Question

Statement 1: If $x+\left(1/x\right)=1$ and $p=x^{4000}+\left(1/x^{4000}\right)$ and q is the digit at unit place in the number $2^{2n}+1,nϵN$ and n > 1, then the value of $p+q=8$.
Statement 2: If $\omega ,{\omega }^2$ are the roots of $x+1/x=-1$, the $x^2+1/x^2=-1,x^3+\left(1/x^3\right)=2$.

• A. Both the statements are true, and Statement 2 is the correct explanation for Statement 1.
• B. Both the statements are true, but Statement 2 is not the correct explanation for Statement 1.
• C. Statement 1 is true and Statement 2 is false.
• D. Statement 1 is false and Statement 2 is true.

Answer 1

The correct option is D Statement 1 is false and Statement 2 is true.

$x+\frac{1}{x}=1$
$\Rightarrow x^2-x+1=0$
$\therefore x=-\omega ,-{\omega }^2$
Now for $x=-\omega$,
$p={\omega }^{4000}+\frac{1}{{\omega }^{4000}}=\omega +\frac{1}{\omega }=-1$
Similarly for $x=-{\omega }^2,p=-1$. For n > 1,
$2^n=4k$
$\therefore 2^{2^n}=2^{4k}=(16{)}^k=$ a number with last digit 6
$\Rightarrow q=6+1=7$
Hence, $p+q=-1+7=6$.