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
Show Hint
- 5
- 3
- 2
- 4
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.
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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