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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, distinct and have values at least 4, is |
Answer» Correct Answer - `k=2` (i) Given, `x^(2)-8kx+1(k^(2)-k+1)=0` `Now, D=64{ k^(2)-k+1}=64 (k-1)gt0 k gt 1` `(ii)-(b)/(2a)gt4implies(8k)/(2)gt4impliesk gt1` `(iii) f(4)ge0` `implies16-32k+16(k^(2)-k+1)ge0` `impliesk^(2)-3k+2ge0` `implies(k-2)(k-1)ge0` `impliesk le1 ork ge2` Hence, `k=3` |
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