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

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