If the function f (x )=x (x+3 )e^ -x2 satisfies all the conditions of Rolle's theorem in [-3 ,0 ], then a root of f^ ' (x )=0 is

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Correct answer: C. (c) According to R.T theorem function should be continuous and differentiable in (-3 ,0 ) f (x )=x (x+3 )e^-x/2=(x^2+3x)e^-x/2f^ ' (x )= (2x+3 )e^-x/2+(x^2+3x)e^-x/2 ( -12 )=0e^-x/2( (2x+3 )- (x^2+3x)2)=0 Here e^-x/2 0 Hence, (x^2-x-6)=0 ⇒x^2-x-6=0 ⇒ (x+2 ) (x-3 )=0 ⇒x=-2 ∈ [-3 ,0 ]

$If the function f (x) = x (x + 3) e^{\frac{- x}{2}} satisfies all the conditions of Rolle's theorem in [- 3, 0], then a root of f^{’} (x) = 0 is$

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