Question

The interval in which `y = x^2e^(-x)` is increasing with respect to `x` is

विकल्प

• `(- oo, oo)`

• `(- 2, 0)`

• `(2, oo)`

• `(0, 2)`

Answer 1

`(0, 2)`

Explanation:

`f(x) = x^2e^(-x)`

`f^'(x) = 2xe^(-x) + x^2(-e^(-x)) = xe^(-x) (2 - x) = e^(-x) x(2 - x)`

Now `e^(-x)` is positive for all x ∈ R

`f^'(x)` = 0 at `x` = 0, 2

`x` = 0, `x` = 2 divide the number line into three disjoint interval Viz `(- oo, 0), (0, 2),(2, oo)`

(a) Interval `(- oo, 0)`

`x` is negative and `(2 - x)` is positive

∴ `f^'(x) = e^(-x) x(2 - x) = (+) (-) (+)` = Negative

⇒ `f` is decreasing in `(- oo, 0)`

(b) Interval `(0, 2)`

∴ `f^'(x) = e^(-x) x(2 - x) = (+) (+) (+)` = Positive

⇒ `f` is increasing in `(0, 2)`

(c) Interval `(2, oo)`

∴ `f^'(x) = e^(-x) x(2 - x) = (+) (+) (-)` = Negative

⇒ `f` is decreasing in the interval `(2, oo)`