Question : If $x+\frac{1}{x}=17,$ what is the value of $\frac{x^{4}+\frac{1}{x^{2}}}{x^{2}-3x+1}\; ?$
Option 1: $\frac{2431}{7}$
Option 2: $\frac{3375}{7}$
Option 3: $\frac{3375}{14}$
Option 4: $\frac{3985}{9}$
Correct Answer: $\frac{2431}{7}$
Solution :
Given: $x+\frac{1}{x}=17$
$⇒x^{2}+1=17x$
$⇒x^{2}-3x+1=14x$
Also, $x^{3}+\frac{1}{x^{3}}=(x+\frac{1}{x})^{3}-3(x)(\frac{1}{x})(x+\frac{1}{x})$
$⇒x^{3}+\frac{1}{x^{3}}=17^{3}-3(1)(17)=4862$
Now, $ \frac{x^{4}+\frac{1}{x^{2}}}{x^{2}-3x+1}$
= $\frac{x^{4}+\frac{1}{x^{2}}}{14x}$
= $\frac{x^{3}+\frac{1}{x^{3}}}{14}$
= $\frac{4862}{14}$
= $\frac{2431}{7}$
Hence, the correct answer is $\frac{2431}{7}$.
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