InterviewSolution
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InterviewSolution
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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. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - 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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