If t is a real number and k = t 2 - t + 1 t 2 + t + 1 , then the system of equations 3 x - y + 4 z = 3 x + 2 y - 3 z = - 2 6 x + 5 y + k z = - 3 for any allowable value of k , has

Question

If t is a real number and k = t 2 - t + 1 t 2 + t + 1 , then the system of equations 3 x - y + 4 z = 3 x + 2 y - 3 z = - 2 6 x + 5 y + k z = - 3 for any allowable value of k , has, a unique solution, infinite solutions, no solution, 2 solutions

MARKS App · Accepted Answer

k = t 2 - t + 1 t 2 + t + 1 ⇒ t 2 k - 1 + t k + 1 + k - 1 = 0 as t ∈ R D ≥ 0 ⇒ k + 1 2 - 4 k - 1 2 ≥ 0 ⇒ k ∈ 1 3 , 3 for the given equation, Δ = 3 - 1 4 1 2 - 3 6 5 k = 7 k + 5 > 0 ∀ k ∈ 1 3 , 3 hence, unique solution

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