Question

If $f\left(x\right)=|x-1|+|x+4|+|x-9|+....+|x-2500|\forall xϵR$, then all the values of $x$ where $f\left(x\right)$ has minimum values lie in

• A. $(600,700)$
• B. $(576,678)$
• C. $(625,678)$
• D. None of these

Answer 1

Answer: (A)

$(600,700)$

$f\left(x\right)=|x-1|+|x-4|+|x-9|+|x-16|+...+|x-2500|$

$f\left(x\right)={\sum }_{k=1}^n|x+{(-1)}^k{k}^2|$

$f\left(x\right)=|50x+{\sum }_{k=1}^{50}{(-1)}^k{k}^2|$

Now, $f\left(x\right)$ will be minimum when the value is from the middle

${\left(25\right)}^2=625$

And this value lies in $(600,700)$

Values for which $f\left(x\right)$ is minimum includes $(600,700)$