The smallest value of k , for which both the roots of the equation x 2 − 8 k x + 16 ( k 2 − k + 1 ) = 0 are real, distinct and have values atleast 4 , is

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(i) Given, x 2 − 8 k x + 16 ( k 2 − k + 1 ) = 0 Now, D = 64 { k 2 − ( k 2 − k + 1 ) } ​ = 64 ( k − 1 ) > 0 ∴ k > 1 16 − 32 k + 16 ( k 2 − k + 1 ) ≥ 0 ⇒ k 2 − 3 k + 2 ≥ 0 ⇒ ( k − 2 ) ( k − 1 ) ≥ 0 ⇒ k ≤ 1 or k ≥ 2 ​ Hence, k = 2

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