Question

Assertion :If $x+\frac{1}{x}=1,A=x^{2000}+\frac{1}{x^{2000}}$ & B be the unit place digit in the number $5^5+1,nϵN$ and n > I, then value of A + B = 5. Reason: If $\omega ,{\omega }^2$ are roots of $x^2+x+1=0$, then $x^2+\frac{1}{x^2}=-1$ and $x^3+\frac{1}{x^3}=2$.

• A. Both (A) & (R) are individually true & (R) is correct explanation of (A).
• B. Both (A) & (R) are individually true but (R) is not the correct (proper) explanation of (A).
• C. (A) is true (R) is false.
• D. (A) is false (R) is true.

Answer 1

The correct option is B Both (A) & (R) are individually true but (R) is not the correct (proper) explanation of (A).

As $x+\frac{1}{x}=1$
$\Rightarrow x^2=x+1=0$
$\Rightarrow x+(\omega +{\omega }^2)x+1=0$
$\Rightarrow (x+\omega )(x+{\omega }^2)=0$
$\Rightarrow x=-\omega ,-{\omega }^2$
for $x=-\omega ,A=x^{2000}+\frac{1}{x^{2000}}={\omega }^2+\frac{1}{{\omega }^2}={\omega }^2+\omega =-1$
for $n>1,5^n=5k$
$\therefore 5^{5n}=5^{5k}=(5^5{)}^k=(3125{)}^k$
$\therefore$ Unit place in $B=5^{5^k}+1$ is $5+1=6$
$\therefore A+B=-1+6=5$
$\therefore$ Assertion (A) is true.
Again $x+\frac{1}{x}=-1$
$\Rightarrow x^2+x+1=0.....(\ast )$
$\Rightarrow x=\omega ,{\omega }^2$ are roots of (*)
for $x=\omega ,x^2+\frac{1}{x^2}={\omega }^2+\frac{1}{{\omega }^2}={\omega }^2+\omega =-1$
and for $x=\omega ,x^3+\frac{1}{x^3}=2$
Similarly for $x={\omega }^3,x^2+\frac{1}{x^2}={\omega }^4+\frac{1}{{\omega }^4}=\omega +{\omega }^2=-1$
and $x={\omega }^2,x^3+\frac{1}{x^3}={\omega }^6+{\omega }^6=1+1=2$
$\therefore$ option b is correct