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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 if is onto

B

one-one if f is one-one

C

continuous if f is continuous

D

differentiable if f is differentiable

Text Solution

Verified by Experts

The correct Answer is:
C
Let `h(x)=|x|. "Then " h:R to R` is continuous many-one and into function.
We have
`hof (x)=h(f(x))=|f(x)|=g(x)`
Since composition of continuous functions is continuous. Therefore, g(x) is continuous if f is continuous
Since, composition of two bijections is a bijection. Here h(x) is many-one. So, g(x) cannot be one-one even if f is one-one. Also, g(x) cannot be onto even if f is onto. We observe that (f)=sin x is everywhere differentiable but |sin x|is not differentiable at `x=n pi, n in Z`. Therefore g(x) need not be differentiable even if f is differentiable
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