Injectivity and Surjectivity of Absolute Trigonometric Function

To determine whether the function $f: \mathbb{R} \to \mathbb{R}$ given by $f(x) = |x|(x - \sin x)$ is one-one (injective) and/or onto (surjective), we analyze its properties step-by-step.

1. Symmetry of the Function

We first check if the function is even or odd: $f(-x) = |-x|(-x - \sin(-x)) = |x|(-x + \sin x) = -|x|(x - \sin x) = -f(x)$

Since $f(-x) = -f(x)$ for all $x \in \mathbb{R}$, $f(x)$ is an odd function, which means it is symmetric about the origin.

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2. Injectivity (One-One)

To check if $f$ is one-one, we examine the monotonicity of $f(x)$ by analyzing its derivative.

For $x > 0$, we have $|x| = x$, so: $f(x) = x(x - \sin x) = x^2 - x \sin x$

Differentiating $f(x)$ with respect to $x$ for $x > 0$: $f'(x) = 2x - (\sin x + x \cos x) = (x - \sin x) + x(1 - \cos x)$

We know that for all $x > 0$:

1. $x - \sin x > 0$
2. $1 - \cos x \ge 0$, which implies $x(1 - \cos x) \ge 0$

Thus, the sum of two non-negative terms (where at least one is strictly positive) gives: $f'(x) > 0 \quad \text{for all } x > 0$

Since $f(x)$ is an odd function, its derivative is an even function, which implies: $f'(x) > 0 \quad \text{for all } x < 0$

At $x = 0$: $f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} \frac{|h|(h - \sin h)}{h} = 0$

Since $f'(x) > 0$ for all $x \neq 0$ and $f(x)$ is continuous on $\mathbb{R}$, $f(x)$ is strictly increasing on the entire real number line $\mathbb{R}$.

Because $f(x)$ is strictly increasing, every horizontal line intersects the graph of $f(x)$ at most once. Hence, $f$ is one-one (injective).

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3. Surjectivity (Onto)

To check if $f$ is onto, we determine the range of $f(x)$.

• $f(x)$ is continuous on $\mathbb{R}$.
• As $x \to \infty$, since $x - \sin x \ge x - 1$, we have: $\lim_{x \to \infty} f(x) = \lim_{x \to \infty} x(x - \sin x) = \infty$
• As $x \to -\infty$, using the odd symmetry of $f$: $\lim_{x \to -\infty} f(x) = -\infty$

By the Intermediate Value Theorem, since $f(x)$ is continuous and takes all values from $-\infty$ to $\infty$, the range of $f$ is $\mathbb{R}$, which is equal to its codomain $\mathbb{R}$.

Hence, $f$ is onto (surjective).

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Conclusion

The function $f$ is BOTH one-one and onto.

Correct Option: (C)