CAT 2023

Slot 3

QA

Question & Solution

It is given that $x^8 + \left(\frac{1}{x}\right)^8 = 47$, which can be written as:

=> $\left(x^4\right)^{^2}+\left(\ \frac{\ 1}{x^4}\right)^{^2}=47$

=> $\left(x^4+\frac{\ 1}{x^4}\right)^{^2}-2\cdot x^4\cdot\frac{1}{x^4}=47$

=> $\left(x^4+\frac{\ 1}{x^4}\right)^{^2}=49$

=> $x^4+\frac{\ 1}{x^4}=7$

Similarly, $x^4+\frac{\ 1}{x^4}=7$ can be expressed as:

=> $\left(x^2\right)^{^2}+\left(\frac{\ 1}{x^2}\right)^{^2}=7$

=> $\left(x^2+\frac{\ 1}{x^2}\right)^{^2}-2\cdot x^2\cdot\frac{1}{x^2}=7$

=> $\left(x^2+\frac{\ 1}{x^2}\right)^{^2}=9$

=> $x^2+\frac{\ 1}{x^2}=3$

By the same logic, we get $x+\frac{1}{x}=\sqrt{\ 5}$

Now, $x^3+\frac{1}{x^3}=\left(x+\frac{1}{x}\right)^{^3}-3\cdot x\cdot\frac{1}{x}\left(x+\frac{1}{x}\right)$

=> $x^3+\frac{1}{x^3}=\left(\sqrt{\ 5}\right)^{^3}-3\sqrt{\ 5}=2\sqrt{\ 5}$

By the same logic, we can say that 

=> $x^9+\frac{1}{x^9}=\left(x^3+\frac{1}{x^3}\right)^{^3}-3\cdot x^3\cdot\frac{1}{x^3}\left(x^3+\frac{1}{x^3}\right)$

=> $x^9+\frac{1}{x^9}=\left(2\sqrt{\ 5}\right)^{^3}-3\left(2\sqrt{\ 5}\right)$

=> $x^9+\frac{1}{x^9}=40\sqrt{\ 5}-6\sqrt{\ 5}=34\sqrt{\ 5}$

The correct option is D