For which values of a, f(x)=a(x+sin x) is an increasing function?

The correct answer is Option (3) → \(a \in [0, \infty)\)

Differentiating:

    \(f'(x) = a(1+\cos x)\)

Since \(\cos x \ge -1\), the factor \(1+\cos x \ge 0\) for every x, and it is 0 only at the separate points \(x = (2n+1)\pi\).

If a > 0, then \(f'(x) \ge 0\) and \(f'(x) = 0\) only at those separate points, so f is increasing (in fact strictly increasing).

If a = 0, then \(f(x) = 0\) for every x, a constant function. A function is increasing when \(x_1 \lt x_2\) gives \(f(x_1) \le f(x_2)\); a constant function satisfies this, so a = 0 is also allowed.

If a < 0, then \(f'(x) \le 0\) and f is decreasing.

So f is increasing exactly for \(a \ge 0\), that is, \(a \in [0, \infty)\).

Why the other options are incorrect

• Option (1) \(a \le 0\) — Incorrect. For every negative a, \(f'(x) \le 0\) and f is decreasing.

• Option (2) \(a \in (0, \infty)\) — Incorrect. It leaves out a = 0, which gives a constant function, and a constant function is increasing in the sense \(f(x_1) \le f(x_2)\). The interval \((0, \infty)\) is the set of values for which f is strictly increasing.

• Option (4) \(a \in R\) (set of real numbers) — Incorrect. It includes negative values of a, for which f is decreasing.