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Question

The function f(x)=x(x+3)e(1/2)x satisfies all the conditions of Rolle's theorem in [–3, 0]. The value of c is

A
\N
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B
-1
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C
-2
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D
-3
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Solution

The correct option is C -2
f(x)=x(x+3)e(1/2)x
f'(x) = (x+3)e(1/2)x+x.e(1/2)x(x+3)e(1/2)x2
As f(x) satisfies all the conditions of rolle's theorem we'll get atleast a point in the interval [-3, 0] where f'(x) will vanish.
f'(x) = 0

(x+3)e(1/2)x+x.e(1/2)x(x+3)e(1/2)x2 = 0
= e(1/2)x(2x+3fracx(x+3)2)
x = -2, 3
In the interval [-3, 0] it'll be x = -2 .


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