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Question Number 213467 by efronzo1 last updated on 06/Nov/24

\underset{⏟}{B}

Commented by mr W last updated on 06/Nov/24

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Commented by mr W last updated on 06/Nov/24

q u e s t i o n i s s t r a n g e .

i f X ⩾ 1, t h e n X ⩾ 2, X ⩾ 3, e t c . a r e a l l

t r u e . t h a t m e a n s f o r X ⩾ a, t h e r e i s

n o m a x i m u m v a l u e o f a .

Commented by A5T last updated on 06/Nov/24

S a y X =∣ x - 1 ∣ + ∣ x - 2 ∣ + ∣ x - 3 ∣ + \dots + ∣ x - 2022 ∣⩾ a^{2}

m u s t b e t r u e f o r a l l x .

I s h o w e d t h a t X ⩾ 1011^{2} a n d e q u a l i t y h o l d s w h e n

x = \frac{2023}{2} .

S o, t h e m a x i m u m a f o r w h i c h X ⩾ a^{2} w i l l b e t r u e

f o r a l l x i s 1011 . S o, i f a > 1011; t h e n X ⩾ a^{2} w i l l

n o l o n g e r b e t r u e f o r x = \frac{2023}{2} f o r i n s t a n c e .

Answered by A5T last updated on 06/Nov/24

∣ x - 1 ∣ + ∣ 2022 - x ∣ + ∣ x - 2 ∣ + ∣ 2021 - x ∣ + \dots +

∣ x - 1011 ∣ + ∣ 1012 - x ∣⩾ 2021 + 2019 + 2017 + \dots + 1

= 1011^{2} \Rightarrow M a x i m u m v a l u e o f a = 1011

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