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 Question:
We are given a trigonometric function $f(x) = \sin^2(7x) - \sin^2(5x)$.
We need to test each of the four options to find which statement is false.

Step 2: Key Formula or Approach: We simplify the function using the standard identity:
\[ \sin^2 A - \sin^2 B = \sin(A + B) \sin(A - B) \] By setting $A = 7x$ and $B = 5x$, we get:
\[ f(x) = \sin(12x) \sin(2x) \] We can now analyze the behavior of the simplified function.

Step 3: Detailed Explanation:

• Let us evaluate Option (D) first:
\[ f\left(\frac{\pi}{12}\right) = \sin\left(12 \cdot \frac{\pi}{12}\right) \sin\left(2 \cdot \frac{\pi}{12}\right) = \sin(\pi) \sin\left(\frac{\pi}{6}\right) = 0 \times \frac{1}{2} = 0 \] So, Option (D) is TRUE.

• Let us evaluate Option (C):
\[ f\left(x + \frac{\pi}{2}\right) = \sin\left(12\left(x + \frac{\pi}{2}\right)\right) \sin\left(2\left(x + \frac{\pi}{2}\right)\right) \] \[ f\left(x + \frac{\pi}{2}\right) = \sin(12x + 6\pi) \sin(2x + \pi) \] Since $\sin(\theta + 6\pi) = \sin\theta$ and $\sin(\theta + \pi) = -\sin\theta$:
\[ f\left(x + \frac{\pi}{2}\right) = \sin(12x) \cdot (-\sin(2x)) = -\sin(12x)\sin(2x) = -f(x) \] \[ f\left(x + \frac{\pi}{2}\right) + f(x) = 0 \] So, Option (C) is TRUE.

• Let us evaluate Option (B):
For $x \in \left(0, \frac{\pi}{48}\right)$:
The argument of the second sine term is $2x \in \left(0, \frac{\pi}{24}\right)$, where $\sin(2x) > 0$.
The argument of the first sine term is $12x \in \left(0, \frac{\pi}{4}\right)$, where $\sin(12x) > 0$.
Since both terms are positive in this interval, their product $f(x) > 0$.
So, Option (B) is TRUE.

• Let us evaluate Option (A):
Since options B, C, and D are mathematically correct, Option (A) must be the false statement.
To confirm, let's examine the derivative on $\left(\frac{3\pi}{2}, 2\pi\right)$.
The derivative $f'(x)$ is given by:
\[ f'(x) = 12 \cos(12x) \sin(2x) + 2 \sin(12x) \cos(2x) \] In the interval $\left(\frac{3\pi}{2}, 2\pi\right)$, $12x$ spans a wide range of $6\pi$ radians (from $18\pi$ to $24\pi$), which covers multiple complete periods of both sine and cosine.
Thus, the derivative $f'(x)$ changes sign multiple times on this interval, meaning $f(x)$ is not monotonically increasing.
Therefore, Option (A) is NOT TRUE.

Step 4: Final Answer:
The statement "f is increasing on $\left(\frac{3\pi}{2}, 2\pi\right)$" is not true.