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# For  interval  x €  [-1.0) , f(x)=x-1  &   for  interval  x€ [0,1] , f(x)= .x^2g(x) =  sin xh(x) = f(|g(x)|) + |f(g(x))|Using   the  above  data, show that h(x) is  discontinuous  at 0  in the interval  [-1,1]  .

85 Points
11 years ago

Dear Vivek,

Remember -                               | x,  x≥0
|x| =
| -x,  x<0       and   [ mod.(x-y) ≥ mod. x - mod. y ]

For x€ [0,1] , f(x)= .x^2 so,  RHL : h(x) =  (mod. sinx )2 + mod. sin2x

=   sin2x + sin2x

Lim h -0 we have h(x) = 0

for  x €  [-1.0) , f(x)=x-1 LHL : h(x) = mod. sinx -1 + mod. ( sinx -1 )

=  - sinx -1 +mod. sinx - mod. 1

=    - sinx -1 - sinx -1

lim h - 0 , we have h(x) = -2

As, LHL not equal to RHL so discontinous at 0

All the best.

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