The smallest value of k, for which ...
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The smallest value of k, for which both the roots of the equation  

x^2 - 8kx + 16(k^2 – k + 1) = 0  

are real and have values at least 4, is  

(A) 1  

(B) 2  

(C) 0  

(D) 4

JEE/Engineering Exams
Maths
Solution
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The smallest value of k, for which both the roots of the equation    x^2 - 8kx + 16(k^2 – k + 1) = 0    are real and have values at least 4, is
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The given equation is  

x^2 - 8kx + 16(k2 - k + 1) = 0  

∵ Both the roots are real and distinct  

∴ D > 0 ⇒ (8k)^2 - 4 x 16 (k^2 - k + 1) > 0

k > 1............   (i)  

∵ Both the roots are greater than or equal to 4  

∴a +β >8 and f (4) ≥ 0  

⇒k >..............................   (ii)  

And 16 - 32k + 16(k2 - k + 1) ≥ 0  

⇒ k^2 - 3k + 2 ≥ 0 ⇒ (k - 1) (k - 2) ≥  0  

k ∈ (-∞ , 1)] ⋃ [ 2,∞ )...................    

(iii)  

Combining (i), (ii) and (iii), we get k ≥ 2 or the smallest value of k = 2. 

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