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] .

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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All the best.

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