If $f(x) = xe^{x(1-x)}$, then $f(x)$ is (a) increasing in $[-\frac{1}{2}, 1]$ (b) decreasing in $\mathbb{R}$ (c) increasing in $\mathbb{R}$ (d) decreasing in $[-\frac{1}{2}, 1]$

1. The theory you need

• Derivative test for monotonicity:
  • If $f'(x) > 0$ on an interval, $f(x)$ is increasing there.
  • If $f'(x) < 0$ on an interval, $f(x)$ is decreasing there.
• Product & chain rules for differentiation, because $f(x) = x \cdot e^{x(1-x)}$ is a product of $x$ and an exponential expression.
• Sign analysis of a quadratic: after simplifying the derivative we will meet a quadratic expression whose sign decides everything.

2. Logical chain a student should follow

1. Differentiate $f(x)$ using product rule:

   $$f'(x) = \left(x\right)' \cdot e^{x(1-x)} + x \cdot \left(e^{x(1-x)}\right)'$$

2. Handle the exponential part with chain rule:

   $$\left(e^{x(1-x)}\right)' = e^{x(1-x)} \cdot (1 - 2x)$$

3. Factor out the common exponential (it is always positive, so its sign will not change the inequality):

   $$f'(x) = e^{x(1-x)} \left[1 + x(1 - 2x)\right]$$

4. Simplify inside the bracket to a quadratic:

   $$1 + x - 2x^2 = -2x^2 + x + 1$$

5. Find roots of that quadratic to know where its sign flips. Roots turn out to be $x = -\dfrac{1}{2}$ and $x = 1$.

6. Check the sign of the quadratic between and outside the roots (because its leading coefficient is negative, it is positive between the roots and negative outside).

7. Combine information to state intervals of increase/decrease and match with given options.

3. Why each step?

• Factoring the exponential early keeps calculations clean.
• Roots of a quadratic split the real line into natural test-intervals.
• Since $e^{\text{something}} > 0$ for all real numbers, the exponential never changes sign and therefore never changes the sign of $f'(x)$; only the quadratic matters.

What is the question about?

We have a magic curve drawn by the rule

$f(x) = x e^{x(1-x)}$

and we must tell when the curve is going uphill (increasing) and when it is going downhill (decreasing).

How do we usually decide that?

Think of the curve like a road. To know if the road is sloping up or down, we look at its slope. In maths the slope is given by the derivative $f'(x)$. If the slope is positive, the road climbs; if the slope is negative, it descends.

Baby steps to check:

1. Find the slope $f'(x)$.
2. See where the slope is positive or negative.
3. Match with the options.

That’s it! We just have to do a little algebra with the derivative.

Step 1 Differentiate $f(x)$

Step 2 Factor sign-deciding part

Since $e^{x(1-x)} > 0$ for all real $x$, the sign of $f'(x)$ is exactly the sign of

Step 3 Find roots of $g(x)$

Using quadratic formula:

Step 4 Sign chart for $g(x)$

Leading coefficient is negative ($-2$), so

• $g(x) > 0$ on the open interval $\bigl(-\tfrac12,\,1\bigr)$,
• $g(x) < 0$ on $(-\infty,\,-\tfrac12)$ and $(1,\,\infty)$,
• $g(x) = 0$ at $x=-\tfrac12$ and $x=1$.

Step 5 Translate to monotonicity of $f$

Therefore

• $f'(x) > 0$ on $\bigl(-\tfrac12,\,1\bigr)$ (\Rightarrow f$ is increasing there).
• $f'(x) < 0$ on $(-\infty,\,-\tfrac12)$ and $(1,\,\infty)$ (\Rightarrow f$ is decreasing there).
• Including the endpoints where $f'(x)=0$ gives the closed interval $\left[-\tfrac12,\,1\right]$ for non-strict increase.

Step 6 Match with options

(a) increasing in $\left[-\dfrac12,\,1\right]$ ✔️ (True)

(b) decreasing in $\mathbb{R}$ ✖️

(c) increasing in $\mathbb{R}$ ✖️

(d) decreasing in $\left[-\dfrac12,\,1\right]$ ✖️

Final Answer: Option (a)