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.
Solve for \( x \):
\( \log_{10}(x^2) = 2 \).
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 $ S $ denote the locus of the point of intersection of the pair of lines $$ 4x - 3y = 12\alpha,\quad 4\alpha x + 3\alpha y = 12, $$ where $ \alpha $ varies over the set of non-zero real numbers. Let $ T $ be the tangent to $ S $ passing through the points $ (p, 0) $ and $ (0, q) $, $ q > 0 $, and parallel to the line $ 4x - \frac{3}{\sqrt{2}} y = 0 $.
Then the value of $ pq $ is