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Home » Shocking Scooby Doo Cyber Chase: Is the Ghost Online? 🤖👻 - jntua results

Shocking Scooby Doo Cyber Chase: Is the Ghost Online? 🤖👻 - jntua results

📅 March 10, 2026 👤 scraface

Mar 10, 2026

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As $ p o q^+ $, $ R o \infty $. As $ p \gg q $, $ R o 0 $. So $ R >...
rac{p+q}{p-q} + rac{p-q}{p+q} \geq 2 \Rightarrow R = \left( rac{p+q}...
ight) - 2 \geq 2 - 2 = 0,
with equality only if $ rac{p+q}{p-q} = rac{p-q}{p+q} $, which impli...
R = rac{x + 1}{x - 1} + rac{x - 1}{x + 1} - 2.
Let $ f(x) = rac{x+1}{x-1} + rac{x-1}{x+1} - 2 $. Combine:
f(x) = rac{(x+1)^2 + (x-1)^2}{(x-1)(x+1)} - 2 = rac{2x^2 + 2}{x^2 -...
ight) = 2\left( rac{2}{x^2 - 1}
ight) = rac{4}{x^2 - 1}.
Since $ x > 1 $, $ x^2 - 1 > 0 $, so $ f(x) > 0 $, and as $ x o 1^+ $...

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