InterviewSolution
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InterviewSolution
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`f(x)=|x|+|x-1|` की `x=0` और `x=1`पर सांतत्य का परीक्षण। कीजिए । |
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Answer» `" "|x|={{:(x",",xge0),(-x",",xlt0):}` और `" "|x-1|={{:(x-1",",xge1),(-(x-1)",",xlt1):}` `therefore f(x)=|x|+|x-1|={{:(1-2x",",xlt0),(" "1",",0lexlt1),(2x-1",",xle1):}` x = 0 पर f (0) = 1 `R.H.L.=underset(xrarr0^(+))(lim)f(x)=underset(hrarr0)(lim)f(0+h)=underset(hrarr0)(lim)1=1` `L.H.L.=underset(xrarr0^(-))(lim)f(x)=underset(hrarr0)(lim)f(0-h)` `=underset(hrarr0)(lim){1-2(-4)}=1+0=1` `because R.H.L.=f(0)=L.H.L.` `therefore f(x),x=0` पर सतत है।| x = 1 पर `f(1)=2(1)-1=1` `R.H.L.=underset(xrarr1^(+))(lim)f(x)=underset(hrarr0)(lim)f(1+h)` `=underset(hrarr0)(lim)2(1+h)-1=2(1+0)-1=1` `L.H.L.=underset(xrarr1^(-))(lim)2(1+h)-1=2(1+0)-1=1` `L.H.L.=underset(xrarr1^(-))(lim)f(x)=underset(hrarr0)(lim)f(1-h)=underset(hrarr0)(lim)(1)=1` `because R.H.L. =f(1) = L.H.L.` `therefore f(x),x=1` पर सतत है | |
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