The function f(x) = (4x³-3x²)/6 - 2sinx + (2x-1)cosx : (1) increases in [1/2,∞) (2) decreases (-∞,1/

Q: Wait, why did the cosine terms just vanish? Did I make a mistake?

You are absolutely on the right track! That cancellation is the 'hidden beauty' of the problem. When you differentiate $(2x-1)\cos x$ using the product rule, you get $2\cos x - (2x-1)\sin x$. When you combine this with the derivative of $-2\sin x$, which is $-2\cos x$, the terms cancel perfectly. It is a deliberate design to reward your patience in calculation.

Q: How do I know for sure that $x - \sin x$ is positive for $x > 0$?

Think of the graphs of $y = x$ and $y = \sin x$. At the origin, they touch, but for any $x > 0$, the line $y = x$ stays strictly above the sine wave. You can also prove this using the Mean Value Theorem or by observing that the derivative of $g(x) = x - \sin x$ is $1 - \cos x$, which is always $\ge 0$. Since $g(0) = 0$ and it is non-decreasing, $g(x)$ must be positive for all $x > 0$.

Q: Why do we care about the sign of $x$ when analyzing $x - \sin x$?

Great observation! The factor $(x - \sin x)$ changes sign at $x = 0$. Since our interval of interest is $[\frac{1}{2}, \infty)$, we are safely in the region where $x > 0$, so we know for certain that $(x - \sin x)$ is positive. Always check if your interval crosses the 'critical points' of your factors!

Q: Can I just pick a test point like $x = 1$ to check if it increases?

Testing a point is a fantastic way to verify your intuition, but it is not a proof! In a JEE Advanced exam, you must show that the derivative is non-negative for the *entire* interval. Use the test point to guide your confidence, but rely on the algebraic sign analysis of the factors to secure the marks.

Q: What if the derivative was zero at some points? Is it still increasing?

Yes! A function is considered 'increasing' if $f'(x) \ge 0$. Even if the derivative hits zero at isolated points (like at $x = 1/2$), the function does not 'turn back' or decrease. It just momentarily levels off. So, the closed interval $[1/2, \infty)$ is perfectly valid.

Question

The function $f(x) = \frac{4x^3-3x^2}{6}-2\sin x + (2x-1)\cos x$ :

Options:

(A) increases in $[\frac{1}{2},\infty)$

(B) decreases $(-\infty,\frac{1}{2}]$

(C) increases in $(-\infty,\frac{1}{2}]$

(D) decreases $[\frac{1}{2},\infty)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the JEE journey. Today, we are going to dismantle a problem that, at first glance, looks like a chaotic mess of polynomials and trigonometry.

You see a function like $$f(x) = \frac{4x^3-3x^2}{6}-2\sin x + (2x-1)\cos x$$ and your instinct might be to panic. But I want you to take a deep breath; in mathematics, complexity is often just a mask for elegance. Let us peel back that mask together.

### The Art of Differentiation

To understand how a function behaves—whether it is climbing the mountain of increase or sliding down the valley of decrease—we must look at its rate of change. We need the derivative, $f'(x)$.

First, the polynomial part: 
$$\frac{d}{dx} \left( \frac{4x^3-3x^2}{6} \right) = \frac{12x^2-6x}{6} = 2x^2 - x$$

Now, the trigonometric dance. We have $-2\sin x$ and the product $(2x-1)\cos x$. Applying the product rule to the latter, we get $2\cos x - (2x-1)\sin x$.

When we combine everything, we get:

$$f'(x) = (2x^2 - x) - 2\cos x + 2\cos x - (2x-1)\sin x$$

Do you see it? The $-2\cos x$ and $+2\cos x$ vanish into thin air! This is the moment where the problem rewards your courage. We are left with:

$$f'(x) = 2x^2 - x - (2x-1)\sin x$$

### Factoring the Soul of the Function

Now that the clutter is gone, look at the expression $2x^2 - x$. We can factor out an $x$ to get $x(2x-1)$.

Suddenly, the entire derivative looks like this:

$$f'(x) = x(2x-1) - (2x-1)\sin x$$

If we factor out the common term $(2x-1)$, we arrive at the heart of the problem:

$$f'(x) = (2x-1)(x - \sin x)$$

This is the 'Aha!' moment. We have reduced a terrifying expression into the product of two simple factors. The behavior of the entire function now depends entirely on the signs of these two factors.

### The Final Analysis

We are interested in the interval $[\frac{1}{2}, \infty)$. Let us analyze our factors in this region:

1. The factor $(2x-1)$: For any $x \ge \frac{1}{2}$, this term is clearly $\ge 0$.

2. The factor $(x - \sin x)$: We know that for all $x > 0$, the line $y = x$ is strictly greater than the curve $y = \sin x$. Since our interval is $[\frac{1}{2}, \infty)$, we are strictly in the positive domain. Thus, $(x - \sin x) > 0$.

When you multiply two positive numbers, the result is positive. Therefore, $f'(x) \ge 0$ for all $x$ in the interval $[\frac{1}{2}, \infty)$.

### Conclusion: The Climb

Because the derivative is non-negative throughout this interval, the function $f(x)$ is **strictly increasing**.

You have successfully navigated the complexity, identified the hidden cancellations, and used the power of factoring to reveal the truth. This is the essence of JEE Advanced mathematics—not just calculation, but the ability to see the underlying structure. Keep this clarity, keep this focus, and you will conquer any problem that comes your way.

Animated Solution for Mathematics - Differentiation: The function $f (x) = \frac{4 x ^{3} - 3 x ^{2}}{6} - 2 sin x + (2 x - 1) cos x$ :

Select Answer:

Visualized Solution

Introduction to Monotonicity

Derivative of Polynomial Terms

Derivative of Trigonometric Terms

Simplifying $f^{'} (x)$

Factoring $f^{'} (x)$

Analyzing $(x - sin x)$

Analyzing $(2 x - 1)$

Interval of Increase

Final Conclusion

The Sigma Insight: Monotonicity

Analyzing the Setup

The Art of Differentiation

Factoring the Soul of the Function

The Final Analysis

Conclusion

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Animated Solution for Mathematics - Differentiation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The function f(x)=64x3−3x2​−2sinx+(2x−1)cosx :

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Introduction to Monotonicity

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