Doubt...jnu

[Shefali]

pls help....

[ViV]

(a)

[Shefali]

how did u solve viv

[Dreyfus]

Its b, convex and continuous in [-1,1]
As f(x)=1 in this interval hence continuous also each pt on f(x) lies on f(x) hence convex and also concave

[Dreyfus]

Viv ...func is discontinuous at -1 and 1

[Shefali]

vaibhav..i didnt get your pt..pls elaborate

[Granpa Simpson]

I think the answer will be b.

[Granpa Simpson]

Vaibhav is right..the function is discontinuous at x=1,..in the interval [-1,1] since mod(x)<=1, so f(x)=1, however in the interval (1,infinity), f(x)=2x since mod(x)>1, so at x=1+, f(x)=2 and at x=1-, f(x)=1, so it is discontinuous at x=1, b is the right answer.

[Shefali]

if the fn is discont at 1 then how is b the right ans..

[Granpa Simpson]

Thanx for pointing out this crucial point Shefali...actually by diverting our focus on the continuity part we both might have missed this crucial point. The function is continuous in (-1,1). The answer in that case will be d.

[Anakin Skywalker]

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[Dreyfus]

Hey subhayu ....f(x) is 1 throughout the interval [-1,1]  ....note that at 1 and -1 also function attains 1 DAT means f is horizontal line in[-1,1]... And every line is both convex and concave too....so it should be b....u  tell me where I m missing?

[Anakin Skywalker]

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[Granpa Simpson]

A function is continuous on the interval [a, b]
if f is continuous on (a, b) and f is continuous from the right at a and f is continuous from
the left at b. So ur right Vaibhav..theres nothing that u are missing out...I was missing out..

[Shefali]

thanxxx guyss

[ViV]

Hmm..It's (b)
My bad..