The equation x + cosx $=$ a has exactly one positive root. Complete set of values of 'a' is
-
(0, 1)
- $(-\infty, 1)$
-
(-1,1)
- $(1,\infty)$
- $(0,\infty)$
Let f(x) = x + cos(x). f'(x) = 1 - sin(x). Since sin(x) <= 1, f'(x) >= 0. The function is strictly increasing. f(0) = 1. For x > 0, f(x) > 1. Thus, for x + cos(x) = a to have a positive root, a must be greater than 1.
The equation x + cosx = a has exactly one positive root when a is strictly greater than 1. To prove this, the function f(x) = x + cosx evaluated at x = 0 gives f(0) = 1. Because the derivative f'(x) = 1 - sinx is always non-negative and strictly positive for all x except isolated points, the function is strictly increasing. Therefore, if a > 1, the horizontal line y = a intersects the increasing curve exactly once for x > 0, giving the complete set of values for a as (1, infinity).