Question

If \[ f(x) = x \cdot e^{x(1-x)}, \] then \( f(x) \) is: 

• A. increasing in \(\mathbb{R}\)
• B. increasing in \((-\frac{1}{2}, 1)\)
• C. decreasing in \(\mathbb{R}\)
• D. decreasing in \([-\frac{1}{2}, 1]\)

Hint:

When an exponential factor appears in \(f'(x)\), check whether it is always positive. If yes, you only need to study the polynomial factor.

Answer 1

The Correct Option is B

Concept:
To study increasing or decreasing nature, we find \(f'(x)\) and check where it is positive or negative. ip

Step 1: Differentiate the function.
\[ f(x)=x e^{x(1-x)} = x e^{x-x^2} \] Using product rule: \[ f'(x)=e^{x-x^2} + x e^{x-x^2}(1-2x) \] \[ f'(x)=e^{x-x^2}\left(1+x-2x^2\right) \] ip

Step 2: Study the sign of \(f'(x)\).
Since \[ e^{x-x^2}>0 \quad \text{for all } x, \] the sign depends on: \[ 1+x-2x^2 \] So, \[ 1+x-2x^2>0 \] \[ 2x^2-x-1<0 \] \[ (2x+1)(x-1)<0 \] Thus, \[ -\frac12 < x < 1 \] ip

Step 3: Conclude the interval of increase.
Therefore, \(f(x)\) is increasing in: \[ \left(-\frac12,1\right) \] ip Hence, the correct answer is:
\[ \boxed{(B)\ \text{increasing in }\left(-\frac12,1\right)} \]