If g(x)=\frac{2 h(x)+|h(x)|}{2 h(x)-|h(x)|} where h(x)=\sin x-\sin ^{n} x, n \in R^{+}, where R^{+} is the set of positive real numbers, and f(x)=\\begin{array}{cc}{[g(x)] &\end{array} , x \in(0, \frac{\pi}{2}) \cup(\frac{\pi}{2}, \pi) ; 3, , x=\frac{\pi}{2}. Where [x] denotes the greatest integer function, then

Correct Answer is : f(x) is continuous and differentiable at x=π2, when n>1

f(x) is continuous and differentiable at x=\frac{\pi}{2}, when 0

f(x) is continuous and differentiable at x=\frac{\pi}{2}, when n > 1

f(x) is continuous but not differentiable at x=\frac{\pi}{2}, when 0

f(x) is continuous but not differentiable at x=\frac{\pi}{2}, when n > 1

JEE Advanced 2020 Full Test 6 Paper 2

Question : 39 of 54

Marks: +1, -0

If g(x)=‌

2h(x)+|h(x)|

2h(x)−|h(x)|

where h(x)=sin‌x−sinnx,n∈R+, where R+ is the set of positive real numbers, and f(x)={

[g(x)],

x∈(0,‌

)∪(‌

,π)

‌

x=‌

Where [x] denotes the greatest integer function, then

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