Question : If $x+ \frac{1}{x} =2$, then the value of $({x}^{99}+ \frac{1}{x^{99}} –2)$ is:
Option 1: –2
Option 2: 0
Option 3: 2
Option 4: 4
Correct Answer: 0
Solution :
Given:
$⇒x+ \frac{1}{x} =2$
By squaring both sides,
$({x}+ \frac{1}{x})^{2} =({2})^{2}$
$⇒{x}^{2}+ \frac{1}{x^{2}}+2×x×\frac{1}{x} =4$
$⇒{x}^{2}+ \frac{1}{x^{2}}=2$
Similarly, we can find,
$({x}+ \frac{1}{x})^{3} =({2})^{3}$
$⇒{x}^{3}+ \frac{1}{x^{3}}+3×x×\frac{1}{x}(x+ \frac{1}{x}) =8$
$⇒{x}^{3}+ \frac{1}{x^{3}}+3×2 =8$
$⇒{x}^{3}+ \frac{1}{x^{3}}=2$
Similarly, we can find the value of,
$⇒{x}^{99}+ \frac{1}{x^{99}}=2$
$\therefore {x}^{99}+ \frac{1}{x^{99}} -2=0$
Hence, the correct answer is 0.
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