Problem No. 50

Algebra Level 5

Find the maximum value of the expression ( x − 4 ) 2 + ( x 3 − 2 ) 2 − ( x − 2 ) 2 + ( x 3 + 4 ) 2 \sqrt{(x-4)^2 + (x^3 - 2)^2 } - \sqrt{ (x-2)^2 + (x^3 + 4)^2} as x x runs through all real numbers.

3 10 3\sqrt{10} 4 10 4\sqrt{10} 5 10 5\sqrt{10} 2 10 2\sqrt{10}

Benedict Dimacutac by Benedict Dimacutac , 21, Philippines

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1 solution

Soldà Federico
Aug 24, 2015

Let a and b two complex numbers: a = ( x − 2 ) + ( x 3 + 4 ) i ⇒ ∣ a ∣ = ( x − 2 ) 2 + ( x 3 + 4 ) 2 a + b = ( x − 4 ) + ( x 3 − 2 ) i ⇒ ∣ a + b ∣ = ( x − 4 ) 2 + ( x 3 − 2 ) 2 S o b = ( a + b ) − a = ( ( x − 4 ) − ( x − 2 ) ) + ( ( x 3 − 2 ) − ( x 3 + 4 ) ) i = − 2 − 6 i a=(x-2)+(x^{3}+4)i \Rightarrow |a|=\sqrt{(x-2)^{2}+(x^{3}+4)^2} \\ a+b=(x-4)+(x^{3}-2)i \Rightarrow |a+b|=\sqrt{(x-4)^{2}+(x^{3}-2)^{2}} \\ So \\ b=(a+b)-a=((x-4)-(x-2))+((x^{3}-2)-(x^{3}+4))i=-2-6i For the Triangle Inequality: ∣ a ∣ + ∣ b ∣ ≥ ∣ a + b ∣ ∣ a + b ∣ − ∣ a ∣ ≤ ∣ b ∣ ( x − 4 ) 2 + ( x 3 − 2 ) 2 − ( x − 2 ) 2 + ( x 3 + 4 ) 2 ≤ ∣ b ∣ ∣ b ∣ = ( − 2 ) 2 + ( − 6 ) 2 = 40 = 2 10 |a|+|b|\geq |a+b| \\ |a+b|-|a|\leq |b| \\ \sqrt{(x-4)^{2}+(x^{3}-2)^{2}}-\sqrt{(x-2)^{2}+(x^{3}+4)^2}\leq |b| \\ |b|=\sqrt{(-2)^{2}+(-6)^2}=\sqrt{40}=\boxed{2\sqrt{10}}

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