If the function f: R → R is defined by f(x) = |x|(x – sinx), then which of the following statements

Q: Wait, why do we need to check the limits at infinity to prove it is onto?

Great question! A function is onto (surjective) if its range equals its codomain. Since our codomain is $\mathbb{R}$, we need to ensure the function covers every value from $-\infty$ to $+\infty$. Because the function is continuous, checking the limits at the boundaries tells us exactly what values it attains. If the limit at $-\infty$ is $-\infty$ and at $+\infty$ is $+\infty$, the Intermediate Value Theorem guarantees it covers all real numbers.

Q: Is it really necessary to split the derivative analysis for $x > 0$ and $x < 0$?

Absolutely. The modulus function $|x|$ creates a 'kink' or a change in definition at $x=0$. While the function is continuous everywhere, its algebraic form changes, so we must treat the positive and negative domains separately to ensure our derivative calculations are rigorous. Skipping this step is a common trap in JEE Advanced.

Q: Why can't we just look at the graph and say it's one-one?

Intuition is a powerful tool, but in JEE Advanced, we need rigorous proof. A graph can be misleading if you don't account for subtle oscillations or flat regions. By calculating the derivative $f'(x)$ and showing it is strictly positive, we mathematically guarantee that the function never 'turns back' on itself, which is the definition of being one-one.

Question

If the function $f: \mathbb{R} 	o \mathbb{R}$ is defined by $f(x) = |x|(x - \sin x)$, then which of the following statements is TRUE?

Options:

(A) $f$ is one-one, but NOT onto

(B) $f$ is onto, but NOT one-one

(C) $f$ is BOTH one-one and onto

(D) $f$ is NEITHER one-one NOR onto

Sigma Percentile · Accepted Answer

### The Beauty of the Modulus Welcome, future engineers! Today, we are going to peel back the layers of a function that might look intimidating at first glance: $f(x) = |x|(x - \sin x)$. In the world of JEE Advanced, we don't just solve problems; we investigate them. We look for the hidden symmetry, the underlying structure, and the elegant logic that binds everything together. Let's embark on this journey. ### Phase 1: The Surgery of the Modulus Whenever you see a modulus function, do not panic. Treat it as a surgeon treats a patient—with precision. The modulus $|x|$ is simply a gatekeeper that splits our function into two distinct worlds: For $x \ge 0$: $$f(x) = x(x - \sin x) = x^2 - x \sin x$$ For $x < 0$: $$f(x) = -x(x - \sin x) = -x^2 + x \sin x$$ By splitting the function, we have removed the ambiguity. We are now ready to analyze the range and the monotonicity of these two pieces. ### Phase 2: The Hunt for 'Onto' To determine if the function is 'onto' (surjective), we must ask: does this function cover every possible value on the y-axis? Since our codomain is $\mathbb{R}$, we need to see if the function spans from $-\infty$ to $+\infty$. Let's look at the limits. As $x o \infty$, the term $x^2(1 - \frac{\sin x}{x})$ dominates. Since $\frac{\sin x}{x}$ vanishes to zero, the function behaves like $x^2$, which shoots to $+\infty$. Conversely, as $x o -\infty$, the function behaves like $-x^2$, which dives to $-\infty$. Because the function is continuous, it must pass through every single value in between. Thus, the range is $\mathbb{R}$, and we have successfully proven the function is **onto**. ### Phase 3: The Derivative and the Proof of One-One Now, for the final boss: is the function 'one-one' (injective)? A function is one-one if it is strictly monotonic—it must always be climbing or always be falling. We need to check the derivative $f'(x)$. For $x > 0$, our derivative is: $$f'(x) = \frac{d}{dx}(x^2 - x \sin x) = 2x - (\sin x + x \cos x)$$ This looks complex, but let's use a bit of algebraic magic. We can rewrite $2x$ as $x + x$: $$f'(x) = (x - \sin x) + (x - x \cos x) = (x - \sin x) + x(1 - \cos x)$$ Look closely at this expression. For $x > 0$, we know that $x > \sin x$, so the first term is positive. We also know that $\cos x \le 1$, so $(1 - \cos x) \ge 0$. Since $x$ is positive, the second term is also non-negative, making the sum strictly positive. When we repeat this logic for $x < 0$, we find the same result: the derivative is strictly positive. This means the function is strictly increasing across the entire real line. A strictly increasing function can never take the same value twice; therefore, it is **one-one**. ### Conclusion We have proven that our function is both one-one and onto. It is a bijection. It is a perfect mapping. I hope this journey through the calculus of this function has shown you that math is not just about formulas—it is about the story the numbers tell. Keep practicing, keep questioning, and keep falling in love with the process!

Animated Solution for Mathematics - Functions: If the function $f : R \to R$ is defined by $f (x) = ∣ x ∣ (x - sin x)$ , then which of the following statements is TRUE?

Select Answer:

Visualized Solution

Analyze the Function

Handling the Modulus

Piecewise Definition Setup

Expanding the Function

Strategy for 'Onto'

Limit at $+ \infty$

Limit at $- \infty$

Range and Surjectivity

Strategy for 'One-One'

Differentiating for $x > 0$

Rearranging the Derivative

Sign Analysis for $x > 0$

Derivative for $x < 0$

Sign Analysis for $x < 0$

Final Conclusion

The Sigma Insight: Classification of Functions

The Beauty of the Modulus

Phase 1

Phase 2

Phase 3

Conclusion

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Animated Solution for Mathematics - Functions: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the function f:R→R is defined by f(x)=∣x∣(x−sinx), then which of the following statements is TRUE?