Question

For any re R.minimum value of x-1|+|2x-1|+|3x-1|+.. |119x-1| is​

Answer 1

Required minimum value is 0

Explanation:

We know that,

| a + b + c + d + ... | ≤ | a | + | b | + | c | + | d | + ...

i.e., | a | + | b | + | c | + ... ≥ | a + b + c + ... |

Now, | x - 1 | + | 2x - 1 | + ... + | 119x - 1 |

≥ | (x - 1) + (2x - 1) + ... + (119x - 1) |

= | (x + 2x + ... + 119x) - (1 + 1 + 1 ... 119 times) |

= | x (1 + 2 + ... + 119) - 119 |

= | x * 119/2 * (1 + 119) - 119 |

= | x * 119/2 * 120 - 119 |

= | 119 (60x - 1) | ..... (1)

∴ the minimum value of (60x - 1) = 0, at x = 1/60

From (1), we get, by putting x = 1/60,

| x - 1 | + | 2x - 1 | + ... | 119x - 1 | ≥ 0

∴ the required minimum value is 0.