نمونه سوالات فصل 6 ریاضی دوازدهم تجربی همراه با جواب تشریحی

(الف

\(\left\{ \begin{array}{l}b = 3\\\\c = 4\end{array} \right. \Rightarrow {a^2} = {b^2} + {c^2} \to 2a = 10\)

(ب

\(2a + 2c = 18\)

\(O( - \frac{a}{2}, - \frac{b}{2}) = ( - 1,1)\,,\,r = \frac{1}{2}\sqrt {{a^2} + {b^2} - 4c} = \sqrt {10} \)

سطح مقطع

\({a^2} = {b^2} + {c^2}c = 4 \Rightarrow F{F^`} = 8\)

الف)

\(O( - \frac{a}{2}, - \frac{b}{2}) = (2, - 1)\,,\,R = \frac{1}{2}\sqrt {{a^2} + {b^2} - 4c} = 3\)

ب)

خیر

\({(0)^2} + {(3)^2} + 2(3) - 4(0) - 4 \ne 0\)

\(\frac{3}{4}\)

(الف

\(\left| \begin{array}{l}a = 5\\\\b = 4\end{array} \right. \to {c^{}} = 25 - 16 \to c = 3 \to {A^`}F = 8\)

(ب

\({S_\Delta } = \frac{1}{2}(5 + 3) \times 4 = 16\)

\(r = \frac{{|3 \times 0 - 4(3) - 3|}}{{\sqrt {{3^2} + {{( - 4)}^2}} }} = 3 \Rightarrow {(x - 0)^2} + {(y - 3)^2} = 9\)

(الف

\(a = \frac{5}{4}c \Rightarrow \frac{{25}}{{16}}{c^2} = 9 + {c^2}\,\,\,F{F^`} = 2c = 8\)

(ب

\(a = 5 \Rightarrow A(1, - 1),A( - 9, - 1)\)

\(r = \frac{{|3 \times 0 - 4 \times 3 - 3|}}{{\sqrt {{3^2} + {{( - 4)}^2}} }} = 3 \Rightarrow {(x - 0)^2} + y - 3{)^2} = 9\)

2

(الف

\(O\left| \begin{array}{l}\frac{{1 + 1}}{2} = 1\\\\\frac{{3 - 5}}{2} = - 1\end{array} \right.F{F^`} = |3 - ( - 5)| = 8 = 2c \to c = 4\)

(ب

\({b^2} = {a^2} - {c^2} = 36 - 16 = 20 \to b = \sqrt {20} \Rightarrow B{B^`} = 2\sqrt {20} \)

\(\begin{array}{l}d = \frac{{|3 \times 2 - 4( - 3) + 2|}}{{\sqrt {{3^2} + {{( - 4)}^2}} }} = 4\\\\R = \sqrt {{3^2} + {{( - 4)}^2}} = 5 \Rightarrow {(x - 2)^2} + {(y + 3)^2} = 25\end{array}\)

\(V = 2(\frac{1}{3}\pi {r^2}h) = 2 \times \frac{1}{3} \times \pi \times {2^2} \times 3 = 8\pi \)

سهمی

\(\begin{array}{l}O(1,3),2a = 10 \Rightarrow a = 5,\frac{c}{a} = 0/8 \Rightarrow c = 4\\\\ \Rightarrow {a^2} = {b^2} + {c^2} \Rightarrow b = \sqrt {25 - 16} = \sqrt 9 = 3\end{array}\)

\(\begin{array}{l}B(1,4)\,{B^1}(5,4) \Rightarrow O(\frac{{1 + 5}}{2},4) = (3,4)\\\\\left. \begin{array}{l}A{A^`} = 8 \Rightarrow 2a = 8 \Rightarrow a = 4\\\\B{B^`} = 4 \Rightarrow 2b = 4 \Rightarrow b = 2\end{array} \right\}{a^2} = {b^2} + {c^2} \Rightarrow c = \sqrt {12} \end{array}\)

\(\begin{array}{l}{x^2} + {y^2} - 2x - 6y + 6 = 0\\\\{a^2} + {b^2} - 4c > 0 \to {( - 2)^2} + {( - 6)^2} - 4(6) = + 16 > 0 \Rightarrow \\\\O\left| \begin{array}{l} + \frac{2}{2}\\\\ + \frac{6}{2}\end{array} \right. \Rightarrow O\left| \begin{array}{l}1\\\\3\end{array} \right.\,,\,r = \frac{1}{2}\sqrt {{a^2} + {b^2} - 4c} = \frac{{\sqrt {16} }}{2} = \frac{4}{2} = 2\end{array}\)

\(\begin{array}{l}3{x^2} + 3{y^2} - 9x + 12y + 24\\\\ = 0{x^2} + {y^2} - 3x + 4y + 8 = 0\\\\{a^2} + {b^2} - 4c > 0 \otimes \to {( - 3)^2} + {(4)^2} - 4 \times 8 = - 1 < 0\end{array}\)

\(\begin{array}{l}\left. \begin{array}{l}r = \frac{1}{2}\sqrt {{{(2 - 1)}^2} + {{(3 - 1)}^2}} = \frac{1}{2}\sqrt {1 + 4} = \frac{{\sqrt 5 }}{2}\\\\{x_0} = \frac{{3 + 1}}{2} = 2\,.\,{y_0} = \frac{{2 + 1}}{2} = \frac{3}{2}\end{array} \right\}\\\\ \Rightarrow {(x - 2)^2} + {(y - \frac{3}{2})^2} = {(\frac{{\sqrt 5 }}{2})^2}\end{array}\)

\(\begin{array}{l}{(x - 3)^2} + {(y - 4)^2} - 16 = 0\\\\0 + {(2 - 4)^2} - 16 = - 12 < 0\end{array}\)

\(\begin{array}{l}OM = \sqrt {O{H^2} + H{M^2}} = \sqrt {{{(\frac{1}{5})}^2} + {4^2}} = \sqrt {\frac{{401}}{{25}}} = \frac{1}{5}\sqrt {401} \\\\{(x - 1)^2} + {(y - 3)^2} = \frac{{401}}{{25}}\end{array}\)

\(\begin{array}{l}(1):{(x + 1)^2} + {(y - 1)^2} - 2 = 0 \Rightarrow {r_1} = \sqrt 2 ,{O_1}( - 1,1)\\\\(2):{x^2} + {y^2} - 4y + b = 0\\\\ \Rightarrow {O_2}\left| \begin{array}{l}0\\\\ + 2\end{array} \right.,{r_2} = \frac{1}{2}\sqrt {{4^2} + {0^2} - 4b} = \frac{1}{2}\sqrt {16 - 4b} = \sqrt {4 - b} \\\\ \Rightarrow O{O^`} = \sqrt {{{(2 - 1)}^2} + {{(0 - ( - 1))}^2}} = \sqrt 2 \\\\O{O^`} = |{r_1} - {r_2}|\\\\ \Rightarrow \left\{ \begin{array}{l} - \sqrt {4 - b} \Rightarrow b = 4\\\\\sqrt 2 = \sqrt {4 - b} - \sqrt 2 \Rightarrow 2\sqrt 2 = \sqrt {4 - b} \end{array} \right.\\\\ \Rightarrow 8 = 4 - b \Rightarrow b = - 4\end{array}\)

\(\begin{array}{l}{x^2} + {y^2} - 2x - 6y + a = 0\\\\r = \frac{1}{2}\sqrt {{{( - 2)}^2} + {{( - 6)}^2} - 4a} = 2 \Rightarrow \sqrt {40 - 4a} = 4\\\\40 - 4a = 16 \Rightarrow a = \frac{{24}}{4} = 6\end{array}\)

\(\left. \begin{array}{l}{M_{OA}} = \frac{{3 - 0}}{{1 - 0}} = 3 \Rightarrow {m_d} = - \frac{1}{3}\\\\A(1,3)\end{array} \right\} \Rightarrow y = - \frac{1}{3}x + \frac{{10}}{3}\)

\(\begin{array}{l}{x^2} + {y^2} - 2x - 3 = 0 \Rightarrow O( + 1,0)\\\\r = \frac{1}{2}\sqrt {{{( - 2)}^2} - 4( - 3)} = 2\\\\2x + y = 1 \Rightarrow OH = \frac{{|2 + 0 - 1|}}{{\sqrt {{2^2} + 1} }} = \frac{1}{{\sqrt 5 }} < r\\\\r - OH = 2 - \frac{{\sqrt 5 }}{5}\\\\r + OH = 2 + \frac{{\sqrt 5 }}{5}\end{array}\)

\(\begin{array}{l}{(x + 4)^2} + {(y - 2)^2} - m = 0\,,\,M( - 3,3)\\\\ \Rightarrow {( - 3 + 4)^2} + {(3 - 2)^2} - m < 0 \Rightarrow 2 - m < 0 \Rightarrow m > 2\end{array}\)

بیضی

مخروط ناقص

کشیده تر

\(\left[ \begin{array}{l}\alpha \\\\\beta - \frac{a}{2}\end{array} \right]\,,\,\left[ \begin{array}{l}\alpha \\\\\beta + \frac{a}{2}\end{array} \right]\)

\(\begin{array}{l}V = \frac{1}{3}\pi {(BH)^2}(AH) + \frac{1}{3}\pi {(BH)^2}({A^`}H)\\\\ = \frac{1}{3}\pi B{H^2}(AH + {A^`}H) = \frac{1}{3}\pi B{H^2}(A{A^`})\\\\ = \frac{1}{3}\pi {(4/8)^2} \times 10 = 76/8\pi \end{array}\)

\(\begin{array}{l}\left. \begin{array}{l}a = 3\\\\b = 2\end{array} \right\} \Rightarrow c = \sqrt {9 - 4} = \sqrt 5 \\\\A(6,2)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{A^`}(0,2)\\\\B(3,4)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{B^`}(3,0)\\\\F(3 + \sqrt 5 ,2)\,\,\,\,\,\,\,F(3 - \sqrt 5 ,2)\end{array}\)

\(\begin{array}{l}\left. \begin{array}{l}A(10,3)\\\\{F^`}(3,3)\\\\O(6,3)\end{array} \right\} \Rightarrow \left| \begin{array}{l}a = 10 - 6 = 4\\\\c = 6 - 3 = 3\end{array} \right. \Rightarrow b = \sqrt {16 - 9} = \sqrt 7 \\\\2a = 9\\\\2b = 2\sqrt 7 \\\\\frac{c}{a} = \frac{3}{4} = 0/75\end{array}\)

کوچکتر