Discussion
Let \(f: \left[\frac{-\pi}{2}, \frac{\pi}{2}\right] \to (-\infty, \infty)\) be given by \(f(x) = \tan(x)\sin(x)\). Then \(f(x)\) is
Answer: C
The keyed answer is an even function.
Replace \(x\) by \(-x\): \(\tan(-x)\sin(-x) = (-\tan x)(-\sin x) = \tan x \sin x\). So \(f(-x) = f(x)\) for every \(x\) in the domain, and \(f\) is even. That also rules out the two odd options and the option calling it 1-1 and odd, because an odd function satisfies \(f(-x) = -f(x)\).
The question is loosely set. Because \(f(-x) = f(x)\), distinct points \(x\) and \(-x\) give the same value, so \(f\) is also not one-one, which makes the first option a true statement too, and it is the more informative of the two. The key names the shorter property, even, and that is the one taught here. The function is also not onto \((-\infty, \infty)\), since \(\tan x \sin x \geq 0\) on the domain, and \(\tan x\) is not defined at the end points \(\pm \frac{\pi}{2}\) of the closed interval.
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development