For the function f(x) = sin x + 3x - (2/π)(x² + x), where x ∈ [0, π/2], consider the following two s

Q: Wait, why do we need the second derivative $f''(x)$ to check if $f'(x)$ is decreasing? Can't we just look at $f'(x)$ directly?

That is a great question! While you *could* try to analyze $f'(x)$ directly, it involves a trigonometric term and a linear term. Differentiating it to find $f''(x)$ is the standard, most robust way to determine the 'slope of the slope.' If $f''(x) < 0$, it guarantees $f'(x)$ is decreasing without needing to graph it.

Q: Why do we evaluate the minimum of $f'(x)$ at $x = \frac{\pi}{2}$? Is it always the right endpoint?

Not always! It depends on the nature of the function. Because we proved in Statement (II) that $f'(x)$ is a *decreasing* function, its value gets smaller as $x$ increases. Therefore, the smallest value must be at the largest possible $x$ in the interval, which is the right endpoint, $\frac{\pi}{2}$.

Q: Is it really safe to assume $1 - \frac{2}{\pi} > 0$? How do I know for sure?

Absolutely! We know that $\pi \approx 3.14$. Therefore, $\frac{2}{\pi} \approx \frac{2}{3.14}$, which is clearly less than $1$. Since you are subtracting a value less than $1$ from $1$, the result must be positive. In JEE problems, always rely on these fundamental inequalities.

Question

For the function $f(x) = \sin x + 3x - \frac{2}{\pi}(x^2 + x)$, where $x \in [0, \frac{\pi}{2}]$, consider the following two statements : (I) $f$ is increasing in $(0, \frac{\pi}{2})$. (II) $f'$ is decreasing in $(0, \frac{\pi}{2})$. Between the above two statements,

Options:

(A) only (II) is true.

(B) only (I) is true.

(C) neither (I) nor (II) is true.

(D) both (I) and (II) are true

Sigma Percentile · Accepted Answer

### The Dance of Derivatives: A Journey into Monotonicity

Welcome, future engineers! Today, we are going to dissect a problem that sits at the very heart of calculus. It is not just about solving for $x$; it is about understanding the 'behavior' of a function.

We are looking at the function $f(x) = \sin x + 3x - \frac{2}{\pi}(x^2 + x)$ on the interval $[0, \frac{\pi}{2}]$. This is a beautiful mix of trigonometry and algebra, and it is our job to determine its monotonicity.

### Phase 1: The First Derivative - The Velocity of Change

To understand if a function is increasing or decreasing, we must look at its rate of change. Think of $f(x)$ as the position of a particle, and $f'(x)$ as its velocity.

If the velocity is positive, the particle moves forward (the function increases). If it is negative, the particle moves backward (the function decreases). Let us find this velocity by differentiating $f(x)$ with respect to $x$:

$$f'(x) = \frac{d}{dx}(\sin x) + \frac{d}{dx}(3x) - \frac{2}{\pi}\frac{d}{dx}(x^2 + x)$$

Applying our standard rules, we get:

$$f'(x) = \cos x + 3 - \frac{2}{\pi}(2x + 1)$$

This expression is our key. It tells us everything about the direction in which $f(x)$ is heading.

### Phase 2: The Second Derivative - The Curvature

Now, Statement (II) asks us about the behavior of $f'(x)$ itself. Is $f'(x)$ increasing or decreasing? To answer this, we need the derivative of the derivative—the second derivative, $f''(x)$.

This is the 'acceleration' of our function. Let us differentiate $f'(x)$:

$$f''(x) = \frac{d}{dx}(\cos x) + \frac{d}{dx}(3) - \frac{2}{\pi}\frac{d}{dx}(2x + 1)$$

The derivative of $\cos x$ is $-\sin x$, the constant $3$ vanishes, and the derivative of $(2x+1)$ is $2$. Thus, we arrive at:

$$f''(x) = -\sin x - \frac{4}{\pi}$$

Look closely at this result. In the interval $(0, \frac{\pi}{2})$, $\sin x$ is always positive, and the term $\frac{4}{\pi}$ is also positive. Therefore, $f''(x) = -(\sin x + \frac{4}{\pi})$ is strictly negative for all $x$ in our domain.

Because the derivative of $f'(x)$ is negative, $f'(x)$ must be a strictly decreasing function. **Statement (II) is confirmed!**

### Phase 3: The Logical Bridge

Now, we tackle Statement (I). We need to know if $f(x)$ is increasing, which requires $f'(x) > 0$ for all $x$ in $(0, \frac{\pi}{2})$.

But how do we prove a function is positive across an entire interval? We use the property we just discovered: $f'(x)$ is decreasing. If a decreasing function's **minimum** value is positive, then the entire function must be positive.

Where does a decreasing function reach its minimum? At the rightmost endpoint of the interval, which is $x = \frac{\pi}{2}$. Let us calculate $f'(\frac{\pi}{2})$:

$$f'\left(\frac{\pi}{2}\right) = \cos\left(\frac{\pi}{2}\right) + 3 - \frac{2}{\pi}\left(2 \cdot \frac{\pi}{2} + 1\right)$$

Since $\cos(\frac{\pi}{2}) = 0$, this simplifies to:

$$3 - \frac{2}{\pi}(\pi + 1) = 3 - \left(\frac{2\pi}{\pi} + \frac{2}{\pi}\right) = 3 - \left(2 + \frac{2}{\pi}\right) = 1 - \frac{2}{\pi}$$

### The Final Victory

We know that $\pi \approx 3.14$, so $\frac{2}{\pi}$ is clearly less than $1$. Therefore, $1 - \frac{2}{\pi} > 0$.

Since the minimum value of $f'(x)$ is positive, $f'(x)$ must be positive for the entire interval. If the velocity is always positive, the function $f(x)$ must be strictly increasing. **Statement (I) is also true!**

We have systematically proven that both statements are correct. Calculus is not just about formulas; it is about building a logical chain where each link supports the next. You have mastered the logic of monotonicity today.

Select Answer:

Visualized Solution

Function and Domain Analysis

Finding the First Derivative $f^{'} (x)$

Finding the Second Derivative $f^{''} (x)$

Analyzing the Sign of $f^{''} (x)$

Verifying Statement (II)

Strategy for Statement (I)

Evaluating $f^{'} (\frac{π}{2})$

Simplifying the Minimum Value

Verifying Statement (I)

Final Conclusion

The Sigma Insight: Monotonicity

The Dance of Derivatives

Phase 1

Phase 2

Phase 3

The Final Victory

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