The function f(x) = x e^(x(1-x)), x ∈ R, is
(A) increasing in (-1/2, 1)
(B) decreasing in (1/2, 2)
(

Q: Wait, why did the inequality sign flip when we divided by $-e^{x-x^2}$?

That is a classic trap! Remember, $e^{x-x^2}$ is always positive. Therefore, $-e^{x-x^2}$ is always negative. When you divide an inequality by a negative number, the direction of the inequality must reverse. It is a fundamental rule of algebra that we often overlook in the heat of the moment.

Q: Do I really need to use the Wavy Curve method? Can't I just test points?

You absolutely can test points, but the Wavy Curve method (or sign scheme) is the gold standard for JEE Advanced. It is faster, less prone to arithmetic errors, and gives you a visual representation of the function's behavior across the entire domain. Master it, and you will save precious minutes in the exam.

Q: Is the exponential term ever zero? Could it affect the critical points?

Excellent question. The exponential function $e^u$ is strictly positive for all real $u$. It never touches the x-axis, so it never equals zero. Consequently, it contributes nothing to the critical points. It only acts as a scaling factor that preserves the sign of the quadratic expression.

Question

The function $f(x) = x e^{x(1-x)}, x \in \mathbb{R}$, is

Options:

(A) increasing in $(-\frac{1}{2}, 1)$

(B) decreasing in $(\frac{1}{2}, 2)$

(C) increasing in $(-1, -\frac{1}{2})$

(D) decreasing in $(-\frac{1}{2}, \frac{1}{2})$

Sigma Percentile · Accepted Answer

### Analyzing the Anatomy of the Function

To begin our journey, we observe the structure of the function:
$$f(x) = x \cdot e^{x-x^2}$$

This is a classic product of two distinct entities: a linear function $u(x) = x$ and an exponential function $v(x) = e^{x-x^2}$.

Whenever you see a product, your mind should immediately jump to the **Product Rule**. It is the fundamental law of differentiation for products, defined as:
$$f'(x) = u'(x)v(x) + u(x)v'(x)$$

Think of it as a dance where each term takes a turn being differentiated while the other remains constant.

### The Calculus Surgery

Let us perform the surgery. We know $u(x) = x$, so its derivative $u'(x)$ is simply $1$.

Now, consider $v(x) = e^{x-x^2}$. This requires the **Chain Rule**, where the derivative of $e^g(x)$ is $e^g(x) \cdot g'(x)$.

Since the derivative of $x-x^2$ is $1-2x$, we find:
$$v'(x) = e^{x-x^2}(1-2x)$$

Assembling these pieces into our Product Rule formula, we get:
$$f'(x) = (1) \cdot e^{x-x^2} + (x) \cdot [e^{x-x^2}(1-2x)]$$

At this stage, we identify the common factor, $e^{x-x^2}$, and factor it out:
$$f'(x) = e^{x-x^2} [1 + x(1-2x)]$$

### The Algebraic Cleanup

Now, let us focus on the bracketed expression: $1 + x(1-2x)$. Expanding this, we obtain $1 + x - 2x^2$.

To make this easier to factor, we rearrange it into standard quadratic form: $-2x^2 + x + 1$. Factoring out a negative sign gives us:
$$-(2x^2 - x - 1)$$

We factor the quadratic $2x^2 - x - 1$ by splitting the middle term. This results in:
$$-(2x+1)(x-1)$$

Putting it all together, our derivative reaches its final, compact form:
$$f'(x) = -e^{x-x^2}(2x+1)(x-1)$$

### The Wavy Curve Analysis

To determine where the function is increasing, we set the condition $f'(x) > 0$:
$$-e^{x-x^2}(2x+1)(x-1) > 0$$

The term $e^{x-x^2}$ is always positive and does not affect the sign. However, because we have a negative sign in front, dividing by $-e^{x-x^2}$ flips the inequality:
$$(2x+1)(x-1) < 0$$

Using the **Wavy Curve method**, we identify the critical points at $x = -\frac{1}{2}$ and $x = 1$.

Testing the intervals, we find that the expression is negative between these two points. Thus, the function is increasing in the interval:
$$\mathbf{\left( -\frac{1}{2}, 1 \right)}$$

Animated Solution for Mathematics - Differentiation: The function $f (x) = x e^{x (1 - x)}, x \in R$ , is

Select Answer:

Visualized Solution

Introduction to the Function $f (x)$

The Product Rule Strategy

Differentiating the Parts

Assembling the Derivative $f^{'} (x)$

Factoring the Exponential Term

Simplifying the Quadratic Expression

Factoring the Quadratic

The Final Form of $f^{'} (x)$

Condition for Increasing Function

Finding Critical Points

The Sign Scheme (Wavy Curve Method)

Final Conclusion

The Sigma Insight: Monotonicity

Analyzing the Anatomy of the Function

The Calculus Surgery

The Algebraic Cleanup

The Wavy Curve Analysis

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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)=xex(1−x),x∈R, is