Problem
Let f : (0, 1) → R be the function defined as f(x) = [4x](x – 1/4)²(x – 1/2) where [x] denotes the greatest integer less than or equal to x. Then which of the following statements is (are) true?
A The function f is discontinuous exactly at one point in (0, 1)
B There is exactly one point in (0, 1) at which the function f is continuous but NOT differentiable
C The function f is NOT differentiable at more than three points in (0, 1)
D The minimum value of the function f is -1/512
Solution
Let’s begin by rewriting f(x) according to the values [4x] takes in the intervals (0, 1/4), [1/4, 1/2), [1/2, 3/4), and [3/4, 1).
\( f(x) = \left\{ \begin{array}{lr} 0 & 0 < x < 1/4 \\
(x-1/4)^2(x-1/2) & 1/4 \le x < 1/2 \\
2(x-1/4)^2(x-1/2) & 1/2 \le x < 3/4 \\
3(x-1/4)^2(x-1/2) & 3/4 \le x < 1
\end{array} \right. \)
The next step is to figure out the continuity at the points where the function changes, i.e. x = 1/4, 1/2, and 3/4. I’ll skip the calculations.
At x = 1/4, f(1/4⁻) = f(1/4⁺) = f(1/4) = 0, implying that the function is continuous here. (Here, f(1/4⁻) denotes the left hand limit of f(x) at x = 1/4.)
At x = 1/2, f(1/2⁻) = f(1/2⁺) = f(1/2) = 0, implying that the function is continuous here as well.
At x = 3/4, f(3/4⁻) = 1/8 but f(3/4⁺) = f(3/4) = 3/16, implying that the function is discontinuous here.
Given the above analysis, option A is true.
Now, let’s figure out the differentiability at the same points. I’ll skip the calculations again.
Considering x = 1/4, f'(1/4⁻) = f'(1/4⁺) = 0, implying that the function is differentiable here.
Considering x = 1/2, f'(1/2⁻) = 1/16, but f'(1/2⁺) = 1/8, implying that the function is not differentiable here.
At x = 3/4, the function isn’t continuous, so it isn’t differentiable as well.
Given the above analysis, option B is true, and option C is false. What about option D?
Given the graph of (x – 1/4)²(x – 1/2), the function f(x) remains negative in the interval (1/4, 1/2), positive in (1/2, 1).
So, its minimum value will be achieved in (1/4, 1/2). To find that, we’ll differentiate f(x), equate it to zero, and find the points of extrema – the regular stuff.
In (1/4, 1/2), f'(x) comes out to be (x – 1/4)(3x – 5/4). Equating it to zero gives x = 5/12 and 1/4.
Neglecting x = 1/4, the minimum value occurs at x = 5/12, and is equal to f(5/12) or -1/432, making option D false.
Comments
An easy problem overall. All we had to do this break the function into piecewise intervals. The rest of the process was simple.
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