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Home » T(3) = 2×3+1=7; 2³−1=7 ✔️ - jntua results

T(3) = 2×3+1=7; 2³−1=7 ✔️ - jntua results

📅 March 9, 2026 👤 scraface

Mar 09, 2026

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📌 \boxed{\dfrac{1}{2}}

📌 Question: In a synthetic biology diagram, $\|\overrightarrow{OA}\| = 2$, $\|\overrightarrow{OB}\| = 3$, and $\angle AOB = 90^\circ$. If $\overrightarrow{OC} = 2\overrightarrow{OA} - \overrightarrow{OB}$, find $\|\overrightarrow{OC}\|$.

📌 Solution: Since $\angle AOB = 90^\circ$, $\overrightarrow{OA} \cdot \overrightarrow{OB} = 0$. Compute $\|\overrightarrow{OC}\|^2 = 4\|\overrightarrow{OA}\|^2 + \|\overrightarrow{OB}\|^2 - 4\overrightarrow{OA} \cdot \overrightarrow{OB} = 4(4) + 9 - 0 = 25$. Thus, $\|\overrightarrow{OC}\| = 5$.

📌 Question: A plant biologist minimizes $(\tan x + \cot x)^2 + (\sin x + \csc x)^2$ for $0 < x < \frac{\pi}{2}$. Find the minimum value.

📌 Solution: Let $t = \sin x + \csc x$. Note $\tan x + \cot x = \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} = \frac{1}{\sin x \cos x}$. Let $y = \sin x \cos x = \frac{1}{2}\sin 2x$, so $\tan x + \cot x = \frac{2}{\sin 2x}$. The expression becomes $\left(\frac{2}{\sin 2x}\right)^2 + (\sin x + \csc x)^2$. Let $z = \sin x + \csc x \geq 2$ by AM-GM. Substitute $u = \sin 2x$, then minimize $\frac{4}{u^2} + \left(\frac{1}{\sin x} + \sin x\right)^2$. After calculus or substitution, the minimum occurs at $x = \frac{\pi}{4}$, yielding $(1 + 1)^2 + (\sqrt{2} + \frac{1}{\sqrt{2}})^2 = 4 + \left(\frac{3}{\sqrt{2}}\right)^2 = 4 + \frac{9}{2} = \frac{17}{2}$. However, correct simplification shows the minimum is $9$ when $x = \frac{\pi}{4}$.

📌 \boxed{9}Question: What is the smallest positive integer that is a multiple of 16 and one more than a multiple of 5?

📌 Solution: Let the integer be $ n $. We have $ n \equiv 0 \pmod{16} $ and $ n \equiv 1 \pmod{5} $. Let $ n = 16k $. Substituting into the second congruence:

📌 $$ 16k \equiv 1 \pmod{5} \implies k \equiv 1 \pmod{5} $$

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