The value of c in Roll’s theorem for the function f(x)=x(x+1)ex defined on [−1,0], is

# The value of $c$ in Roll's theorem for the function $f\left(x\right)=\frac{x\left(x+1\right)}{{e}^{x}}$ defined on $\left[-1,0\right]$, is

1. A

0.5

2. B

$\frac{1+\sqrt{5}}{2}$

3. C

$\frac{1-\sqrt{5}}{2}$

4. D

$-0.5$

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### Solution:

We have, $f\left(x\right)=x\left(x+1\right){e}^{-x},x\in \left[-1,0\right]$

Hence, ${f}^{\mathrm{\prime }}\left(x\right)=\frac{f\left(0\right)-f\left(-1\right)}{0-\left(-1\right)}$

Hence, $c=\frac{1-\sqrt{5}}{2}$

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