Playing with graphs - (2)

Algebra Level 5

{ ∣ y ∣ = e − x − 0.5 0.5 ( ∣ x ∣ + ∣ y ∣ ) + ∣ ∣ x ∣ − ∣ y ∣ 2 ∣ ≤ 2 \begin{cases} |y|=e^{-x}-0.5 \\ 0.5(|x|+|y|)+\left|\dfrac{|x|-|y|}{2}\right|\leq 2 \end{cases}

Find the area between the above curves.

This is a submission to Problem Writing Party 2016.
14 + 2 ln ⁡ 2 14+2\ln2 16 + 2 ln ⁡ 2 16+2\ln2 14 + 3 ln ⁡ 2 14+3\ln2 14 + 2 ln ⁡ 4 14+2\ln4

Shivam Jadhav by Shivam Jadhav , 22, India

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1 solution

Tom Engelsman
Apr 20, 2019

The latter inequality is actually a square of side length 4 4 defined by:

− 2 + ∣ x ∣ + ∣ y ∣ 2 ≤ ∣ x ∣ − ∣ y ∣ 2 ≤ 2 − ∣ x ∣ + ∣ y ∣ 2 ⇒ ∣ x ∣ ≤ 2 , ∣ y ∣ ≤ 2 ⇒ x , y ∈ [ − 2 , 2 ] -2 + \frac{|x|+|y|}{2} \le \frac{|x|-|y|}{2} \le 2 - \frac{|x|+|y|}{2} \Rightarrow |x| \le 2, |y| \le 2 \Rightarrow x, y \in [-2,2]

which has an area of 16. 16. The desired area between ∣ y ∣ = e − x − 1 2 |y| = e^{-x} - \frac{1}{2} and this square can never exceed 16 16 . The only choice that satisfies this necessary and sufficient condition is 14 + 2 l n ( 2 ) . \boxed{14 + 2 ln(2)}.

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