Question : If $(x+\frac{1}{x})=–2$, then the value of $(x^7+\frac{1}{x^7})$ is:
Option 1: 1
Option 2: –1
Option 3: 0
Option 4: –2
Correct Answer: –2
Solution :
Given: $(x+\frac{1}{x})=–2$
⇒ $(x+\frac{1}{x})^2=(–2)^2$
⇒ $x^2+\frac{1}{x^2}+2=4$
⇒ $x^2+\frac{1}{x^2}=2$
So, $(x^4+\frac{1}{x^4})=(x^2+\frac{1}{x^2})^2-2=2^2-2$
⇒ $(x^4+\frac{1}{x^4})=4-2$
⇒ $(x^4+\frac{1}{x^4})=2$
Also, $(x^3+\frac{1}{x^3})=(x+\frac{1}{x})^3-3(x+\frac{1}{x})=(–2)^3–3×–2$
⇒ $(x^3+\frac{1}{x^3})=–8+6$
⇒ $(x^3+\frac{1}{x^3})=–2$
So, $(x^7+\frac{1}{x^7})= [x^3+\frac{1}{x^3}][x^4+\frac{1}{x^4}]–[x+\frac{1}{x}]=-2×2-(-2)=-2$
Hence, the correct answer is –2.
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