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Statement-1: If |f(x)| le |x| for all x...

Statement-1: If `|f(x)| le |x|` for all `x in R` then `|f(x)|` is continuous at 0. Statement-2: If `f(x)` is continuous then `|f(x)|` is also continuous.

A

1

B

2

C

3

D

4

Text Solution

Verified by Experts

The correct Answer is:
A
Let f be continuous, at x-a, Then for every `epsilon gt 0` there exist `delta gt 0` such that
`|f(x)-f(a)| lt in "whenever" |x-a| lt delta`
`Rightarrow ||f(x)-|f(a)|| lt |f(x)-f(a)| lt in "whenever"|x-a| lt delta`
`Rightarrow |f|(x)-|f|(a) lt in "whenever" |x-a| lt delta`
`Rightarrow |f|` is continuous at x=a
So, statement-2 is true.
Now, `|f(x)| le |x|`
`Rightarrow ,|f(0)| le 0" "["Replacing x by 0"]`
`Rightarrow f(0)=0`
`therefore |f(x)| le |x|`
`Rightarrow |f(x)-f(0)| lt |x-0|" "[therefore f(0)=0]`
`Rightarrow |f(x)-f(0)| lt in "whenever" |x-0| lt delta ( in)`
`Rightarrow |f(x)|` is continuous at x=0 [Using statement-2]
Hence both the statements are true and statement-2 is a correct explanation for statement-1,
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