A) 0
B) 1
C) \[\frac{1-\sqrt{3}}{2}\]
D) \[\frac{\sqrt{3}-1}{2}\]
Correct Answer: D
Solution :
In the neighborhood for\[x=\frac{2\pi }{3}\], we have \[\cos x<0\]and\[\sin x>0\] \[\therefore \] \[y=-\cos x+\sin x\] \[\Rightarrow \] \[\frac{dy}{dx}=\sin x+\cos x\] \[\Rightarrow {{\left( \frac{dy}{dx} \right)}_{\left( x=\frac{2\pi }{3} \right)}}=\sin \frac{2\pi }{3}+\cos \frac{2\pi }{3}\] \[=\frac{\sqrt{3}-1}{2}\]
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