Question:
Let \(f(x)=\left|x-x_{1}\right|+\left|x-x_{2}\right|\), where \(x_{1}\) and \(x_{2}\) are distinct real numbers. Then the number of points at which \(f(x)\) is minimum, is :
Let \(f(x)=\left|x-x_{1}\right|+\left|x-x_{2}\right|\), where \(x_{1}\) and \(x_{2}\) are distinct real numbers. Then the number of points at which \(f(x)\) is minimum, is :
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|x - x1| + |x - x2| is the sum of distances from x to two fixed points; find where it is smallest.
Updated On: Nov 12, 2026
- 1
- 2
- 3
- more than 3
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The Correct Option is D
Solution and Explanation
Let \(x_1<x_2\). For \(x<x_1\), \(f(x)=x_1+x_2-2x\), which decreases; for \(x>x_2\), \(f(x)=2x-x_1-x_2\), which increases.
For \(x_1\le x\le x_2\), \(f(x)=(x-x_1)+(x_2-x)=x_2-x_1\), a constant.
The minimum value \(x_2-x_1\) is therefore attained at every point of the interval \([x_1,x_2]\), which has infinitely many points, so more than 3.
Answer: (4) more than 3
For \(x_1\le x\le x_2\), \(f(x)=(x-x_1)+(x_2-x)=x_2-x_1\), a constant.
The minimum value \(x_2-x_1\) is therefore attained at every point of the interval \([x_1,x_2]\), which has infinitely many points, so more than 3.
Answer: (4) more than 3
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