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

A
$2 - \sqrt{5}$
B
$5 + \sqrt{5}$
C
$3 - \sqrt{5}$
D
$\sqrt{5} - 3$

Solution:

For $x \neq \sqrt{5}$

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

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

Solution:

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