Q.The values of $a$ for which the function $f(x) = \sin x - ax + b$ increases on $\mathbb{R}$ are ______.

Concept: Derivative Sign Analysis — a function increases on $\mathbb{R}$ when its derivative is non-negative for all $x$.

1. Differentiate: $f'(x) = \cos x - a$.
2. For $f$ to increase on $\mathbb{R}$, we need $f'(x) \ge 0$ for every real $x$, i.e., $\cos x - a \ge 0$ for all $x$. …

The key idea is simple: a function increases (strictly or non-strictly) on an interval when its derivative is never negative there. For the whole real line, we need $f'(x) \ge 0$ for every $x \in \mathbb{R}$.

Let’s see why this works. The derivative $f'(x)$ tells us the slope of the tangent at each point. If the slope is always at least zero, the function never goes downhill — it either rises or stays flat. That’s exactly what “increases on $\mathbb{R}$” means (non-decreasing, to be precise; many exam problems use “increases” to mean “does not decrease”).

Now, $f(x) = \sin x - ax + b$. Differentiate:

The constant $b$ vanishes — it only shifts the graph vertically, which has no effect on whether the function rises or falls.

So the condition becomes:

Equivalently:

This is a “for all $x$” statement. It means $a$ must be less than or equal to every value that $\cos x$ can take. In other words, $a$ must be a lower bound for the set $\{\cos x : x \in \mathbb{R}\}$.

What is the smallest value $\cos x$ ever reaches? It’s $-1$. So the condition “$a \le \cos x$ for all $x$” is equivalent to:

$a \le \min_{x \in \mathbb{R}} \cos x = -1$ …

Method: Finding Parameter Values for Which a Function Is Monotonic on All of $\mathbb{R}$

When a question asks "for what values of $a$ does $f$ increase/decrease everywhere," it is really asking you to turn a for-all-$x$ inequality into a bound on the parameter — a distinct skill from ordinary sign-chart problems, which only need to hold on one interval at a time.

Steps

Step 1: Differentiate, keeping the parameter symbolic.

Write $f'(x)$ in terms of $x$ and the parameter (call it $a$).

Step 2: State the monotonicity requirement as an inequality that must hold for EVERY real $x$.

• For $f$ increasing (non-decreasing) on $\mathbb{R}$: $f'(x)\ge 0$ for all $x\in\mathbb{R}$.
• For $f$ decreasing on $\mathbb{R}$: $f'(x)\le 0$ for all $x\in\mathbb{R}$.

Step 3: Isolate the parameter on one side of the inequality.

Rearrange so the parameter sits alone, leaving a bounded expression in $x$ on the other side, e.g.

where $h(x)$ is built from the trig/bounded part of $f'(x)$.

Step 4: Use the known range of the bounded expression. …

Common Mistakes

Mistake 1: Concluding $a\le 1$ instead of $a\le -1$

Why it's wrong: The condition is $a\le\cos x$ for every real $x$, which means $a$ must be less than or equal to the SMALLEST value $\cos x$ ever takes (which is $-1$), not its largest value ($1$). Students often reflexively use $\cos x\le 1$ and answer $a\le 1$, which is the wrong bound entirely. Correct approach: since the inequality must hold for all $x$, take the minimum of $\cos x$ over $\mathbb{R}$, which is $-1$, giving $a\le -1$.

Mistake 2: Trying to also constrain the constant $b$ …