If f is a continuous function and x3−(5+1)x2+(5−2+f(x))x+25−5f(x)=0satisfies for x∈R then f(5) is equal to

# Ifis a continuous function and ${x}^{3}-$$\left(\sqrt{5}+1\right){x}^{2}+\left(\sqrt{5}-2+f\left(x\right)\right)x+2\sqrt{5}-\sqrt{5}f\left(x\right)=0$satisfies for $x\in \mathbf{R}$ then $f\left(\sqrt{5}\right)$ is equal to

1. A

$2-\sqrt{5}$

2. B

$5+\sqrt{5}$

3. C

$3-\sqrt{5}$

4. D

$\sqrt{5}-3$

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### Solution:

For $x\ne \sqrt{5}$

$\begin{array}{l}f\left(x\right)=-\frac{{x}^{3}-\left(\sqrt{5}+1\right){x}^{2}+\left(\sqrt{5}-2\right)x+2\sqrt{5}}{x-\sqrt{5}}\\ =-\frac{\left(x+1\right)\left(x-2\right)\left(x-\sqrt{5}\right)}{x-\sqrt{5}}\\ =-\left(x+1\right)\left(x-2\right)\\ f\left(\sqrt{5}\right)=\underset{x\to \sqrt{5}}{lim} f\left(x\right)=-\left(1+\sqrt{5}\right)\left(\sqrt{5}-2\right)\\ =-\left(3-\sqrt{5}\right)=\sqrt{5}-3\end{array}$  