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Question

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be defined as follows:
$$f(x) = \left(\frac{|x|}{2} - x\right)\left(x - \frac{|x|}{2}\right)$$
Which of the following statements is/are true?

To determine the correct statements about the function \(f(x) = \left(\frac{|x|}{2} - x\right)\left(x - \frac{|x|}{2}\right)\), we need to analyze the function's behavior, its derivative, and continuity. Let's proceed step-by-step:

1. Understanding the Function:

The given function is \(f(x) = \left(\frac{|x|}{2} - x\right)\left(x - \frac{|x|}{2}\right)\). Notice the use of absolute value, which suggests the function might behave differently for different ranges of \(x\):

  • For \(x \geq 0\)\(|x| = x\).
  • For \(x < 0\)\(|x| = -x\).

2. Case Analysis:

  • Case 1: \(x \geq 0\)
    Substituting \(|x| = x\), the function becomes: 
\[f(x) = \left(\frac{x}{2} - x\right)\left(x - \frac{x}{2}\right) = \left(-\frac{x}{2}\right)\left(\frac{x}{2}\right) = -\frac{x^2}{4}\]

  • This is a downward-opening parabola with its vertex (maximum point) at \(x = 0\).
  • Case 2: \(x < 0\)
    Substituting \(|x| = -x\), the function becomes: 
\[f(x) = \left(\frac{-x}{2} - x\right)\left(x - \frac{-x}{2}\right) = \left(-\frac{3x}{2}\right)\left(\frac{x}{2}\right) = -\frac{3x^2}{4}\]

  • This is also a downward-opening parabola with its vertex (maximum point) at \(x = 0\).

3. Derivative and Differentiability:

Calculate the derivative \(f'(x)\) for each of the cases:

  • For \(x \geq 0\):
     
\[f'(x) = \frac{d}{dx}\left(-\frac{x^2}{4}\right) = -\frac{x}{2}\]
  • For \(x < 0\):
     
\[f'(x) = \frac{d}{dx}\left(-\frac{3x^2}{4}\right) = -\frac{3x}{2}\]

We need to verify the derivative at \(x = 0\) and ensure continuity:

  • From the right: 
\[\lim_{x \to 0^+} f'(x) = \lim_{x \to 0^+} -\frac{x}{2} = 0\]
  • From the left: 
\[\lim_{x \to 0^-} f'(x) = \lim_{x \to 0^-} -\frac{3x}{2} = 0\]

Both side derivatives approach the same limit as \(x \to 0\), indicating \(f'(x)\) is continuous at \(x = 0\). However, \(f(x)\) is non-differentiable at \(x = 0\) due to the change in behavior.

Conclusion:

  • The function \(f(x)\) has a local maximum at \(x = 0\).
  • The derivative \(f'(x)\) is continuous over \(\mathbb{R}\) but not differentiable at \(x = 0\).

Therefore, the correct statements are:

  • \(f\) has a local maximum.
  • \(f'\) is continuous over \(\mathbb{R}\).
  • \(f'\) is not differentiable over \(\mathbb{R}\).
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Important Questions from Maxima & Minima

  1. Which of the following statements is false about convex minimization problem?

  2. For what value of 'x' will the function y = x2 - 4x have the maximum or minimum value?

  3. For a right-angled triangle, if the sum of the lengths of the hypotenuse and a side is kept constant, in order to have a maximum area of the triangle, the angle between the hypotenuse and the side is

  4. The optimum value of the function f(x) = x2 – 4x + 2 is

  5. As \(\rm x\) varies from \(\rm −1\ to \ +3\), which one of the following describes the behaviour of the function \(\rm f(x) = x^3 – 3x^2 + 1\)?

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