The function f(x)=x eˣ⁽¹⁻ˣ⁾, x ∈ ℝ, is :

Solution

Maths sir · 3 steps

Differentiate, set aside the exponential because it is always positive, and the sign rests on a quadratic.

1. Product rule. Write f(x) = x e^(x - x²). Then f'(x) = e^(x-x²) + x (1-2x) e^(x-x²) = e^(x-x²)(1 + x - 2x²).

   f'(x) = - e^(x-x²) (2x+1)(x-1)

2. Read the sign. Since e^(x-x²) > 0 for every x, f'(x) > 0 exactly when (2x+1)(x-1) < 0, that is, for -1/2 < x < 1. So f increases on (-1/2, 1) and decreases on (-∞, -1/2) and on (1, ∞).

   Option	Interval	Sign of f'	Claim holds?
   (a)	(-1/2, 1)	positive throughout	yes, increasing
   (b)	(1/2, 2)	positive up to 1, negative after	no, not monotonic
   (c)	(-1, -1/2)	negative	no, it decreases
   (d)	(-1/2, 1/2)	positive	no, it increases

3. So. The only correct statement is (a): f is increasing in (-1/2, 1).

The idea to keep. For x eᵍ⁽ˣ⁾, factor out the exponential. It is always positive, so the leftover polynomial alone decides the sign of f'. Then check whether each interval sits inside one sign region.

Answer from the NTA final answer key. Solution by PrepWiser.