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Home » Therefore, the remainder is \(\boxed0\). - jntua results

Therefore, the remainder is \(\boxed0\). - jntua results

📅 March 9, 2026 👤 scraface

Mar 09, 2026

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At \(v = 2\), \(g = 8 / (1)^2 = 8\)
Wait — this suggests decreasing, but 440 > 8, so yes, decreases.
But we seek the maximum value. Since \(g(v)\) decreases from near infi...
Thus, \((\sec x + \csc x)^2\) is unbounded on \(0 < x < \frac{\pi}{2}\...
But wait — is that correct?
Let \(x \to 0^+\), \(\sec x \to 1\), \(\csc x \to \infty\), so sum \(\...
Hence, there is **no maximum value**; it spans \((8, \infty)\).
But let’s verify at \(x = \frac{\pi}{4}\):
\(\sec \frac{\pi}{4} = \sqrt{2}\), \(\csc \frac{\pi}{4} = \sqrt{2}\),...
But as \(x \to 0^+\), \(f(x) \to \infty\), so no maximum.

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