Question : If $x=2+\sqrt3$, then the value of $\frac{x^{2}-x+1}{x^{2}+x+1}$ is:
Option 1: $\frac{2}{3}$
Option 2: $\frac{3}{4}$
Option 3: $\frac{4}{5}$
Option 4: $\frac{3}{5}$
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Correct Answer: $\frac{3}{5}$
Solution :
$x=2+\sqrt{3}$
$\frac{1}{x}=2–\sqrt{3}$ (by rationalisation)
So, $x+\frac{1}{x}=2+\sqrt{3}+2–\sqrt{3}$
⇒ $x+\frac{1}{x}=2+2$
$\therefore x+\frac{1}{x}=4$
Now, $\frac{x^2–x+1}{x^2+x+1}$
Dividing both the numerator and the denominator by $x$, we get,
$=\frac{x–1+\frac{1}{x}}{x+1+\frac{1}{x}}$
$=\frac{(x+\frac{1}{x})–1}{(x+\frac{1}{x})+1}$
$=\frac{4–1}{4+1}$
$=\frac{3}{5}$
Hence, the correct answer is $\frac{3}{5}$.
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