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Letf:[-oo,0]->[1,oo) be defined as f(x) ...

Let`f:[-oo,0]->[1,oo)` be defined as `f(x) = (1+sqrt(-x))-(sqrt(-x) -x)`, then

A

injective but not surjective

B

injective as well as surjective

C

neither injective nor surjective

D

surjective but injective

Text Solution

Verified by Experts

The correct Answer is:
B
`f(x)=(1+sqrt(-x))-(-sqrt(-x)-x)`
`=1-x, x le0`
Then `1-x ge 1`.
Hence function is injective and surjective.
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