11 Replies
@Apu
Note for OP
+solved @user to close the thread when your doubt is solved. Mention the user who helped you solve the doubt. This will be added to their stats.so you need to find the range of the function
Range of acosx+bsinx = [-√(a²+b²),+√(a²+b²)]
sinx-sqrt3 cosx can be written as 2(cos60sinx-sin60cosx)
Then add one to that.
Option (a) I think.
which is 2sin(x-60)
so the fucntion is 2sin(x-60)+1
yeah option a
$$sinx - \sqrt{3} cosx + 1$$
$$2(\frac{1}{2}sinx + \frac{ \sqrt{3}}{2} cosx) + 1$$
$$2\cdot sin(x+\pi/3) + 1$$
Therefore $[-1,3]$
Keshav
+solved @Keshav
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