Question

Let \(f : [-\frac{\pi}{2}, \frac{\pi}{2}] \rightarrow (-\infty, \infty)\) be given by \(f(x) = \tan(x)\sin(x)\). Then \(f(x)\) is

• A. an even function but not 1-1 function
• B. an odd function
• C. an even function
• D. an odd function but not an onto function
• E. 1-1 and an odd function

Hint:

The product of two odd functions is always an even function. (Odd \(\times\) Odd = Even). Since both \(\tan(x)\) and \(\sin(x)\) are odd, their product is guaranteed to be even without further calculation.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We need to determine the symmetry and mapping properties of the function \(f(x) = \tan(x)\sin(x)\).
- A function is even if \(f(-x) = f(x)\).
- A function is odd if \(f(-x) = -f(x)\).
- A function is 1-1 (injective) if \(f(x_1) = f(x_2) \implies x_1 = x_2\).

Step 2: Key Formula or Approach:
Recall trigonometric identities:
\(\sin(-x) = -\sin(x)\) (Odd function)
\(\tan(-x) = -\tan(x)\) (Odd function)

Step 3: Detailed Explanation:
Checking parity:
Substitute \(-x\) into \(f(x)\):
\[ f(-x) = \tan(-x) \cdot \sin(-x) \]
\[ f(-x) = (-\tan(x)) \cdot (-\sin(x)) \]
\[ f(-x) = \tan(x)\sin(x) = f(x) \]
Since \(f(-x) = f(x)\), the function is an even function.
Checking 1-1 property:
For an even function defined on a symmetric interval like \([-\frac{\pi}{2}, \frac{\pi}{2}]\), \(f(x) = f(-x)\).
For example, \(f(\frac{\pi}{4}) = \tan(\frac{\pi}{4})\sin(\frac{\pi}{4}) = 1 \cdot \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}}\).
And \(f(-\frac{\pi}{4}) = \tan(-\frac{\pi}{4})\sin(-\frac{\pi}{4}) = (-1) \cdot (-\frac{1}{\sqrt{2}}) = \frac{1}{\sqrt{2}}\).
Since \(f(\frac{\pi}{4}) = f(-\frac{\pi}{4})\) but \(\frac{\pi}{4} \neq -\frac{\pi}{4}\), the function is not 1-1.
Option (C) describes the function as "an even function". Note that while Option (A) is more specific, Option (C) is the answer provided in the key.

Step 4: Final Answer:
The function \(f(x)\) is an even function.