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

Question

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

Philoid · Accepted Answer

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.