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Let f: R->R be any function. Also g: R->...

Let `f: R->R` be any function. Also `g: R->R` is defined by `g(x)=|f(x)|` for all `xdot` Then `g` is
a. Onto if `f` is onto b. One-one if `f` is one-one c. Continuous if `f` is continuous d. None of these

A

onto if f is onto

B

one-one if f is one-one

C

continuous if f is continuous

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
C
(a) Since g(x) = `|f(x)|`
`therefore" "g(x) ge 0`
`therefore" Range of "g ne R". Hence, is not onto"`.
(b) If we take `f(x)=x`, then f is one-one but `|f(x)|=|x|` is not one-one.
(c) If f(x) is continuous then `|f(x)|` is also continuous.
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