The smallest positive integral value of $p$ for which the function $f(x) = 6px - p\sin 4x - 5x - \sin 3x$ is monotonic increasing and has no critical points on $R$ is:
Step-by-Step Solution
Key Concept: For f(x) to be monotonically increasing with no critical points, f'(x) ≥ 0 for all x ∈ ℝ. The derivative f'(x) = 6p - 4p cos 4x - 5 - 3 cos 3x must remain non-negative by finding the minimum value using the range of cosine functions [-1, 1].
We have $f'(x) = 6p - 4p\cos 4x - 5 - 3\cos 3x = 4p(1-\cos 4x) + 2(p-4) + 3(1-\cos 3x)$. For $f'(x) \geq 0$ for all $x \in \mathbb{R}$, we need $p > 4$ and check boundary conditions. The least integral value is $p = 5$.
Correct Answer: 5