Question

Assertion :If $x+\frac{1}{x}=1$ and $p=x^{4000}+\frac{1}{x^{4000}}$ and $q$ is the digit at unit place in the number $2^{2^n}+1,n\in N$ and $n>1$, then the value of $p+q=8$. Reason: If $\omega ,{\omega }^2$ are the roots of $x+\frac{1}{x}=-1$, then $x^2+\frac{1}{x^2}=-1,x^3+\frac{1}{x^3}=2.$

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Assertion is incorrect and Reason is correct

Answer 1

The correct option is D Assertion is incorrect and Reason is correct

Let $x=\cos \theta +i\sin \theta$
$\frac{1}{x}=\overline{x}=\cos \theta -i\sin \theta$
Hence, $x+\frac{1}{x}=2\cos \theta =1$
$\cos \theta =\frac{1}{2}$
$\theta =\frac{\pi }{3}$
So,$x^{4000}+{\frac{1}{x}}^{4000}$
$=2\cos \left(4000\frac{\pi }{3}\right)$
$=-1$
Now consider $n=2$ in $2^{2^n}+1$
$=2^4+1$
$=17$
Hence, $q=7$
$p+q=-1+7$
$=6$
$\ne 8$
Hence assertion is wrong.