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)`, is

Options

A. 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`

Answer

A

Explanation

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`