If fx-x ex1-x- then fx isA- increasing on -1-2- 1-B- decreasing on RC- increasing on RD- decreasing on -1-2- 1-

The correct option is A increasing on [ 1/2,1 ]f(x)=xe^x(1 x) ⇒ f'^(x)=e^x(1 x)+(1 2x)x e^x(1 x) = e^x(1 x)(2x^2 x 1)= e^x(1 x)(2x+1)(x 1) ∴ f(x) is increasi ...

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Byju's Answer

Standard XII

Mathematics

Monotonically Increasing Functions

If fx=x ex1-x...

Question

If $f (x) = x e^{x (1 - x)}$ , then f(x) is

increasing on

[\frac{- 1}{2}, 1]

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decreasing on R

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increasing on R

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decreasing on

[\frac{- 1}{2}, 1]

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Solution

The correct option is A increasing on $[\frac{- 1}{2}, 1]$
$f (x) = x e^{x (1 - x)}$
$\Rightarrow f^{' (} x) = e^{x (1 - x)} + (1 - 2 x) x e^{x (1 - x)}$
$= - e^{x (1 - x)} (2 x^{2} - x - 1) = - e^{x (1 - x)} (2 x + 1) (x - 1)$

$∴ f (x)$ is increasing on $[\frac{- 1}{2}, 1]$

Suggest Corrections

If f(x)=xex(1−x), then f(x) is

The correct option is A increasing on [−12,1]f(x)=xex(1−x) ⇒f′(x)=ex(1−x)+(1−2x)x ex(1−x) =−ex(1−x)(2x2−x−1)=−ex(1−x)(2x+1)(x−1) ∴f(x) is increasing on [−12,1]

If $f (x) = x e^{x (1 - x)}$ , then f(x) is

The correct option is A increasing on $[\frac{- 1}{2}, 1]$
$f (x) = x e^{x (1 - x)}$
$\Rightarrow f^{' (} x) = e^{x (1 - x)} + (1 - 2 x) x e^{x (1 - x)}$
$= - e^{x (1 - x)} (2 x^{2} - x - 1) = - e^{x (1 - x)} (2 x + 1) (x - 1)$

$∴ f (x)$ is increasing on $[\frac{- 1}{2}, 1]$