The maximum value of the function \[ f(x) = 3\sin^{12}x + 4\cos^{16}x \] is ?
Step 1: Bounds of \(\sin^{12}x\) and \(\cos^{16}x\).
Because \(-1 \le\sin x\le 1\) and \(-1 \le\cos x\le 1\), we have \(0\le \sin^{12}x \le 1\) and \(0\le \cos^{16}x \le 1\).
Step 2: Identifying maximum of \(3\sin^{12}x + 4\cos^{16}x\).
We suspect the maximum occurs when one of \(\sin x\) or \(\cos x\) is 1 or 0, because powers flatten any partial values. Indeed, test: \[ \sin x=1 \implies \cos x=0 \quad\Rightarrow\quad f(x)=3\cdot 1 +4\cdot 0=3. \] \[ \sin x=0 \implies \cos x=1 \quad\Rightarrow\quad f(x)=3\cdot 0 +4\cdot 1=4. \] No intermediate combination of \(\sin x,\cos x\) in \((0,1)\) would yield a sum exceeding 4, due to the high exponents diminishing partial values significantly. Hence the maximum is \(\boxed{4}\).
The number of solutions of the equation $ \cos 2\theta \cos \left( \frac{\theta}{2} \right) + \cos \left( \frac{5\theta}{2} \right) = 2 \cos^3 \left( \frac{5\theta}{2} \right) $ in the interval \(\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right ]\) is:
Observe the following data given in the table. (\(K_H\) = Henry's law constant)
| Gas | CO₂ | Ar | HCHO | CH₄ |
|---|---|---|---|---|
| \(K_H\) (k bar at 298 K) | 1.67 | 40.3 | \(1.83 \times 10^{-5}\) | 0.413 |
The correct order of their solubility in water is
For a first order decomposition of a certain reaction, rate constant is given by the equation
\(\log k(s⁻¹) = 7.14 - \frac{1 \times 10^4 K}{T}\). The activation energy of the reaction (in kJ mol⁻¹) is (\(R = 8.3 J K⁻¹ mol⁻¹\))
Note: The provided value for R is 8.3. We will use the more precise value R=8.314 J K⁻¹ mol⁻¹ for accuracy, as is standard.