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

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.

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

$\frac{1-\sqrt5}{2}$
$\text{f}(\text{x})=\frac{\text{x}(\text{x}+1)}{\text{e}^{\text{x}}}$ defined on $[-1, 0]$
$\Rightarrow f(-1) = 0$ also $f(0) = 0$
Now, $f(x) = e^{-x}(x^2 + x)$
$\Rightarrow f'(x) = e^{-x}(2x + 1) - (x^2 + x)e^{-x}$
$\Rightarrow f'(x) = e^{-x}(2x + 1 - x^2 + x)$
$\Rightarrow f'(x) = e^{-x}(-x^2 + x - 1)$
$\Rightarrow f'(x) = 0$
$\Rightarrow e^{-x}(-x^2 + x - 1) = 0$
$\Rightarrow -x^2 + x - 1 = 0$
$\Rightarrow x^2 - x + 1 = 0$
$\Rightarrow\text{x}=\frac{1\pm\sqrt5}{2}$
As, $\text{x}\in[-1,0]$
$\text{x}=\frac{1-\sqrt5}{2}$

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