Question

Consider the function $f : \mathbb{R} \to \mathbb{R}$ defined by $f(x) = \sin^2(7x) - \sin^2(5x)$. Which of the following statements is NOT TRUE?

• A. $f$ is increasing on $\left(\frac{3\pi}{2}, 2\pi\right)$.
• B. $f(x) > 0$, for all $x \in \left(0, \frac{\pi}{48}\right)$.
• C. $f\left(x + \frac{\pi}{2}\right) + f(x) = 0$, for all $x \in \mathbb{R}$.
• D. $f\left(\frac{\pi}{12}\right) = 0$.

Hint:

The identity $\sin^2 A - \sin^2 B = \sin(A+B)\sin(A-B)$ is a highly efficient way to simplify differences of squared trigonometric functions.
Always test the easiest options first (like evaluating simple points or checking basic period properties) to eliminate options quickly.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We simplify the trigonometric expression using the identity $\sin^{2}A - \sin^{2}B = \sin(A+B)\sin(A-B)$.

Step 2: Detailed Explanation:
$f(x) = \sin(7x+5x)\sin(7x-5x) = \sin(12x)\sin(2x)$.
$\bullet$ Check (D): $f(\pi/12) = \sin(12 \cdot \pi/12)\sin(2 \cdot \pi/12) = \sin(\pi)\sin(\pi/6) = 0 \times 0.5 = 0$. (True)
$\bullet$ Check (C): $f(x + \pi/2) = \sin(12x + 6\pi)\sin(2x + \pi) = \sin(12x) \cdot (-\sin 2x) = -f(x)$. Thus, $f(x+\pi/2) + f(x) = 0$. (True)
$\bullet$ Check (B): In $(0, \pi/48)$, $12x \in (0, \pi/4)$ and $2x \in (0, \pi/24)$. Both sines are positive, so $f(x) > 0$. (True)
$\bullet$ Check (A): In $(3\pi/2, 2\pi)$, $12x$ spans an interval of $6\pi$. This means the term $\sin(12x)$ will oscillate through three complete periods, changing signs many times. The function cannot be monotonically increasing. (NOT True)

Step 3: Final Answer:
Statement (A) is false.