IPMAT Indore 2022 (SA) - Given that f(x)=|x|+2|x-1|+|x-2|+|x-4|+|x-6|+2|x-10|, x (-∞, ∞) the minimum value of f(x) is _________. | PYQs + Solutions | AfterBoards
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IPMAT Indore 2022

Algebra
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Modulus

Medium

Given that f(x)=∣x∣+2∣x−1∣+∣x−2∣+∣x−4∣+∣x−6∣+2∣x−10∣,x∈(−∞,∞) f(x)=|x|+2|x-1|+|x-2|+|x-4|+|x-6|+2|x-10|, x \in(-\infty, \infty) the minimum value of f(x)f(x) is _________.

Entered answer:

Correct Answer: 26
How to find critical point? Equate the stuff within the modulus to 00. \newline ∣x∣→x=0|x| \rightarrow \boxed{x=0} \newline ∣x−1∣→x−1=0→x=1|x-1| \rightarrow x-1=0 \rightarrow \boxed{x =1}
How to find the median of critical points? Note: Repeat the critical point by the number of times the term appears.
In our question, the points are: 0,1,1,2,4,6,10,100, 1, 1, 2, 4, 6, 10, 10. Hence, median =2,4= 2, 4.
Minimum Value (multiple same terms)

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Minimum Value (multiple same terms)

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