Question

|x - 11 + 1x – 2| + |– 31+...+|x – 2019| + |– 2020.
Determine the greatest interval of real numbers [a, b] on which
the given expression has a constant value k for all x € [a, b].
What is the value of k?​

Answer 1

SOLUTION

GIVEN

The given expression is

$\sf{ |x - 1| + |x - 2| + |x - 3| + .. + |x - 2019| + |x - 2020| }$

TO DETERMINE

The greatest interval of real numbers [a,b] on which the given expression has a constant value k for all x ∈ [a,b]

The value of k

EVALUATION

The given expression is

$\sf{ |x - 1| + |x - 2| + |x - 3| + .. + |x - 2019| + |x - 2020| }$

There are even number ( 2020 ) of modulus in the expression

First we consider the expression

$\sf{ |x - 1| + |x - 2| }$

Then

$\sf { |x - 1| + |x - 2| } = \begin{cases} & \sf{ \: \: \: \: - (x - 1) - (x - 2) = - 2x + 3 \: \: \: \: when \: x <1 } \\ \\ & \sf{ (x - 1) - (x - 2) = 1 \: \: \: \: \: when \:1 \leqslant x \leqslant 2} \\ \\ & \sf{ \: \: \: \: (x - 1) + (x - 2) = 2x - 3 \: \: \: \: when \: x <2 } \end{cases}$

So above expression has constant value when 1 ≤ x ≤ 2

Now we consider the given expression

$\sf{ |x - 1| + |x - 2| + |x - 3| + .. + |x - 2019| + |x - 2020| }$

By same argument as explained above we can say that the given expression has constant value when 1010 ≤ x ≤ 1011

For 1010 ≤ x ≤ 1011 the given expression becomes

$\sf{ |x - 1| + |x - 2| + |x - 3| + .. + |x - 2019| + |x - 2020| }$

$\sf{ = |x - 1| + |x - 2| + |x - 3| + .. + |x - 1009| + |x - 1010| + | x - 1011 | + ..+ |x - 2019| + |x - 2020| }$

$\sf{ = (x - 1) + (x - 2)+ (x - 3)+ .. + (x - 1009) + (x - 1010) - ( x - 1011) + .. - (x - 2019) - (x - 2020) }$

$\sf{ = (- 1) + ( - 2)+ ( - 3)+ .. + ( - 1009) + ( - 1010) + 1011 + .. + 2019 + 2020 }$

$\sf{ = - (1 + 2+ 3+ .. + 1009 + 1010) + 1011 + .. + ( 2019 + 2020 ) }$

$\sf{ = (1 + 2+ 3+ .. + 1009 + 1010 + 1011 + .. + 2019 + 2020 ) - 2 \times (1 + 2+ 3+ .. + 1009 + 1010) }$

$\displaystyle \sf{ = \frac{2020}{2} (1 + 2020) - 2 \times \frac{1010}{2}(1 + 1010) }$

$\displaystyle \sf{ = 1010 \times 2021 -1010 \times 1011 }$

$\displaystyle \sf{ = 1010 \times( 2021 - 1011) }$

$\displaystyle \sf{ = 1010 \times 1010 }$

$\displaystyle \sf{ = 1020100 }$

FINAL ANSWER

The required interval = [ 1010 , 1011 ]

The required value of k = 1020100

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