Saved Bookmarks
| 1. |
16. Find the domain and range of the real function, defined by \( f(x)=\frac{x^{2}}{\left(1+x^{2}\right)} \). Show that \( f \) is many-one.Not through graphical method. |
|
Answer» f (x) = \(\frac {x^2}{1+x^2}\) Since, \(x^2 \geq 0\) \(\Rightarrow\) \(1+ x^2 \geq 1 > 0\) \(\therefore\) Domain of function f (x) is R. Also, \(x^2 + 1 > x^2\) \(\Rightarrow\) \(\frac {x^2}{x^2+1} < 1\) Also \(x^2 \geq 0 \) & \(x^2 + 1 \geq 0\) \(\therefore\) \(\frac {x^2}{x^2+1}\geq0\) Hence, 0 \(\leq \frac {x^2}{x^2+1} < 1\) \(\therefore\) Range of function f (x) is (0,1) \(\because\) f (-1) = \(\frac {(-1)^2}{(-1)^2 +1} = \frac 12\) (\(\because\) f (-1) = f(1)) f (1) = \(\frac 1{1+1} = \frac 12\) (\(\because\) f is many one function) |
|