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Question

The smallest value of k, for which both the roots of the equation x28kx+16(k2k+1) = 0 are real, distinct and have values at least 4,is


A

1

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B

2

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C

0

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D

3

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Solution

The correct option is B

2


α and β 4

x28kx+16(k2k+1)=0

(i) D > 0

64 k264(k2k+1) > 0

k - 1 > 0 k > 1

(ii) 8k2 > 4 k > 1

(iii) f(4) 0

16 - 32k + 16 (k2k+1)0

k23k+20k1 or k 2

k [2,)

least value of k = 2


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