نمونه سوالات فصل 4 حسابان دوازدهم همراه با جواب تشریحی

الف)

\(\mathop {\lim }\limits_{x \to 2} \frac{{f(x) - f(2)}}{{x - 2}} = {f^`}(2) = 2\)

ب)

\({m_A} > {m_B}\)

الف) خیر

ب) بله، در تمام نقاط بازه \(( - \infty ,2)\)  مشتق پذیر است.

پ)

\(\begin{array}{l}x \ge 2\,:\,f(x) = \sqrt {x - 1} + 2 \to f`(x) = \frac{1}{{2\sqrt {x - 1} }}\\\\ \to f_ + ^`(2) = \frac{1}{2}\end{array}\)

\(f(x) = {x^3} - x \to {f^`}(x) = 3{x^2} - 1\)

\(g(x) = {(\frac{{2x - 1}}{{x + 1}})^3}(\frac{{2(x + 1) - 1(2x - 1)}}{{{{(x + 1)}^2}}}\)

الف)

\(\frac{{h(4) - h(2)}}{{4 - 2}} = \frac{{96 - 64}}{2} = 16\)

ب)

\({h^`}(t) - 8t + 40 = 16 \to t = 3\)

\(\begin{array}{l}{f^`}(2) = 0 \to {f^`}(x) = 3{x^2} + 2bx \Rightarrow b = - 3\\\\f(2) = 1 \to 8 + ( - 12) + d = 1 \Rightarrow d = 5\end{array}\)

الف) A

ب) C

\(\begin{array}{l}f_ + ^`(1) = \mathop {\lim }\limits_{x \to {1^ + }} \frac{{f(x) - f(1)}}{{x - 1}} = \mathop {\lim }\limits_{x \to {1^ + }} \frac{{|{x^2} - 1|}}{{x - 1}} = 2\\\\ \Rightarrow y - 0 = 2(x - 1)\end{array}\)

\({f^`}(x) = \frac{{ - 2(x + 4) - 1( - 2x + 3)}}{{{{(x + 4)}^2}}}\)

\({g^`}(x) = \frac{3}{{2\sqrt {3x + 1} }}({x^2} + 2x) + (2x + 2)(\sqrt {3x + 1} )\)

\(\frac{{f(25) - f(0)}}{{25 - 0}} = \frac{{85 - 50}}{{25}} = \frac{7}{5}\)

\({(fg)^`}(2) = {f^`}(2)g(2) + f(2){g^`}(2) = 5 \times + 3( - 6) = 22\)

\(f(x) = \left\{ \begin{array}{l}ax + 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,x < 0\\\\{x^2} + 3x + 1\,\,\,\,x \ge 0\end{array} \right.\)

\({f^`}(x) = \frac{{\frac{{9(x + 1) - 1(9x - 2)}}{{{{(x + 1)}^2}}}}}{{2\sqrt {\frac{{9x - 2}}{{x + 1}}} }}\)

\(\begin{array}{l}\frac{{f}}{{t}} = \frac{{f(5) - f(0)}}{{5 - 0}} = \frac{{30 - 10}}{5} = 4\\\\{f^`}(t) = 2t - 1 \to {f^`}(2) = 2(2) - 1 = 3\end{array}\)

\(\left| \begin{array}{l}f_ + ^`( - 1) = \mathop {\lim }\limits_{x \to - {1^ + }} \frac{{{x^2} + 3 - 1}}{{x - ( - 1)}} = - 2\\\\f_ - ^`( - 1) = \mathop {\lim }\limits_{x \to - {1^ - }} \frac{{2x + 6 - 4}}{{x - ( - 1)}} = 2\end{array} \right. \Rightarrow f_ + ^`( - 1) \ne f_ - ^`( - 1)\)

\({f^`}(x) = 5{({x^2} + 2x - 1)^4}(2x + 2)\)

\({g^`}(x) = (\frac{3}{{2\sqrt {3x + 2} }})({x^2} + 1) + (\sqrt {3x + 2} (3{x^2})\)

\(3{f^`}(1) + 2{g^`}(1) = 9 + 10 = 19\)

\(\begin{array}{l}\frac{{f(5) - f(0)}}{{5 - 0}} = 4\\\\{f^`}(t) = 2t - 1 = 4 \to t = \frac{5}{2}\end{array}\)

\(\begin{array}{l}\frac{{f(4) - f(3)}}{{4 - 3}} = 1/5 \Rightarrow B(3,22/5)\\\\\frac{{f(5) - f(4)}}{{5 - 4}} = 1/5 \Rightarrow C(5,25/5)\end{array}\)

\({f^`}(x) = (\frac{{9\sqrt x - \frac{1}{{2\sqrt x }}(9x - 2)}}{{{{(\sqrt x )}^2}}}\)

\({g^`}(x) = (6x){(2x - 5)^2} + (3)(2){(2x - 5)^2}(3{x^2} - 4)\)

الف)

\(\frac{{h}}{{r}} = \frac{{h(8) - h(5)}}{{8 - 5}} = \frac{{0 - (75)}}{{9 - 5}} = - 25\)

ب)

\({h^`}(t) = - 10t + 40 = 35 \Rightarrow t = 0/5\)

\(\begin{array}{l}f(x) = \frac{{x - 2}}{{x + 3}} \to f(x) = 0 \Rightarrow x - 2 = 0 \Rightarrow x = 2 \Rightarrow (2,0)\\\\{f^`}(x) = \frac{{(x + 3) - (x - 2)}}{{{{(x + 3)}^2}}} = \frac{5}{{{{(x + 3)}^2}}} \Rightarrow {f^`}(2) = \frac{5}{{{5^2}}} = \frac{1}{5}\\\\y - 0 = \frac{1}{5}(x - 2) \Rightarrow y = \frac{1}{5}x - \frac{2}{5}\end{array}\)

\(\begin{array}{l}\mathop {\lim }\limits_{x \to {1^ + }} f(x) = \mathop {\lim }\limits_{x \to {1^ - }} f(x) = f(1) \Rightarrow 3 + a = b - 1\,(1)\\\\ \Rightarrow a - b = - 4\\\\f_ + ^`(1) = f_ - ^`(1) \Rightarrow 6x + 3a{x^2} = bb - 3a = 6\,(2)\\\\\left\{ \begin{array}{l}a = - 4\\\\ - 3a = + 6\end{array} \right. - 2a = 2 \Rightarrow a = - 1 \Rightarrow b = 3\end{array}\)

\(\begin{array}{l}\mathop {\lim }\limits_{x \to {{( - 3)}^ + }} \frac{{f(x) - f( - 3)}}{{x - ( - 3)}} = \mathop {\lim }\limits_{x \to {{( - 3)}^ + }} \frac{{[x] - 0}}{1} = - 3\\\\\mathop {\lim }\limits_{x \to {{( - 3)}^ - }} \frac{{f(x) - f( - 3)}}{{x - ( - 3)}} = \mathop {\lim }\limits_{x \to {{( - 3)}^ - }} \frac{{[x] - 0}}{1} = - 4\end{array}\)

نیست

بی نهایت

\(\begin{array}{l}f(x) = - \frac{2}{{{x^2}}} + 4{x^4} \Rightarrow {f^`}(x) = \frac{{ + 2 \times 2x}}{{{{({x^2})}^2}}} + 16{x^3}\\\\ \Rightarrow {f^`}(x) = \frac{4}{{{x^3}}} + 16{x^3}\\\\{f^{``}}(x) = \frac{{ - 4 \times 3{x^2}}}{{{{({x^2})}^2}}} + 48{x^2} \Rightarrow {f^{``}}(x) = \frac{{ - 12}}{{{x^4}}} + 48{x^2}\\\\ \Rightarrow {f^{``}}(1) = - 12 + 48 = 36\end{array}\)

\(\begin{array}{l}f(x) = \frac{{\sqrt[3]{x}}}{{5{x^2} + 8x}}\\\\ \Rightarrow {f^`}(x) = \frac{{(\frac{1}{{3\sqrt[3]{{{x^2}}}}})(5{x^2} + 8x) - \sqrt[3]{x}(10x = 8)}}{{{{(5{x^2} + 8x)}^2}}}\end{array}\)

\(\begin{array}{l}f(x) = \frac{1}{x}{(2\sqrt x - 1)^4} \Rightarrow \\\\{f^`}(x) = \frac{{ - 1}}{{{x^2}}}{(2\sqrt x - 1)^4} + \frac{1}{x} \times 4{(2\sqrt x - 1)^3}(\frac{1}{{\sqrt x }})\end{array}\)

\(\begin{array}{l}\left. \begin{array}{l}f(x) = {x^2} - 1\\\\g(x) = \frac{{\sqrt x }}{x} \Rightarrow {g^`}(x) = \frac{{\frac{x}{{2\sqrt x }} - \sqrt x }}{{{x^2}}} = \frac{{ - x}}{{2{x^2}\sqrt x }} = \frac{{ - 1}}{{2x\sqrt x }}\end{array} \right\}\\\\ \Rightarrow {(gof)^`}(x) = {f^`}(x){g^`}(f(x))\\\\ = (2x)(\frac{{ - 1}}{{2({x^2} - 1}}) = \frac{{ - x}}{{\sqrt {({x^2} - 1)} }}\end{array}\)

\(f(x) = \frac{{\sqrt {3x} }}{{\sqrt {3 + {x^3}} }} \Rightarrow {f^`}(x) = \frac{{\frac{{ + 3}}{{2\sqrt {3x} }} \times \sqrt {3 + {x^3}} - \sqrt {3x} \times \frac{{3{x^2}}}{{2\sqrt {3 + {x^3}} }}}}{{3 + {x^3}}}\)

\(\begin{array}{l}{(2f - g)^`}(3) = 2{f^`}(3) - {g^`}(3) = 2 \times 2 - 5 = 4 - 5 = - 1\\\\h(x) = f(x),g(x) \Rightarrow {h^`}(x) = {f^`}(x)g(x) + f(x){g^`}(x)\end{array}\)

\(\begin{array}{l}{f^`}(x) = \frac{{\frac{1}{{\sqrt {{x^2} + 9} }}(x - 1) - \sqrt {{x^2} + 9} }}{{{{(x - 1)}^2}}}\\\\ \Rightarrow {f^`}(4) = \frac{{\frac{{4 \times 3}}{5} - 5}}{{{{(3)}^2}}} = \frac{{ - 13}}{{45}}\\\\ \Rightarrow \frac{7}{6} - ( - \frac{{13}}{{45}}) = \frac{{105}}{{90}} + \frac{{26}}{{90}} = \frac{{131}}{{90}}\end{array}\)

\(\begin{array}{l}y = \frac{{ - 4{x^3}}}{{\sqrt {3x + 2} }} \Rightarrow {y^`}\\\\ = \frac{{( - 12{x^2})\sqrt {3x + 2} - ( - 4{x^3})(\frac{3}{{2\sqrt {3x + 2} }})}}{{{{(\sqrt {3x + 2} )}^2}}}\\\\ = \frac{{( - 12{x^2})(3x + 2) + 6{x^3}}}{{(3x + 2)\sqrt {3x + 2} }} = \frac{{ - 30{x^3} - 24{x^2}}}{{(3x + 2)\sqrt {3x + 2} }}\end{array}\)

\(\begin{array}{l}y = ({x^2} - 1){(x + 2)^2}\\\\ \Rightarrow {y^`} = (2x){(x + 2)^2} + ({x^2} - 1) \times (2 \times (x + 2))\\\\ = 2(x + 2)[{x^2} + 2x + {x^2} - 1]\\\\ = 2(x + 2)(2{x^2} + 2x - 1)\end{array}\)