Q.Show that for $a \geq 1$, $f(x) = \sqrt{3}\,\sin x - \cos x - 2ax + b$ is decreasing in $\mathbb{R}$.

Step 1: Differentiate: $f'(x) = \sqrt3\cos x + \sin x - 2a$.

Step 2: Combine the trig terms: $\sqrt3\cos x + \sin x = 2\cos\left(x - \dfrac{\pi}{6}\right)$ (since $R=\sqrt{3+1}=2$, $\tan\phi = \frac{1}{\sqrt3} \Rightarrow \phi = \frac{\pi}{6}$), so

$f'(x) = 2\cos\left(x-\frac{\pi}{6}\right) - 2a.$ …

Setting up

To show $f$ is decreasing on all of $\mathbb{R}$, it's enough to show $f'(x) \le 0$ for every $x \in \mathbb{R}$ (with equality never holding on a whole interval).

Step 1 — Differentiate

Step 2 — Combine the trigonometric terms into a single wave

Write $\sqrt3\cos x + \sin x$ as $R\cos(x - \phi)$, where

Matching coefficients: $R\cos\phi = \sqrt3$ and $R\sin\phi = 1$. So

Hence

Step 3 — Write the derivative compactly

Step 4 — Bound it using the range of cosine

For every real $x$, $-1 \le \cos\left(x - \dfrac{\pi}{6}\right) \le 1$, so

Step 5 — Apply the given condition $a \ge 1$ …

Method: Proving Monotonicity of a Function Using the Bounded Range of a Trigonometric Combination

Use this whenever a function's derivative reduces to something of the form $a\sin x + b\cos x - (\text{linear-in-}a\text{-term})$, and you must show the derivative keeps one sign for all real $x$, typically under a stated condition on a parameter.

Steps

Step 1: Differentiate $f(x)$ term by term to obtain $f'(x)$.

Step 2: Combine the sine and cosine terms into a single sinusoid using the auxiliary-angle ("$R$-form") identity.

For $a\sin x + b\cos x$, write it as $R\sin(x+\phi)$ or $R\cos(x-\phi)$ where

Either the sine or cosine form works — they are algebraically equivalent — but rewriting is essential because a single sinusoid has a known, fixed range, whereas the original two-term sum does not obviously.

Step 3: Use the bounded range of sine/cosine, $[-1, 1]$, to bound the whole derivative.

Since $-1 \le \sin(\cdot) \le 1$ (or the cosine form), the sinusoidal part is squeezed between $-R$ and $R$, so

$f'(x) \le R - (\text{constant term}) \quad \text{(for a "decreasing" proof; reverse the inequality for "increasing").}$ …

Common Mistakes

Mistake 1: Sign error differentiating $-\cos x$

A student sometimes writes $\dfrac{d}{dx}(-\cos x)=-\sin x$ instead of $+\sin x$, forgetting the two negatives (from $-\cos x$ and from $\dfrac{d}{dx}\cos x=-\sin x$) cancel. Why it's wrong: this flips the sign of one whole term in $f'(x)$, which would wreck the sign analysis that follows. Correct approach: differentiate carefully term by term — $\dfrac{d}{dx}(-\cos x)=-(-\sin x)=\sin x$.

Mistake 2: Errors combining $\sqrt3\cos x+\sin x$ into a single sinusoid

Miscomputing the amplitude (e.g. using $R=\sqrt3+1$ instead of $R=\sqrt{(\sqrt3)^2+1^2}=2$) or picking the wrong phase angle gives an incorrect bound on $f'(x)$, which can break the whole argument that $a\ge1$ forces $f'(x)\le0$. Correct approach: carefully match coefficients when writing $\sqrt3\cos x+\sin x$ as $R\cos(x-\phi)$ or $R\sin(x+\phi)$, and double-check with a test value like $x=0$. …