The value of c in Rolle's theorem for the function f(x)= x(x+1) e^ x defined on [-1, 0] is :

The value of c in Rolle's theorem for the function f(x)= x(x+1) e…

MCQ

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

A
$0.5$
B
$\frac{1+\sqrt5}{2}$
✓
$\frac{1-\sqrt5}{2}$
D
$-0.5$

✓

Answer

Correct option: C.

$\frac{1-\sqrt5}{2}$

$f(x)=\frac{x(x+1)}{e^x} $ defined on $[-1,0]$
$\Rightarrow f(-1)=0 $ also $f(0)=0$
Now, $ f(x)=e^{-x}\left(x^2+x\right)$
$\Rightarrow f^{\prime}(x)=e^{-x}(2 x+1)-\left(x^2+x\right) e^{-x}$
$\Rightarrow f^{\prime}(x)=e^{-x}\left(2 x+1-x^2+x\right)$
$\Rightarrow f^{\prime}(x)=e^{-x}\left(-x^2+x-1\right)$
$\Rightarrow f^{\prime}(x)=0$
$\Rightarrow e^{-x}\left(-x^2+x-1\right)=0$
$\Rightarrow-x^2+x-1=0$
$\Rightarrow x^2-x+1=0$
$\Rightarrow x=\frac{1 \pm \sqrt{5}}{2}$
As $ x \in[-1,0]$
$x=\frac{1-\sqrt{5}}{2}$

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