If x+1/x=1, λ=x^4000+1x^4000 and μ be the digit at the unit place of the number 2^2n+1, where n∈N and n1, then the value of λ + μ is equal to

Question

If x+1/x=1, λ=x^4000+1x^4000 and μ be the digit at the unit place of the number 2^2n+1, where n∈N and n>1, then the value of λ + μ is equal to

MARKS Content Team · Accepted Answer

Correct answer: 6. ∵ x+1/x=1⇒x^2-x+1=0 ⇒ x + ω x + ω^2 = 0 ∴ x = - ω ,⁡ - ω^2 , where ω and ω^2 are complex cube roots of unity if x=-ω, then λ = - ω^(4 0 0 0) + 1/(- ω^(4 0 0 0)) = ω^(4 0 0 0) + 1/(ω^(4 0 0 0)) ⇒ λ = ω + 1/(ω) = (ω^2 + 1)/(ω) = (- ω)/(ω) = - 1 ∵ ω^3=1 and 1+ω+1ω^2=0 and also for x = - ω^2 , ⁡ λ = - 1 Now, for n>1, ⁡2^n=4k, ⁡k∈N⁡ ∴ 2^2n=2^4k=16^k=a number with last digit 6 ∴ μ = the digit at the unit place of the number 2^2n+1 i.e. μ=6+1=7 Hence, λ+μ=-1+7=6

If $x + \frac{1}{x} = 1$ , $λ = x^{4000} + \frac{1}{x^{4000}}$ and $μ$ be the digit at the unit place of the number $2^{2^{n}} + 1,$ where $n \in N$ and $n > 1,$ then the value of $λ + μ$ is equal to

Solution

More Complex Number Questions

If x+1x=1, λ=x4000+1x4000 and μ be the digit at the unit place of the number 22n+1, where n∈N and n>1, then the value of λ + μ is equal to

Solution

More Complex Number Questions

If $x + \frac{1}{x} = 1$ , $λ = x^{4000} + \frac{1}{x^{4000}}$ and $μ$ be the digit at the unit place of the number $2^{2^{n}} + 1,$ where $n \in N$ and $n > 1,$ then the value of $λ + μ$ is equal to