IPMAT 2022 - Quantitative Aptitude Question 6
This is a detailed step-by-step solution for IPMAT 2022 - Quantitative Aptitude Question 6. Understand the concept, common mistakes, and expert tips to solve similar questions in IIM IPMAT exam.
IPMAT 2022 - Quantitative Aptitude
f(x)= |x| + 2|x – 1| + |x – 2| + |x – 4| + |x – 6| +2|x – 10| x ∈ (– ∞, ∞)
Case 1 ? x ≤ 0
f(x)= –x – 2(x – 1) – (x – 2) – (x – 4) – (x – 6) – 2 (x – 10)
= – x – 2x + 2 – x + 2 – x + 4 – x + 6 – 2x + 20
= – 8x + 34
Minimum value =34
Case 2 ? 0 < x ≤ 1
f(x) = x – 2(x – 1) – (x – 2) – (x – 4) – (x – 6) – 2(x – 10)
= x – 2x + 2 – x + 2 – x + 4 – x + 6 – 2x + 20
= – 6x + 34
= 28
Case 3 ? 1 < x ≤ 2
f(x) = x – 2(x – 1) – (x – 2) – (x – 4) – (x – 6) – 2(x – 10)
= x + 2x – 2 – x + 2 – x + 4 – x + 6 – 2x + 20
= – 2x + 30
Minimum value at x = 2
= – 2(2) + 30
= 26
Case 4 ? 2 < x ≤ 4
f(x) = x + 2(x – 1) – (x – 2) – (x – 4) – (x – 6) – 2(x – 10)
= x + 2x – 2 – x – 2 – x + 4 – x + 6 – 2x + 20
= 26
Minimum value at f(x) is 26
Case 4 ? 4 < x ≤ 6
f(x) = x + 2(x – 1) + (x – 2) + (x – 4) – (x – 6) – 2(x – 10)
= x + 2x – 2 + x – 2 + x – 4 – x + 6 – 2x + 20
= 2x + 18
Minimum value at x = 4
= 26
Case 6 ? 6 < x ≤ 10
f(x) = x + 2(x – 1) + (x – 2) + x – 4 + x – 6 – 2x + 20
= x + 2x – 2 + x – 2 + x – 4 + x – 6 – 2x + 20
= 4x + 6
Minimum value at x = 6
= 24 + 6
= 30
Case 7 ? x > 10
= x + 2x – 2 + x – 2 + x – 4 + x – 6 + 2x – 20
= 8x – 34
Minimum value at x = 10
= 80 – 34
= 46
So the minimum value of f(x) in all cases combined = 26