \begin{array}{l}\min (2, - 21)\\\\\max (5,60)\end{array} \right.\end{array}\)

\(\begin{array}{l}MO = \sqrt {{{(3 - x)}^2} + {{(0 - y)}^2}} \\\\MO = \sqrt {{{(3 - x)}^2} + ( - \sqrt {3x + 9{)^2}} \,,\,x \ge - 3} \\\\ \Rightarrow MO = \sqrt {9 + {x^2} - 6x + 3x + 9} = \sqrt {{x^2} - 3x + 18} \\\\{(MO)^`} = \frac{{2x - 3}}{{2\sqrt {{x^2} - 3x + 18} }} = 0 \Rightarrow x = \frac{3}{2}\end{array}\)

\(\begin{array}{l}P = 2(x + y) = 50 \Rightarrow x = 25 - y\,(1)\\\\S = xyS = (25 - y)y = 25y - {y^2}\\\\{S^`} = 25 - 2y = 0 \Rightarrow y = \frac{{25}}{2}\\\\S = 25y - {y^2} = 25 \times \frac{{25}}{2} - {(\frac{{25}}{2})^2}\\\\ = \frac{{625}}{2} - \frac{{352}}{4} = \frac{{625}}{4} = 156/25\end{array}\)

\(\begin{array}{l}P = z + 2y + \frac{{x\pi }}{2} = 4 \Rightarrow y = 2 - \frac{x}{2} - \frac{{x\pi }}{4}\,(1)\\\\S = xy + \pi {(\frac{x}{2})^2} \times \frac{1}{2} = xy + \frac{{\pi {x^2}}}{8}\,(2)\\\\S = x(2 - \frac{x}{2} - \frac{{x\pi }}{4}) + \frac{{\pi {x^2}}}{8}\\\\ = 2x - {x^2}\frac{\pi }{4} + {x^2}\frac{\pi }{8} = 2x - \frac{{{x^2}}}{2} - {x^2}\frac{\pi }{8}\\\\{S^`} = 2 - x - \frac{\pi }{4}x = 0 \Rightarrow x = \frac{2}{{1 + \frac{\pi }{4}}} = \frac{8}{{4 + \pi }}\\\\y = 2 - \frac{x}{2} - \frac{\pi }{4}x = 2 - x(\frac{1}{2} + \frac{\pi }{4}) = 2 - \frac{2}{{4 + \pi }}(2 + \pi )\\\\ = 2(1 - \frac{{2 + \pi }}{{4 + \pi }}) = \frac{4}{{4 + \pi }}\end{array}\)

\(\begin{array}{l}n(t) = \frac{{3t}}{{3{t^3} + 8}} \Rightarrow {n^`}(t) = \frac{{3(2{t^3} + 8) - 3t(6{t^2})}}{{{{(2{t^3} + 8)}^2}}} = 0\\\\6{t^3} + 24 - 18{t^3} = 0 \Rightarrow - 12{t^2} + 24 = 0\\\\ \Rightarrow {t^3} = \frac{{24}}{{12}} \Rightarrow {t^3} = 2 \Rightarrow t = \sqrt[3]{2}\end{array}\)

\(y = - {x^3} + 3x - 3 \Rightarrow {y^`} = - 3{x^2} + 3 = 0 \Rightarrow x = \pm 1\)

\(\begin{array}{l}f(x) = ({x^3} - 28)\sqrt x \\\\ \Rightarrow {f^`}(x) = (3{x^2})\sqrt x + ({x^3} - 28)\frac{1}{{2\sqrt x }} = 0\\\\{f^`}(x) = \frac{{3{x^2} \times 2x + {x^3} - 28}}{{2\sqrt x }} = 0\left\{ \begin{array}{l}y{x^3} - 28 = 0 \Rightarrow x = \sqrt[3]{4}\\\\\sqrt x = 0 \Rightarrow x = 0\end{array} \right.\end{array}\)

\(\begin{array}{l}S = 2x \times y = 2xy\\\\S = 2x(2 - {x^2})4x - 2{x^3} \Rightarrow {S^`} = 4 - 6{x^2} = 0\\\\ \Rightarrow {x^2} = \frac{2}{3} \Rightarrow x = \pm \sqrt {\frac{2}{3}} y = 2 - {x^2} = \frac{4}{3}\\\\{S_{\max }} = 2xy = 2 \times \sqrt {\frac{2}{3}} \times \frac{4}{3} = \frac{8}{3}\sqrt {\frac{2}{3}} \end{array}\)

 :نقاط بحرانی \(({x_4},{x_5})/{x_7}/{x_6}/{x_5}/{x_4}/{x_3}/{x_2}/{x_1}\)

MAX :نسبی \({x_6}/{x_2}\)

MIN :نسبی \({x_7}/({x_4},{x_5}]\)

MIN :مطلق \(({x_4},{x_5}]\)

\(\begin{array}{l}f(x) = 2{x^3} - 12x + 3{x^2} \Rightarrow {f^`}(x) = 6{x^2} + 6x - 12 = 0\\\\ \Rightarrow {x^2} + x - 2 = 0\,\,\,x = - 2, + 1\end{array}\)

\(\left. \begin{array}{l}f( - 2) = 20\\\\f(1) = - 7\\\\f( - 3) = + 9\\\\f( + 3) = 45\end{array} \right\} \Rightarrow \left| \begin{array}{l}(3,45)\\\\(1, - 7)\end{array} \right.\)

\(\begin{array}{l}\left. \begin{array}{l}c = ab\\\\a + 2b = 60 \Rightarrow a = 60 - 2b\end{array} \right\} \Rightarrow c = (60 - 2b)b = 60b - 2{b^2}\\\\{c^`} = 0 \Rightarrow 60 - 4b = 0 \Rightarrow b = 15 \Rightarrow a = 30\end{array}\)

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\(\begin{array}{l}f(x) = {x^3} + b{x^2} + d1 = 8 + 4b + d\\\\ \Rightarrow 4b + d = - 7\\\\{f^`}(x) = 3{x^2} + 2bx = 012 + 4b = 0\\\\ \Rightarrow b = - 34( - 3) + d = - 7 \Rightarrow d = 5\end{array}\)

\(\begin{array}{l}f(x) = \sqrt {4 - {x^2}} \Rightarrow {f^`}(x) = \frac{{ - 2x}}{{2\sqrt {4 - {x^2}} }} = \frac{{ - x}}{{\sqrt {4 - {x^2}} }}\\\\x = 0 \Rightarrow (0,2)\end{array}\)

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

\(\begin{array}{l}h(x) = \sqrt[3]{x} \Rightarrow {h^`}(x) = \frac{1}{{3\sqrt[3]{{{x^2}}}}}\\\\(0,0)\end{array}\)

\(\begin{array}{l}{f^`}(x) = 6{x^2} + 6x - 12 \Rightarrow {f^`}(x) = {x^2} + x - 2 = 0\\\\ \Rightarrow \left\{ \begin{array}{l}x = - 2 \notin [ - 1,3]\\\\x = 1\end{array} \right.\\\\f(1) = - 7\,,\,f( - 1) = 13\,,\,f(3) = 45\end{array}\)

\(\begin{array}{l}{f^`}(x) = - 4{x^3} + a - 4 + a = 0 \Rightarrow a = 4\\\\f(1) = 2 \Rightarrow - 1 + 4 + b = 2 \Rightarrow b = - 1\end{array}\)

\({f^`}(x) = 1 - \frac{4}{{{{(x + 1)}^2}}}\left\{ \begin{array}{l}x = - 3\\\\x = 1\end{array} \right. \Rightarrow \left\{ \begin{array}{l}f(0) = 5\\\\f(1) = 4\\\\f(2) = \frac{{13}}{2}\end{array} \right.\)

\(\begin{array}{l}D = R\,\,{f^`}(x) = 3{x^2} - 3\\\\{f^`}(x) = 0 \to \left\{ \begin{array}{l}x = 1\\\\x = - 1\end{array} \right.\\\\\max = \left\{ \begin{array}{l}f(3) = 19\\\\f(\frac{{ - 3}}{2}) = \frac{{17}}{{18}}\\\\f( - 1) = 3\end{array} \right.\\\\\min = f(1) = - 1\end{array}\)

\(\begin{array}{l}{D_f} = R\,\,,{f^`}(x) = \frac{{2x}}{{3\sqrt {{{({x^2} - 1)}^2}} }}\\\\ \Rightarrow \left\{ \begin{array}{l}{f^`}(x) = 0 \Rightarrow x = 0\\\\{f^`}(x) \otimes \Rightarrow x = \pm 1\end{array} \right. \Rightarrow \left\{ {0,1, - 1} \right\}\end{array}\)

\({f^`}(x) = \frac{{ - {x^2} + 1}}{{{{({x^2} + 1)}^2}}} \Rightarrow - {x^2} + 1 = 0 \Rightarrow x = \pm 1 \Rightarrow \left\{ { - 1,1} \right\}\)

\(\begin{array}{l}2x + y = 60 \Rightarrow y = 60 - 2x \Rightarrow xy\\\\ = x(60 - 2x) = 60x - 2{x^2}\\\\ \Rightarrow x = \frac{{ - b}}{{2a}} = \frac{{60}}{{40}} = 15 \Rightarrow y = 60 - 30 = 30\end{array}\)