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Answer:

Option (c)

Hint:

You must know about Rolle’s Theorem.

Given: $f(x)=\frac{x(x+1)}{e^{x}}$defined on [ -1, 0]

Solution:

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

$\Rightarrow \; \; \; \; \; f(x)=\frac{x^{2}+x}{e^{x}}$

$\Rightarrow \; \; \; \; \; f^{'}(x)=\frac{e^{x}\left ( 2x+1 \right )-\left ( x^{2}+x \right )\left ( e^{x} \right )}{e^{2x}}$

$\Rightarrow \; \; \; \; \; f^{'}(x)=\frac{\left ( 2x+1 \right )-\left ( x^{2}+x \right )}{e^{x}}$

$\Rightarrow \; \; \; \; \; f^{'}(c)=\frac{\left ( 2c+1 \right )-\left ( c^{2}+c \right )}{e^{c}}$

Using Rolle’s Theorem,

$f^{'}(c)=0$

$\Rightarrow \; \; \; \; \; \frac{\left ( 2c+1 \right )-\left ( c^{2}+c \right )}{e^{c}}=0$

$\Rightarrow \; \; \; \; \; 2c+1-c^{2}-c=0$

$\Rightarrow \; \; \; \; \; c^{2}+c+1=0$

$\Rightarrow \; \; \; \; \; c^{2}-c-1=0$

$\Rightarrow \; \; \; \; \; c=\frac{-(-1)\pm \sqrt{(-1)^{2}-4(1)(-1)}}{2(1)}$                          $\left [ \because c=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a} \right ]$

$\Rightarrow \; \; \; \; \; c=\frac{1\pm \sqrt{1+4}}{2}$

$\Rightarrow \; \; \; \; \; c=\frac{1\pm \sqrt{5}}{2}$

but $\; \; \; \; \; c=\frac{1- \sqrt{5}}{2}\in \left [ -1,0 \right ]$

Hence option (c) is correct.

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