Relations Question 4
Question: The function $ f:(-\infty ,-1)\to ( 0,e^{5} ] $ defined by $ f(x)={e^{{x^{3-3x+2}}}} $ is
Options:
A) many-one and onto
B) many-one and into
C) one-one and onto
D) one-one and into
Show Answer
Answer:
Correct Answer: D
Solution:
$ f(x)={e^{x^{3}-3x+2}} $
Let $ g(x)=x^{3}-3x+2; $
$ g’(x)=3x^{2}-3=3(x^{2}-1) $
$ \ge 0 $ for $ x\in (-\infty ,-1] $
Therefore, $ f(x) $ is increasing function hence $ f(x) $ is one-one.
Now, the range of $ f(x) $ is $ (0,e^{4}] $ .
But co-domain is $ (0,e^{5}] $ . Hence., $ f(x) $ is an into function.
BETA
