Show that for a ≥ 1, f (x) = √3 sin X – cos X – 2ax + b is decreasing in R.

Given: f (x) = √3 sin X – cos X – 2ax + b


To show: the given function is decreasing in R.


Explanation: Given f (x) = √3 sin X – cos X – 2ax + b


Applying first derivative with respect to x, we get



Applying the sum rule of differentiation, we get



Taking the constant terms out we get



But the derivative of sin X = cos x and that of cos x = -sin x, so


f' (x) = √3cos x-(-sin x)-2a


f' (x) = √3cos X+sin X-2a


Multiplying and dividing RHS by 2, we get



But and , substituting these values in above equation, we get



But cos(A-B) = cos Acos B+sin A.sin B, substituting this in above equation, we get



Now, we know cos x always belong to [-1, 1] for a≥1


,


And we know, if f’(x)≤0, then f(x) is decreasing function.


Hence the given function is decreasing function in R.


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