IPMAT 2020 - Quantitative Aptitude Question 5
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IPMAT 2020 - Quantitative Aptitude
f(x) = |3 – x| + |2 + x| + |5 – x|
x ≤ – 2
= 3 – x + (–(x + 2)) + 5 – x
= 3 – x – x – 2 + 5 – x
= 6 – 3x
Minimum value x = –2
= 6 – 3(–2)
= 6 + 6
= 12
– 2 < x ≤ 3
= 3 – x + 2 + x + 5 – x
= 10 – x
Minimum value at x = 3
= 10 – 3
= 7
3 < x ≤ 5
= – (3 – x) + (x + 2) + (5 – x)
= – 3 + x + x + 2 + 5 – x
= 4 + x
Minimum value at x = 3
= 4 + 3 = 7
x > 5
– (3 – x) + x + 2 + (– (5 – x)
– 3 + x + x +2 – 5 + x
3 × 5 – 6 = 9
So the minimum value of given expression is 7