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Let r be a real number and `n in N` be such that the polynomial `2x^(2) + 2x + 1` divides the polynomial `(x + 1)^(n) - r`. Then `(n,r)` can be-(A) `(4000, 4^(1000))` (B) `(4000, 1/4^(1000))` (C) `(4^(1000),1/4^1000)` (D) `(4000,1/4000)`A. `(4000, 4^(1000))`B. `(4000, (1)/(4^(1000)))`C. `(4^(1000),(1)/(4^(1000)))`D. `(4000, (1)/(4000))` |
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Answer» Correct Answer - B `2x^(2) + 2x + 1 = 0` `x = (-1+i)/(2), (-1-i)/(2)` x satisfies `(x + 1)^(n) - r = 0` `((-1+- i)/(2) + 1)^(n) - r = 0` `((1+-i)/(2))^(n) -r = 0` `((1)/(sqrt(2)))^(n) ((1+i)/(sqrt(2)))^(n) = r` `((1)/(sqrt(2)))^(n) (e^(+-( ipi)/(4)))^(n)= r` RHS = real LHS = real only when n = multiply of 4 `n = 4000` `r = ((1)/(sqrt(2)))^(4000) = 1/(4^(1000))` |
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