نمونه سوالات فصل 5 حسابان یازدهم همراه با جواب تشریحی

\(\begin{array}{l}f(x) = \frac{{|x|}}{x} \Rightarrow f(x) = \left\{ \begin{array}{l}\frac{x}{x}\,\,\,\,x > 0\\\\\frac{{ - x}}{x}\,\,x < 0\end{array} \right.\\\\ \Rightarrow f(x) = \left\{ \begin{array}{l}1\,\,\,\,\,\,\,x > 0\\\\ - 1\,\,\,x < 0\end{array} \right.\\\\\mathop {\lim }\limits_{x \to {0^ + }} f(x) = 1\,\,,\,\,\mathop {\lim }\limits_{x \to {0^ - }} f(x) = - 1 \Rightarrow \mathop {\lim }\limits_{x \to 0} f(x) \otimes \end{array}\)

\(\begin{array}{l}\mathop {\lim }\limits_{x \to {0^ + }} f(x) = {(0)^2} + 2 = 2\,\,,\,\,\mathop {\lim }\limits_{x \to {0^ - }} f(x) = - 2(0) - 2 = - 2\\\\ \Rightarrow \mathop {\lim }\limits_{x \to 0} f(x) \otimes \end{array}\)

\(\begin{array}{l}\mathop {\lim }\limits_{x \to {2^ + }} f(x) = - 2 + 2 = 0\,\,,\,\,\mathop {\lim }\limits_{x \to 2} f(x) = 2 - 3 = - 1\\\\ \Rightarrow \mathop {\lim }\limits_{x \to 2} f(x) \otimes \end{array}\)

\(\mathop {\lim }\limits_{x \to {0^ - }} f(x) = \mathop {\lim }\limits_{x \to {0^ + }} f(x) = 0 \Rightarrow \mathop {\lim }\limits_{x \to 0} f(x) = 0\)

(الف

\(\mathop {\lim }\limits_{x \to {2^ + }} f(x) = 0 = 0\)

(ب

\(\mathop {\lim }\limits_{x \to {2^ - }} f(x) = 0 \otimes \)

(پ

\(\mathop {\lim }\limits_{x \to 2} f(x) = 0 \otimes \)

(ت

\(f(2) = \sqrt {2 - 2} = 0\)

\(\begin{array}{l}\mathop {\lim }\limits_{x \to {0^ + }} f(x) = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{(1 - \cos x)(1 + \cos x)}}{{1 - \cos x}} = 2\\\\f(0) = a + 1 = 2 \Rightarrow a = 1\\\\\mathop {\lim }\limits_{x \to {0^ - }} [x + 2] + b = \mathop {\lim }\limits_{x \to {0^ - }} [x] + 2 + b = - 1 + 2 + b = 2 \Rightarrow b = 1\end{array}\)

\(\begin{array}{l}\mathop {\lim }\limits_{x \to {1^{}}} f(x) = \mathop {\lim }\limits_{x \to {1^ - }} \frac{{\sqrt x - 1}}{{x - 1}} \times \frac{{\sqrt x + 1}}{{\sqrt x + 1}} = \mathop {\lim }\limits_{x \to {1^ - }} \frac{1}{{\sqrt x + 1}} = \frac{1}{2}\\\\\mathop {\lim }\limits_{x \to {1^ + }} f(x) = \mathop {\lim }\limits_{x \to {1^ + }} [x] + a = 1 + a\\\\\mathop {\lim }\limits_{x \to {1^ + }} f(x) = \mathop {\lim }\limits_{x \to {1^ - }} f(x) = f(1) \Rightarrow 1 + a = \frac{1}{2} \Rightarrow a = - \frac{1}{2}\end{array}\)

\(\begin{array}{l}\left. \begin{array}{l}\mathop {\lim }\limits_{x \to {1^ + }} {x^2} + 2ax + 2 = 1 + 2a + 22 = 3 + 2a = f(1)\\\\\mathop {\lim }\limits_{x \to {1^ - }} a|x| + 1 = a + 1\end{array} \right\}\\\\ \Rightarrow 3 + 2a = a + 1 \Rightarrow a = - 2\end{array}\)

الف) نادرست

ب) درست

\(\begin{array}{l}f(1) = \mathop {\lim }\limits_{x \to 1} f(x)\\\\f(1) = a + 2\\\\\mathop {\lim }\limits_{x \to 1} f(x) = \mathop {\lim }\limits_{x \to 1} \frac{{(x - 1)(x + 1)}}{{(x - 1)(x - 2)}} = - 2 \Rightarrow a + 2 = - 2\\\\ \Rightarrow a = - 4\end{array}\)

\(\begin{array}{l}\left. \begin{array}{l}\mathop {\lim }\limits_{x \to {{00}^ + }} f(x) = \mathop {\lim }\limits_{x \to {0^ + }} ({x^2} - 2) = {(0)^2} + 2 = 2\\\\\mathop {\lim }\limits_{x \to {0^ - }} f(x) = \mathop {\lim }\limits_{x \to {0^ - }} ( - 2x + 2) = - 2(0) + 2 = 2\end{array} \right\}\\\\ \Rightarrow \mathop {\lim }\limits_{x \to 0} f(x) = f(0)\\\\\mathop {\lim }\limits_{x \to {x_0}} f(x) = f({x_0})\end{array}\)

\(\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt[3]{{28 - x}}}}{{(x + 1)}} = \frac{{\sqrt[3]{{27}}}}{2} = \frac{3}{2}\)

\(\begin{array}{l}\mathop {\lim }\limits_{x \to 3} \frac{{\sqrt {x + 1} - 2}}{{{x^2} - 9}} \times \frac{{\sqrt {x + 1} + 2}}{{\sqrt {x + 1} + 2}}\\\\ = \mathop {\lim }\limits_{x \to 3} \frac{{(x - 3)}}{{(x - 3)(x + 3)(\sqrt {x + 1} + 2)}} = \frac{1}{{24}}\end{array}\)

\(\begin{array}{l}\mathop {\lim }\limits_{x \to 2} \frac{{x - \sqrt {x + 2} }}{{{x^2} - 4}} \times \frac{{x + \sqrt {x + 2} }}{{x + \sqrt {x + 2} }} = \mathop {\lim }\limits_{x \to 2} \frac{{{x^2} - x - 2}}{{({x^2} - 4)(x + \sqrt {x + 2} )}}\\\\ = \mathop {\lim }\limits_{x \to 2} \frac{{(x + 1)}}{{(x + 2)(x + \sqrt {x + 2} }} = \mathop {\lim }\limits_{x \to 2} \frac{{x + 1}}{{(x + 2)(x + \sqrt {x + 2} )}}\\\\ = \frac{3}{{4 \times 4}} = \frac{3}{{16}}\end{array}\)

\(\begin{array}{l}\mathop {\lim }\limits_{x \to 3} \frac{{\sqrt[3]{{x + 5}} - 2}}{{{x^2} - 9}} \times \frac{{\sqrt[3]{{{{(x + 5)}^2}}} + 2\sqrt[3]{{x + 5}} + 4}}{{\sqrt[3]{{{{(x + 5)}^2} + 2\sqrt[3]{{x + 5}} + 4}}}}\\\\\mathop {\lim }\limits_{x \to 3} \frac{1}{{(x + 3)(\sqrt[3]{{{{(x + 5)}^2} + 2\sqrt[3]{{x + 5}} + 4}})}}\\\\ = \mathop {\lim }\limits_{x \to 3} \frac{1}{{(x + 3)\sqrt[3]{{{{(x + 5)}^2} + 2\sqrt[3]{{x + 5}} + 4}}}} = \frac{1}{{6 \times 12}} = \frac{1}{{72}}\end{array}\)

\(\begin{array}{l}\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt x - 1}}{{\sqrt {x + 8} - 3}} \times \frac{{\sqrt x + 1}}{{\sqrt x + 1}} \times \frac{{\sqrt {x + 8} + 3}}{{\sqrt {x + 8} + 3}}\\\\ = \mathop {\lim }\limits_{x \to 1} \frac{{(\sqrt {x + 8} ) + 3}}{{(\sqrt x + 1)}} = \frac{6}{2} = 3\end{array}\)

\(\begin{array}{l}\left\{ \begin{array}{l}x = 1 \Rightarrow x - 1 = 0\\\\x = 5 \Rightarrow x - 5 = 0\end{array} \right.(x - 1)(x - 5) = 0\\\\ \Rightarrow {x^2} - 6x + 5 = 03{x^2} - 18x + 15 = 0\\\\ \Rightarrow \left\{ \begin{array}{l} - a = - 18 \Rightarrow a = 18\\\\2b - 1 = 15 \Rightarrow b = 8\end{array} \right.\end{array}\)

\(\begin{array}{l}\mathop {\lim }\limits_{x \to \frac{{5\pi }}{6}} 2\sin x + 1 = 2\sin (\frac{{5\pi }}{6}) + 1 = 2 \times \sin (\pi - \frac{\pi }{6}) + 1\\\\ = 2 \times \sin (\frac{\pi }{6}) + 1 = 2 \times \frac{1}{2} + 1 = 2\end{array}\)

\(\mathop {\lim }\limits_{x \to 0} \sqrt x \Rightarrow \left\{ \begin{array}{l}\mathop {\lim }\limits_{x \to {0^ + }} \sqrt x = 0\\\\\mathop {\lim }\limits_{x \to {0^ - }} \sqrt x = \otimes \end{array} \right. \Rightarrow \mathop {\lim }\limits_{x \to 0} \sqrt x \otimes \)

\(\begin{array}{l}\left\{ \begin{array}{l}x = - 3 \Rightarrow x + 3 = 0\\\\x = 7 \Rightarrow x - 7 = 0\end{array} \right.(x + 3)(x - 7) = 0\\\\ \Rightarrow {x^2} - 4x - 21 = 0\\\\2{x^2} - 8x - 42 = 0 \Rightarrow \left\{ \begin{array}{l}a = - 8\\\\b = - 43\end{array} \right.\end{array}\)

\(\mathop {\lim }\limits_{x \to 8} \frac{{x + 7}}{{\sqrt {x + 1} }} = \frac{{8 + 7}}{{\sqrt {8 + 1} }} = \frac{{15}}{3} = 5\)

\(\mathop {\lim (2 + [x])}\limits_{x \to {3^ - }} = 2 + 2 = 4\)

\(\mathop {\lim }\limits_{x \to 4} \frac{{{x^2} - 16}}{{{x^2} - 5x + 4}} = \mathop {\lim }\limits_{x \to 4} \frac{{(x + 4)}}{{(x - 1)}} = \mathop {\lim }\limits_{x \to 4} \frac{{x + 4}}{{x - 1}} = \frac{8}{3}\)

\(\begin{array}{l}\mathop {\lim }\limits_{x \to \frac{{3\pi }}{2}} \frac{{1 + {{\sin }^3}x}}{{{{\cos }^2}x}} = \mathop {\lim }\limits_{x \to \frac{{3\pi }}{2}} \frac{{(1 - \sin x + {{\sin }^2}x)}}{{(1 - \sin x)}}\\\\ = \mathop {\lim }\limits_{x \to \frac{{3\pi }}{2}} \frac{{1 - \sin x + {{\sin }^2}x}}{{1 - \sin x}} = \frac{{1 + 1 + 1}}{{1 + 1}} = \frac{3}{2}\end{array}\)

(الف

\(\begin{array}{l}\mathop {\lim }\limits_{x \to 1} (2f(x) - g(x)) = 2\mathop {\lim }\limits_{x \to 1} f(x) - \mathop {\lim }\limits_{x \to 1} g(x)\\\\ = 2( - 1) - 3 = - 5\end{array}\)

(ب

\(\begin{array}{l}\mathop {\lim }\limits_{x \to - 2} (f(x) - 3g(x)) = \mathop {\lim }\limits_{x \to - 2} f(x) - 3\mathop {\lim }\limits_{x \to - 2} g(x)\\\\ = 1 - 3(1) = - 2\end{array}\)

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

\(\begin{array}{l}\mathop {\lim }\limits_{x \to 1} \frac{{|x + 2| - |4x - 1|}}{{{x^2} - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{|x + 2| - |4x - 1|}}{{{x^2} - 1}}\frac{{ - 3}}{{(x + 1)}}\\\\ = \frac{{ - 3}}{2}\end{array}\)

(الف

\(x \ne 1 \Rightarrow h(x) = \frac{{{x^2} - 1}}{{x - 1}} = \frac{{(x + 1)}}{1} \Rightarrow h(x) = x + 1\)

(ب

\(\begin{array}{l}h(1) = 3\\\\\mathop {\lim }\limits_{x \to 1} h(x) = 2\end{array}\)

\(\mathop {\lim }\limits_{x \to 1} \frac{{|x + 1|}}{{x - 1}} = 1 \otimes \)

\(\begin{array}{l}\mathop {\lim }\limits_{x \to 3} \frac{{{x^2} - 3x}}{{{x^2} - 9}} = \frac{0}{0}\\\\\mathop {\lim }\limits_{x \to 3} \frac{{x(x - 3)}}{{(x - 3)(x + 3)}} = \frac{x}{{x + 3}} = \frac{1}{2}\end{array}\)

الف) حد f+g تابع وجود ندارد.

ب) حد f+g تابع وجود ندارد.

\(\begin{array}{l}\mathop {\lim }\limits_{x \to 1} \frac{{{x^2} - x}}{{(x + 2)(x - 1)(x + \sqrt x )}}\\\\ = \mathop {\lim }\limits_{x \to 1} \frac{{x(x - 1)}}{{(x + 2)(x - 1)(x + \sqrt x )}} = \frac{{ + 1}}{6}\end{array}\)

\(\log _2^x = t \Rightarrow \mathop {\lim }\limits_{t \to 1} \frac{{t - \frac{1}{t}}}{{2(t - 1)}}\mathop {\lim }\limits_{t \to 1} \frac{{1 + \frac{1}{{{t^2}}}}}{2} = \frac{2}{2} = 1\)

\(\mathop {\lim }\limits_{x \to {{( - 2)}^ + }} (x) = 1\)

الف)

2

ب)

2-

پ)

1

ت)

حد ندارد

\(\mathop {\lim }\limits_{x \to {1^ + }} f(x) - \mathop {\lim }\limits_{x \to {1^ - }} f(x) + f(3) = 0 - 1 + 2 = 1\)