Question

Let $f(x) = 3 \sin^4 x + 10 \sin^3 x + 6 \sin^2 x - 3, x \in [-\frac{\pi}{6}, \frac{\pi}{2}]$. Then, $f$ is :

• A. increasing in $(-\frac{\pi}{6}, 0)$
• B. decreasing in $(0, \frac{\pi}{2})$
• C. increasing in $(-\frac{\pi}{6}, \frac{\pi}{2})$
• D. decreasing in $(-\frac{\pi}{6}, 0)$

Hint:

Using a substitution like \(t = \sin x\) turns a complex trigonometric derivative into a polynomial derivative, which is much easier to factor and sign-analyze.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The monotonicity of a function is determined by the sign of its first derivative \(f'(x)\).
Since \(f(x)\) is expressed in powers of \(\sin x\), we can use substitution to simplify the derivative analysis.
Step 2: Key Formula or Approach:
1. Let \(t = \sin x\).
2. \(f'(x) = \frac{df}{dt} \cdot \frac{dt}{dx}\).
Step 3: Detailed Explanation:
Let \(h(t) = 3t^4 + 10t^3 + 6t^2 - 3\), where \(t = \sin x\).
For \(x \in [-\pi/6, \pi/2]\), \(t \in [-1/2, 1]\).
\(h'(t) = 12t^3 + 30t^2 + 12t = 6t(2t^2 + 5t + 2)\).
Factoring the quadratic: \(h'(t) = 6t(2t + 1)(t + 2)\).
Now, \(f'(x) = h'(t) \cdot \cos x\).
In the given domain, \(\cos x \ge 0\) and \((t + 2)>0\).
The sign of \(f'(x)\) depends on \(6t(2t + 1)\).
Interval 1: \(x \in (-\pi/6, 0) \Rightarrow t \in (-1/2, 0)\).
Here, \(t<0\) and \(2t + 1>0\), so \(h'(t)<0 \Rightarrow f'(x)<0\). (Decreasing)
Interval 2: \(x \in (0, \pi/2) \Rightarrow t \in (0, 1)\).
Here, \(t>0\) and \(2t + 1>0\), so \(h'(t)>0 \Rightarrow f'(x)>0\). (Increasing)
Step 4: Final Answer:
\(f\) is decreasing in \((-\frac{\pi}{6}, 0)\).