InterviewSolution
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InterviewSolution
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Let [x] denote the greatest integer less than or equal to x and g (x) be given by`g(x)={{:(,[f(x)],x in (0","pi//2) uu (pi//2","pi)),(,3,x=(pi)/(2)):}` `"where", f(x)=(2(sin x-sin^(n)x)+|sinx-sin^(n)x|)/(2(sinx-sin^(n)x)-|sinx-sin^(n)x|),n in R^(+)` then at `x=(pi)/(2),g(x)`, isA. continuous and differentiable when `n gt 1`B. continuous and differentiable when `0 lt n lt 1`C. continuous but not differentiable when `n gt 1`D. continuous but not differentiable when `0 lt n lt 1` |
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Answer» Correct Answer - A Clearly, `0 lt sin x lt 1"for all "x in (0,pi//2)uu (pi//2,pi)` `CASE-I, "When n" gt1` In this case, we have `sin x gt sin^(n)x"for all "x in (0,pi//2) uu (pi//2,pi)` `Rightarrow sinx-sin^(n)x gt 0"for all "x in (0,pi//2)uu(pi//2,pi)` `Rightarrow |sin x -sin^(n)x|=sin x-sin^(n)x"for all "x in(0,pi//2) uu (pi//2,pi)` `therefore f(x)=(2(sin x sin^(n)x)+(sinx-sin^(n)x))/(2(sin x-sin^(n)x)-(sin x-sin^(n)x))=3` `Rightarrow [f(x)]=3"for all "x in (0,pi//2) uu (pi//2,pi)` ltrbgt Thus, we have `g(x)=3"for all "x in (0,pi)` Clearly, it is continuous and differential at `x=pi//2` CASE-II When `0 lt n lt 1` In this case, we have `sin x lt sin^(n) x "for all "x in (0,pi//2) uu (pi//2,pi)` `Rightarrow sin x-sin^(n)x lt 0"for all "x in(0,pi//2) uu (pi//2,pi)` `Rightarrow |sin x-sin^(n)x|=-(sinx-sin^(n)x)"for all "x in (0,pi//2)uu (pi//2,pi)` `therefore f(x)=(2(sin x-sin^(n)x)-(sin x-sin^(n)x))/(2(sin x-sin^(n)x)+(sinx-sin^(n)x))=(1)/(3)` `Rightarrow [f(x)]=0"for all "x in (0,pi//2)uu (pi//2,pi)` Thus, we have `g(x)={{:(,0,"for all"x in (0","pi//2)uu (pi//","pi)),(,3,"for "x =pi//2):}` Clearly, it is discontinuous and hence non-differerntiable also at `x=pi//2` |
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