Saved Bookmarks
| 1. |
The range of the function f(x) = cot-1(x2/(x2 + 1)) is (a, b), find the value of (b/a + 2). |
|
Answer» f(x) = cot-1(\(\frac{x^2}{x^2+1}\)) \(\because\) x2 + 1 > x2 ⇒ \(\frac{x^2}{x^2+1}<1\) Also \(\frac{x^2}{x^2+1}\geq0\) (\(\because\) x2 \(\geq\) 0 & x2 + 1 > 0) ⇒ 0 \(\leq\) \(\frac{x^2}{x^2+1}<1\) \(\because\) cot-1x is a decreasing function. \(\therefore\) cot-1 1 < cot-1(\(\frac{x^2}{x^2+1}\)) \(\leq\) cot-10 ⇒ \(\frac{\pi}4\) < f(x) \(\leq\) \(\frac{\pi}2\) \(\therefore\) Range of f(x) is ( \(\frac{\pi}4\), \(\frac{\pi}2\)]. \(\therefore\) a = π/4 & b = π/2 \(\therefore\) b/a + 2 = \(\frac{\pi/2}{\pi/4}\) + 2 = 4/2 + 2 = 4 |
|