Q.If $f(x) = a(x - \cos x)$, $x \in \mathbb{R}$ is a strictly decreasing function, then 'a' lies in which of the following intervals?
(A) $\{0\}$
(B) $(0, \infty)$
(C) $(-\infty, 0)$
(D) $(-\infty, \infty)$

Concept: Monotonicity Condition — A function is strictly decreasing when its derivative is negative for all $x$ in its domain.

Step 1: Differentiate $f(x) = a(x - \cos x)$:

Step 2: For strict decrease, $f'(x) < 0$ for all $x \in \mathbb{R}$. Since $1 + \sin x \ge 0$ (minimum 0, maximum 2), the sign of $f'(x)$ depends entirely on $a$. …

The key idea is the Monotonicity Condition: a differentiable function is strictly decreasing on $\mathbb{R}$ if and only if its derivative is $\le 0$ everywhere and is zero only at isolated points (not on any interval). Here, we just need $f'(x) < 0$ for all $x$ except possibly at isolated points where it can be zero.

1. Find the derivative.

   $f(x) = a(x - \cos x)$

   Differentiate term by term:

   $f'(x) = a(1 - (-\sin x)) = a(1 + \sin x)$.

2. Analyze the factor $(1 + \sin x)$.

   Since $\sin x \in [-1, 1]$, we have $1 + \sin x \in [0, 2]$.

   It is zero when $\sin x = -1$, i.e., at $x = -\frac{\pi}{2} + 2n\pi$ for $n \in \mathbb{Z}$. These are isolated points — the derivative vanishes at them, but not on any interval.

3. Apply the strictly decreasing condition.

   For $f$ to be strictly decreasing, we need $f'(x) \le 0$ for all $x$, and $f'(x) = 0$ only at isolated points.

   Since $1 + \sin x \ge 0$, the sign of $f'(x)$ is entirely determined by $a$:

   • If $a > 0$, then $f'(x) \ge 0$ (non-negative), so $f$ is non-decreasing — not strictly decreasing. …

Method: Finding the Values of a Parameter for Which a Function Is Strictly Monotonic

This method applies whenever a function contains an unknown constant (commonly called $a$, $k$, or $\lambda$) and you must find which values of that constant make the function strictly increasing or strictly decreasing on its whole domain.

Steps

Step 1: Differentiate the function, treating the parameter as a constant

Find $f'(x)$ as usual — the parameter will appear as a multiplying constant in the derivative, since it doesn't depend on $x$.

Step 2: Isolate the part of the derivative that depends on $x$

Rewrite $f'(x)$ as (parameter) $\times$ (an expression in $x$). Determine the range of values that the $x$-dependent expression itself can take — trigonometric or squared expressions are especially common, since their range is bounded and known (e.g. $\sin x \in [-1,1]$, so an expression like $1+\sin x$ is always $\geq 0$).

Step 3: Impose the required sign on the whole derivative …

Common Mistakes

Mistake 1: Differentiation sign error on $-\cos x$

$\dfrac{d}{dx}(-\cos x) = \sin x$, not $-\sin x$. A student who mis-differentiates gets $f'(x)=a(1-\sin x)$ instead of the correct $a(1+\sin x)$, changing the whole sign analysis and leading to the wrong interval for $a$.

Mistake 2: Requiring $f'(x)<0$ at every single point instead of allowing isolated zeros …