جواب کار در کلاس صفحه 52 درس 2 ریاضی و آمار دهم انسانی (تابع)

الف

\(\begin{array}{l}A = \left\{ { - 1\,,\,1\,,\,2\,,\,3\,,\,4} \right\}\\\\{R_f} = \left\{ { - 1\,,\,0\,,\,1\,,\,3} \right\}\end{array}\)

ب

\(\begin{array}{l}A = \left\{ {0\,,\,1\,,\,2\,,\,3\,,\,4} \right\}\\\\{R_f} = \left\{ 2 \right\}\\\\ \Rightarrow f(x) = 2\end{array}\)

پ

\(\begin{array}{l}A = \left\{ { - 2\,,\, - 1\,,\,0\,,\,1\,,\,2\,,\,3\,,\,4} \right\}\\\\{R_f} = \left\{ {0\,,\,1\,,\,2\,,\,3} \right\}\\\\ \Rightarrow f(x) = x + 2\end{array}\)

ت

\(\begin{array}{l}A = \left[ { - 1\,,\,4} \right]\\\\{R_f} = \left[ { - 1\,,\,4} \right]\\\\ \Rightarrow f(x) = x\end{array}\)

الف

\(\begin{array}{l}f:A \to B\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{R_f} = \left\{ {2\,,\,3\,,\,4\,,\,5\,,\,6} \right\}\\\\f(x) = x + 4\,\,\,\,\,\,\,A = \left\{ {2\,,\, - 2\,,\, - 1\,,\,0\,,\,1} \right\}\end{array}\)

ب

\(\begin{array}{l}f = \left\{ {(1\,,\, - 1)\,,\,(2\,,\, - 2)\,,\,(3\,,\, - 3)\,,\,(4\,,\, - 4)\,,\,(\sqrt 2 \,,\, - \sqrt 2 )} \right\}\\\\A = \left\{ {1\,,\,2\,,\,3\,,\,4\,,\,\sqrt 2 } \right\}\\\\{R_f} = \left\{ { - 1\,,\, - 2\,,\, - 3\,,\, - 4\,,\, - \sqrt 2 } \right\}\end{array}\)

پ

\(\begin{array}{l}A = \left\{ {7\,,\,5\,,\,4\,,\, - 1\,,\,0} \right\}\\\\{R_f} = \left\{ {4\,,\,2\,,\,1\,,\, - 4\,,\, - 3} \right\}\end{array}\)

ت

\(\begin{array}{l}A = \left\{ {1\,,\,2\,,\,3\,,\,4\,,\,5} \right\}\\\\{R_f} = \left\{ 1 \right\}\end{array}\)

ث

\(\begin{array}{l}A = \left\{ {1\,,\,2\,,\,3\,,\,\frac{1}{2}\,,\,\sqrt 2 ,\,0} \right\}\\\\{R_f} = \left\{ {1\,,\,2\,,\,3\,,\,\frac{1}{2}\,,\,\sqrt 2 ,\,0} \right\}\end{array}\)