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

```
7 years ago

## Answers : (1)

85 Points
```										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

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```
7 years ago
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