گام به گام تمرین صفحه 52 درس 2 حسابان یازدهم (تابع)

الف \({D_f} = \mathbb{R} - \left\{ 2 \right\}\)

ب \({D_f} = \mathbb{R}\)

پ \({D_f} = \mathbb{R} - \left\{ { - 4\;,\;3} \right\}\)

ت \({D_f} = \left[ { - \frac{1}{3}\;,\;\infty } \right)\)

ث \({D_f} = \left[ {0\;,\;\infty } \right)\)

ج \({D_f} = \left( { - \infty \;,\;8} \right]\)

با قرینه کردن تابع f نسبت به محور x ، تابع g بدست می آید:

الف 

 \(\left\{ \begin{array}{l}{D_f} = \mathbb{R}\\{R_f} = \left( { - \;\infty \;,\; - 2} \right]\; \cup \left( {0\;,\; + \infty } \right)\end{array} \right.\)

ب 

 \(\left\{ \begin{array}{l}{D_f} = \left[ {2\;,\; + \infty } \right)\\{R_f} = \left[ {5\;,\; + \infty } \right)\end{array} \right.\)

پ 

 \(\left\{ \begin{array}{l}{D_f} = \left[ { - 2\;,\; + \infty } \right)\\{R_f} = \left[ {0\;,\;\sqrt 2 } \right]\; \cup \left( {0\;,\;2} \right)\end{array} \right.\)

ت 

 \(\left\{ \begin{array}{l}{D_f} = \mathbb{R}\\{R_f} = \mathbb{R}\end{array} \right.\)

الف تابع است.

ب تابع نیست.

پ تابع است.

ت تابع نیست.

ث تابع نیست.

ج تابع است.

الف

\(\begin{array}{l}f\left( x \right) = \frac{{255x}}{{100 - x}} \Rightarrow x = 50\\\\ \Rightarrow f\left( {50} \right) = \frac{{255\left( {50} \right)}}{{100 - 50}} = 255\end{array}\)

255 میلیون تومان

ب

\({D_f} = \left[ {0\;,\;100} \right)\)

الف 

 \(\begin{array}{l} - 2 \le x <  - 1 \Rightarrow \left[ x \right] =  - 2 \Rightarrow f\left( x \right) =  - 2 + 1 =  - 1\\ - 1 \le x < 0 \Rightarrow \left[ x \right] =  - 1 \Rightarrow f\left( x \right) =  - 1 + 1 = 0\\0 \le x < 1 \Rightarrow \left[ x \right] = 0 \Rightarrow f\left( x \right) = 0 + 1 = 1\\1 \le x < 2 \Rightarrow \left[ x \right] = 1 \Rightarrow f\left( x \right) = 1 + 1 = 2\\2 \le x < 3 \Rightarrow \left[ x \right] = 2 \Rightarrow f\left( x \right) = 2 + 1 = 3\end{array}\)

ب 

 \(\begin{array}{l} - 2 \le x <  - 1 \Rightarrow \left[ x \right] =  - 2 \Rightarrow f\left( x \right) =  - 2 + 1 =  - 1\\ - 1 \le x < 0 \Rightarrow \left[ x \right] =  - 1 \Rightarrow f\left( x \right) =  - 1 + 1 = 0\\0 \le x < 1 \Rightarrow \left[ x \right] = 0 \Rightarrow f\left( x \right) = 0 + 1 = 1\\1 \le x < 2 \Rightarrow \left[ x \right] = 1 \Rightarrow f\left( x \right) = 1 + 1 = 2\\2 \le x < 3 \Rightarrow \left[ x \right] = 2 \Rightarrow f\left( x \right) = 2 + 1 = 3\end{array}\)

\(\begin{array}{l}y = \left[ x \right] - 3\\ - 1 \le x < 0 \Rightarrow \left[ x \right] =  - 1 \Rightarrow f\left( x \right) =  - 4\\0 \le x < 1 \Rightarrow \left[ x \right] = 0 \Rightarrow f\left( x \right) =  - 3\\1 \le x < 2 \Rightarrow \left[ x \right] = 1 \Rightarrow f\left( x \right) =  - 2\\2 \le x < 3 \Rightarrow \left[ x \right] = 2 \Rightarrow f\left( x \right) =  - 1\\3 \le x < 4 \Rightarrow \left[ x \right] = 3 \Rightarrow f\left( x \right) = 0\\\\y = \left[ {x - 3} \right]\\ - 1 \le x < 0 \Rightarrow  - 4 \le x - 3 <  - 3 \Rightarrow \left[ {x - 3} \right] =  - 4 \Rightarrow f\left( x \right) =  - 4\\0 \le x < 1 \Rightarrow  - 3 \le x - 3 <  - 2 \Rightarrow \left[ {x - 3} \right] =  - 3 \Rightarrow f\left( x \right) =  - 3\\1 \le x < 2 \Rightarrow  - 2 \le x - 3 <  - 1 \Rightarrow \left[ {x - 3} \right] =  - 2 \Rightarrow f\left( x \right) =  - 2\\2 \le x < 3 \Rightarrow  - 1 \le x - 3 < 0 \Rightarrow \left[ {x - 3} \right] =  - 1 \Rightarrow f\left( x \right) =  - 1\\3 \le x < 4 \Rightarrow 0 \le x - 3 < 1 \Rightarrow \left[ {x - 3} \right] = 0 \Rightarrow f\left( x \right) = 0\end{array}\)

این دو تابع با هم برابرند.

الف

\(\begin{array}{l}n\left( t \right) = \frac{{9500t - 2000}}{{4 + t}} \Rightarrow \\n\left( 5 \right) = \frac{{9500\left( 5 \right) - 2000}}{{4 + 5}} = \left[ {5055/\bar 5} \right] = 5055\end{array}\)

ب

\(\begin{array}{l}n\left( t \right) = \frac{{9500t - 2000}}{{4 + t}} \Rightarrow 5500 = \frac{{9500t - 2000}}{{4 + t}} \Rightarrow \\22000 + 5500t = 9500t - 2000\\ \Rightarrow 4000t = 24000 \Rightarrow t = 6\end{array}\)