The function f is discontinuous exactly at the point in (0,1)
There is exactly one point in (0,1) at which the function f is continuous but not differentiable
the function f is not differentiable at more than three points in (0,1)
The minimum value of the function f is\(-\frac{1}{512}\)
To analyze the function \( f(x) = [4x](x-\frac{1}{4})^2(x-\frac{1}{2}) \), where \( [x] \) denotes the greatest integer less than or equal to \( x \), follow these steps:
1. Check where \( [4x] \) may change value within \( (0,1) \). Notice \( [4x]=n \) where \( n \) is an integer, specifically \( n \in \{1, 2, 3\} \) because \( 0 < 4x < 4 \).
2. Points where changes occur are \( x = \frac{1}{4}, \frac{1}{2}, \frac{3}{4} \). Evaluate continuity at these points:
3. Therefore, \( f \) is not continuous at these three points.
1. The problem asks to find exactly one point where the function is continuous but not differentiable:
Thus, function \( f \) is continuous at this domain but not differentiable.
Let \( K \) be an algebraically closed field containing a finite field \( F \). Let \( L \) be the subfield of \( K \) consisting of elements of \( K \) that are algebraic over \( F \).
Consider the following statements:
S1: \( L \) is algebraically closed.
S2: \( L \) is infinite.
Then, which one of the following is correct?
Let $ P(x_1, y_1) $ and $ Q(x_2, y_2) $ be two distinct points on the ellipse $$ \frac{x^2}{9} + \frac{y^2}{4} = 1 $$ such that $ y_1 > 0 $, and $ y_2 > 0 $. Let $ C $ denote the circle $ x^2 + y^2 = 9 $, and $ M $ be the point $ (3, 0) $. Suppose the line $ x = x_1 $ intersects $ C $ at $ R $, and the line $ x = x_2 $ intersects $ C $ at $ S $, such that the $ y $-coordinates of $ R $ and $ S $ are positive. Let $ \angle ROM = \frac{\pi}{6} $ and $ \angle SOM = \frac{\pi}{3} $, where $ O $ denotes the origin $ (0, 0) $. Let $ |XY| $ denote the length of the line segment $ XY $. Then which of the following statements is (are) TRUE?