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] .

Question

For interval x &euro; [-1.0) , f(x)=x-1 & for interval x&euro; [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] .

Anil Pannikar AskiitiansExpert-IITB · Answer

Dear Vivek, Remember - | x, x&ge;0 |x| = | -x, x<0 and [ mod.(x-y) &ge; mod. x - mod. y ] For x&euro; [0,1] , f(x)= .x^2 so, RHL : h(x) = (mod. sinx ) 2 + mod. sin 2 x = sin 2 x + sin 2 x Lim h -0 we have h(x) = 0 for x &euro; [-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 Please feel free to ask your queries here. We are all IITians and here to help you in your IIT JEE preparation. All the best. Win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian. Now you score 5+15 POINTS by uploading your Pic and Downloading the Askiitians Toolbar respectively : Click here to download the toolbar.. Askiitians Expert Anil Pannikar IIT Bombay

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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] .

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] .

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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] .

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] .

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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] .