JEE Advance - Mathematics (2009 - Paper 2 Offline - No. 18)
The smallest value of $$k$$, for which both the roots of the equation
$$${x^2} - 8kx + 16\left( {{k^2} - k + 1} \right) = 0$$$
are real, distinct and have values at least 4, is
Answer
2
Explanation
We have $${x^2} - 8kx + 16({k^2} - k + 1) = 0$$
$$D > 0 \Rightarrow k > 1$$ ..... (1)
$${{ - b} \over {2a}} > 4 \Rightarrow {{8k} \over 2} > 4$$
$$ \Rightarrow k > 1$$ ..... (2)
Now, $$f(4) \ge 0 \Rightarrow 16 - 32k + 16({k^2} - k + 1) \ge 0$$
$${k^2} - 3k + 2 \ge 0$$
$$k \le 1 \cup k \ge 2$$ ..... (3)
Using Eqs. (1), (2) and (3), we get $${k_{\min }} = 2$$.
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