Question:
If \(\int \frac{sinx}{sin4x}\,dx = αlog\frac{1+sinx}{1-sinx}+βlog\frac{1+\sqrt{2}sinx}{1-\sqrt{2}sinx}+c\), then the value of \(32(α+β^2) =\)
If \(\int \frac{sinx}{sin4x}\,dx = αlog\frac{1+sinx}{1-sinx}+βlog\frac{1+\sqrt{2}sinx}{1-\sqrt{2}sinx}+c\), then the value of \(32(α+β^2) =\)
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If volume loss is proportional to surface area, the radius decreases at a constant rate.
Updated On: Oct 1, 2026
- \(5\)
- \(-1\)
- \(9\)
- \(-3\)
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The Correct Option is D
Solution and Explanation
Step 1: Setup:
Volume \(V = \frac43\pi r^3\) and surface area \(S = 4\pi r^2\). The condition is \(\frac{dV}{dt} = -kS\).
Step 2: Simplify:
\(\frac{dV}{dt} = 4\pi r^2\frac{dr}{dt}\). So \(4\pi r^2\frac{dr}{dt} = -k\cdot4\pi r^2\), giving \(\frac{dr}{dt} = -k\). The radius shrinks at a constant rate.
Step 3: Use the data:
The radius falls from 3 cm to 1 cm in 4 months, a drop of 2 cm, so the rate is 0.5 cm per month. To lose the full 3 cm at this rate takes \(\frac{3}{0.5} = 6\) months.
Final Answer:
The mothball evaporates completely in 6 months, option (A). \[ \boxed{6\text{ months}} \]
Volume \(V = \frac43\pi r^3\) and surface area \(S = 4\pi r^2\). The condition is \(\frac{dV}{dt} = -kS\).
Step 2: Simplify:
\(\frac{dV}{dt} = 4\pi r^2\frac{dr}{dt}\). So \(4\pi r^2\frac{dr}{dt} = -k\cdot4\pi r^2\), giving \(\frac{dr}{dt} = -k\). The radius shrinks at a constant rate.
Step 3: Use the data:
The radius falls from 3 cm to 1 cm in 4 months, a drop of 2 cm, so the rate is 0.5 cm per month. To lose the full 3 cm at this rate takes \(\frac{3}{0.5} = 6\) months.
Final Answer:
The mothball evaporates completely in 6 months, option (A). \[ \boxed{6\text{ months}} \]
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