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in Continuity and Differentiability by (29.1k points)
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The value of c in Rolle’s theorem for the function f(x) = \(\frac{x(x+1)}{e^x}\) defined on [–1, 0] is

A. 0.5

B. \(\frac{1+\sqrt{5}}{2}\)

C. \(\frac{1-\sqrt{5}}{2}\)

D. –0.5

1 Answer

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by (28.7k points)
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Best answer

Correct answer is C.

\(\text{f(x)}=\frac{\{\text{x}(\text{x + 1})\}}{e^\text{x}}\)

Differentiating the function with respect to ‘x’,

Using Sridharacharya Formula, In a general equation, ax2 + bx + c = 0

So the required value of c = \(\frac{1-\sqrt{5}}{2}\) Є (-1, 0).

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