Functions Question 711
Question: Let $ A=R-{3},B=R-{1}, $ and let $ f:A\to B $ be defined by $ f(x)=\frac{x-2}{x-3}f $ is
Options:
A) Not one-one
B) Not onto
C) Many-one and onto
D) One-one and onto
Show Answer
Answer:
Correct Answer: D
Solution:
[d] Let $ x_1,x_2\in A $ and let $ f(x_1)=f(x_2) $ Or $ \frac{x_1-2}{x_1-3}=\frac{x_2-2}{x_2-3} $ Or $ x_1=x_2 $ so, f is one-one. To find whether f is onto or not, first let us find the range of f. Let $ y=f(x)=\frac{x-2}{x-3}orx=\frac{3y-2}{y-1} $ X is defined if $ y\ne 1, $ i.e, the range of f is $ R-{1} $ Which is also the co-domain of f. Also, for no value of y, x can be 3 i.e., if we put 3 $ =x=\frac{3y-2}{y-1}then3y-3=3y-2or-3=-2 $ Which is not possible. Hence, f is onto.
BETA
