Q.The function $f(x) = 4\sin^3 x - 6\sin^2 x + 12\sin x + 100$ is strictly:
(A) increasing in $\left(\pi, \dfrac{3\pi}{2}\right)$
(B) decreasing in $\left(\dfrac{\pi}{2}, \pi\right)$
(C) decreasing in $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$
(D) decreasing in $\left(0, \dfrac{\pi}{2}\right)$

Concept: Monotonicity of Trigonometric Functions — we check the sign of $f'(x)$ in each interval.

Let $t = \sin x$. Then $f(x) = 4t^3 - 6t^2 + 12t + 100$.

Differentiate:

The quadratic $\sin^2 x - \sin x + 1$ has discriminant $1 - 4 = -3 < 0$ with a positive leading coefficient, so it is $> 0$ for all $x$. Hence the sign of $f'(x)$ depends only on $\cos x$.

• $\cos x > 0$ on $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ → $f'(x) > 0$ → strictly increasing.
• $\cos x < 0$ on $\left(\frac{\pi}{2}, \frac{3\pi}{2}\right)$ → $f'(x) < 0$ → strictly decreasing. …

Intuition

To classify where a function increases or decreases we look at the sign of its first derivative: $f' > 0$ means strictly increasing, $f' < 0$ means strictly decreasing. Here $f$ is built out of $\sin x$, so its derivative will carry a $\cos x$ factor — and that factor will control everything.

Differentiate

Using the chain rule term by term:

Every term has $12\cos x$, so factor it out:

Show the quadratic factor is always positive

Write $t = \sin x$ and look at $t^2 - t + 1$. Its discriminant is

and the leading coefficient is positive, so the quadratic has no real roots and stays positive for every $t$. Therefore $\sin^2 x - \sin x + 1 > 0$ for all $x$.

Sign of $f'(x)$ on the intervals

• $\cos x > 0$ on $\left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)$ ⇒ $f'(x) > 0$ ⇒ $f$ strictly increasing. …

Method: Determining Monotonicity of a Composite Trigonometric Function

Apply this whenever a function is built by substituting $\sin x$ (or $\cos x$) into a polynomial, and you must determine on which interval it increases or decreases.

Steps

Step 1: Differentiate using the chain rule.

Treat the polynomial in $\sin x$ as a composite function: differentiate the polynomial with respect to $\sin x$, then multiply by $\frac{d}{dx}\sin x=\cos x$.

Step 2: Factor out any common terms to simplify $f'(x)$.

Every term in the derivative will typically carry a common factor (here $\cos x$) — pull it out so $f'(x)$ becomes a product of a trig factor and a polynomial-in-$\sin x$ factor.

Step 3: Check whether the polynomial factor is always positive (or always negative). …

Common Mistakes

Mistake 1: Forgetting the chain-rule factor of $\cos x$ when differentiating powers of $\sin x$.

Why it's wrong: differentiating $\sin^2 x$ and dropping the trailing $\cos x$ multiplier gives a completely wrong derivative. Correct approach: always write $\frac{d}{dx}\sin^n x = n\sin^{n-1}x \cdot \cos x$, keeping the chain-rule factor.

Mistake 2: Assuming the polynomial-in-$\sin x$ factor could be zero or negative without checking the discriminant.

Why it's wrong: skipping the discriminant check risks missing a hidden sign change inside that factor, making the overall sign analysis incomplete. Correct approach: explicitly compute the discriminant of the quadratic factor and confirm it is negative (with a positive leading coefficient) before declaring it "always positive." …