جواب تمرین صفحه 115 درس 5 ریاضی دهم (تابع)

\(\begin{array}{*{20}{l}}{f\left( x \right) = 3x - 1}\\{f\left( 0 \right) = - 1\;\;\:,\;\;\:f\left( {\frac{1}{2}} \right) = \frac{1}{2}\;\;\:,\;\;\:f\left( 5 \right) = 14}\\{R = \left\{ { - 1\:,\:\frac{1}{2}\:,\:14} \right\}}\\{f = \left\{ {\left( {0\:,\: - 1} \right)\:,\:\left( {\frac{1}{2}\:,\:\frac{1}{2}} \right)\:,\:\left( {5\:,\:14} \right)} \right\}}\end{array}\)

اگر دامنه R باشد، داریم:

الف

نادرست؛ زیرا: \(D = \mathbb{R}\;\;,\;\;R = \left[ {\left. { - 1\:,\:\infty } \right)} \right.\)

ب

نادرست؛ زیرا: \(R = \left[ {\left. { - \frac{1}{3}\:,\:\infty } \right)} \right.\)

پ

درست.

ت

نادرست.  

الف

\(y = {x^2} - 3\)

\(\begin{array}{*{20}{l}}{D = \left[ { - 2\:,\:3} \right]\:}\\{R = \left[ {0\:,\:6} \right]{\mkern 1mu} }\end{array}\)

ب

\(y = - {x^2} + 2\)

\(\begin{array}{*{20}{l}}{D = \left[ { - 2\:,\:3} \right]\:}\\{R = \left[ { - 7\:,\:2} \right]{\mkern 1mu} }\end{array}\)

پ

\(y = \left| x \right|\)

\(\begin{array}{*{20}{l}}{D = \left[ { - 2\:,\:3} \right]\:}\\{R = \left[ {0\:,\:3} \right]{\mkern 1mu} }\end{array}\)

ت

\(y = - \left| x \right|\)

\(\begin{array}{*{20}{l}}{D = \left[ { - 2\:,\:3} \right]\:}\\{R = \left[ { - 3\:,\:0} \right]{\mkern 1mu} }\end{array}\)

ث

\(y = {\left( {x + 1} \right)^2}\)

\(\begin{array}{*{20}{l}}{D = \left[ { - 2\:,\:3} \right]\:}\\{R = \left[ {0\:,\:16} \right]{\mkern 1mu} }\end{array}\)

ج

\(y = {\left( {x + 1} \right)^2}\)

\(\begin{array}{*{20}{l}}{D = \left[ { - 2\:,\:3} \right]\:}\\{R = \left[ {0\:,\:\frac{5}{2}} \right]{\mkern 1mu} }\end{array}\)

چ

\(y = \left| {x - 2} \right|\)

\(\begin{array}{*{20}{l}}{D = \left[ { - 2\:,\:3} \right]\:}\\{R = \left[ {0\:,\:4} \right]{\mkern 1mu} }\end{array}\)

ح

\(y = - {\left( {x + 2} \right)^2}\)

\(\begin{array}{*{20}{l}}{D = \left[ { - 2\:,\:3} \right]\:}\\{R = \left[ { - 25\:,\:0} \right]{\mkern 1mu} {\mkern 1mu} }\end{array}\)

خ

\(y = - \left| x \right| - 2\)

\(\begin{array}{*{20}{l}}{D = \left[ { - 2\:,\:3} \right]\:}\\{R = \left[ { - 5\:,\: - 2} \right]{\mkern 1mu} }\end{array}\)

د

\(y = {\left( {x - 2} \right)^2} + 3\)

\(\begin{array}{*{20}{l}}{D = \left[ { - 2\:,\:3} \right]\:}\\{R = \left[ {3\:,\:17} \right]{\mkern 1mu} }\end{array}\)

ذ

\(y = \left| x \right| - 2\)

\(\begin{array}{*{20}{l}}{D = \left[ { - 2\:,\:3} \right]\:}\\{R = \left[ { - 2\:,\:1} \right]{\mkern 1mu} }\end{array}\)

ر

\(y = {\left( {x + \frac{1}{2}} \right)^2} - 3\)

\(\begin{array}{*{20}{l}}{D = \left[ { - 2\:,\:3} \right]\:}\\{R = \left[ { - 3\:,\:\frac{{37}}{4}} \right]{\mkern 1mu} }\end{array}\)

\(\begin{array}{l}f\left( x \right) = \left\{ {\begin{array}{*{20}{l}}\begin{array}{l} - 2\;\;\:\;\;\:\;\;\:\:\:;x \le - 3\\\end{array}\\\begin{array}{l}5x + 13\;\;\:\:\:{\mkern 1mu} ; - 3 < x \le - 2\\\end{array}\\\begin{array}{l} - \frac{3}{4}x + \frac{3}{2}\;\;\:; - 2 < x \le 2\\\end{array}\\{4x - 8\;\;\:\;\;\:{\mkern 1mu} ;2 < x \le 3}\end{array}} \right.\\\\f\left( {\sqrt 5 } \right) = 4\sqrt 5 - 8\\\\f\left( 6 \right) \otimes \\\\f\left( 3 \right) = 4\\\\f\left( {\frac{1}{2}} \right) = \frac{{21}}{8}\\\\f\left( 0 \right) = \frac{3}{2}\\\\f\left( { - \frac{5}{2}} \right) = \frac{1}{2}\end{array}\)

\(\begin{array}{*{20}{l}}\begin{array}{l}f\left( x \right) = ax + b\\\\ \Rightarrow \left\{ {\begin{array}{*{20}{l}}\begin{array}{l}f\left( 4 \right) = 3 = 4a + b\\\end{array}\\{f\left( 0 \right) = 3 = b \Rightarrow a = 0}\end{array}} \right. \Rightarrow f\left( x \right) = 3\\\end{array}\\{ \Rightarrow f\left( { - 4} \right) = f\left( { - 1} \right) = 3}\end{array}\)

\(f\left( x \right) = \left\{ {\begin{array}{*{20}{l}}\begin{array}{l}2,000\;\;\:\;\;\:\;\;\:\:\:;0 < x \le 3\\\end{array}\\{1,000x - 1,000\;\;\:\:\:{\mkern 1mu} ;3 < x}\end{array}} \right.\)

الف

ب

به این دلیل که هیچ یک از مؤلفه های اول آن تکرار نشده است. یا به عبارتی دیگر، هر مؤلفه اول، فقط و فقط یک مقدار از برد را دارا است.

\(\begin{array}{*{20}{l}}{D = \mathbb{N}}\\{R = \left\{ {5\:,\:8\:,\:11\:,\:14\:,\:17\:,\:19\:,\: \cdots } \right\}}\end{array}\)

\(\begin{array}{*{20}{l}}\begin{array}{l}f\left( x \right) = a{x^2} + bx + c\\\end{array}\\\begin{array}{l}\left( {1{\mkern 1mu} ,{\mkern 1mu} - 2} \right) \Rightarrow - {\mkern 1mu} 2 = a{\left( 1 \right)^2} + b\left( 1 \right) + c \Rightarrow a + b + c = - 2\\\end{array}\\\begin{array}{l}\left( {2{\mkern 1mu} ,{\mkern 1mu} - 1} \right) \Rightarrow - 1 = a{\left( 2 \right)^2} + b\left( 2 \right) + c \Rightarrow 4a + 2b + c = - 1\\\end{array}\\\begin{array}{l}\left( {0{\mkern 1mu} ,{\mkern 1mu} 1} \right) \Rightarrow 1 = a{\left( 0 \right)^2} + b\left( 0 \right) + c \Rightarrow c = 1\\\end{array}\\\begin{array}{l}\left\{ {\begin{array}{*{20}{l}}\begin{array}{l}a + b = - 3\\\end{array}\\{4a + 2b = - 2}\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{l}}\begin{array}{l}a = 2\\\end{array}\\{b = - 5}\end{array}} \right.\\\\ \Rightarrow y = 2{x^2} - 5x + 1 \Rightarrow y = 2{\left( {x - \frac{5}{4}} \right)^2} - \frac{{13}}{8}\end{array}\end{array}\)

\(\begin{array}{*{20}{l}}{D = \mathbb{R}}\\{R = \left[ {\left. { - \frac{{13}}{8}\:,\:\infty } \right)} \right.}\end{array}\)