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Function f(x) = |cos x| is

(A) Continuous only in [0, π/2]
(B) Continuous only in [−π/2, π/2]
(C) Continuous only in [−π, π]
(D) None
asked in Set Theory & Algebra by Veteran (70.4k points) | 148 views
answer is C?

continuous function 

@Pavan No, ans D)

@Debasish yes, then what?
it is continuous every where from graph we can say clearly
but it takes mod,

what is difference between cos x and |cos x| ?

see in the graph any where any certain point has two values no and any point function value is not defined no so the function is continuous this is a example of discontinuous https://en.wikipedia.org/wiki/Classification_of_discontinuities#/media/File:Discontinuity_jump.eps.png at x0 different values are there

1 Answer

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Best answer

Below are the graphs for f(x) = cos x  & f(x) = |cos x|      Clearly both are continuous. every where.

 

 

If you want to do manually,

f(x) = |cos x|
f(x) =  cos x  when  cos x > 0  i.e when x is in between  [−π/2, π/2]  
f(x) =  −cos x when  cos x <0  ie  when x is in between  [π/2, −π/2]

Now do (A) (B) (C)
You will see all are continuous.
Answer is D because of the word "ONLY".
It is continuous every where.

answered by Boss (7.6k points)
selected by

@srestha  Source of this question is Virtual Gate. Right ?

might be @Ahwan. I cannot remember exactly


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