Question:
Let $ f : \mathbb{R} \rightarrow \mathbb{R} $ be a function defined by $ f(x) = ||x+2| - 2|x|| $. If m is the number of points of local maxima of f and n is the number of points of local minima of f, then m + n is
Let $ f : \mathbb{R} \rightarrow \mathbb{R} $ be a function defined by $ f(x) = ||x+2| - 2|x|| $. If m is the number of points of local maxima of f and n is the number of points of local minima of f, then m + n is
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To find the number of local maxima and minima, determine the critical points of the function by finding where the derivative is zero or undefined. Then analyze the behavior of the function around these critical points using the first or second derivative test. For absolute value functions, consider the points where the expressions inside the absolute value signs change sign.
Updated On: Mar 25, 2026
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- 3
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The Correct Option is B
Approach Solution - 1
To solve for the number of points of local maxima and minima of the function \( f(x) = \left||x+2| - 2|x|\right| \), we must first understand how the function behaves across the real numbers.
- Consider the function \( f(x) = \left||x+2| - 2|x|\right| \).
- To evaluate this function, we need to analyze the expression based on the properties of the absolute value function.
- We have: \( |x + 2| \) and \( |x| \). We need to evaluate them for different intervals of \( x \) because the expressions will change depending on whether \( x \) is positive or negative.
- Analyze the intervals:
- For \( x \geq 0 \):
- \(|x + 2| = x + 2\) and \(|x| = x\).
- Therefore, \( f(x) = \left|x + 2 - 2x\right| = |2 - x|\).
- For \( x < 0 \):
- \(|x + 2| = -(x + 2) = -x - 2\) and \(|x| = -x\).
- Thus, \( f(x) = \left|-x - 2 - 2(-x)\right| = \left|-x - 2 + 2x\right| = |x - 2|\).
- For \( x \geq 0 \):
- Combine results:
- For \( x \geq 0 \), \( f(x) = |2 - x|\).
- For \( x < 0 \), \( f(x) = |x - 2|\).
- Determine critical points:
- Since both \( |2 - x| \) and \( |x - 2| \) represent linear functions reflected about their respective points, they reach local extrema when \( x = 2 \).
- Since these are linear sections, they only change direction (from increasing to decreasing or vice versa) at the points where \( x + 2 = 0 \), \( x = 0 \), and a transition in sign, primarily at the endpoints of intervals.
- Conclusion:
- At \( x = -2 \), \( x = 0 \), and \( x = 2 \), the behavior of the function changes due to directional changes in the linear components.
- Each of these critical points either represents a local maximum or a minimum.
Thus, counting all such critical points, the number of points of local maxima and minima is 3.
The correct answer is therefore 3.
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Approach Solution -2
\( f(x) = ||x+2| - 2|x|| \) Critical points are \( 0, -2, -\frac{2}{3} \)
No. of maxima = 1 No. of minima = 2 m = 1, n = 2 m + n = 1 + 2 = 3
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