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ISI2016-DCG-7

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Any quadratic function of the form $ax^2+bx+c$ will be $> 0$ if $a > 0 $ and $b^2-4ac < 0$. From the given function $a > 0$.

Now $ b^2-4ac = [-2(4k-1)]^2 - 4.1.(15k^2-2k-7) = k^2-6k+8=(k-2)(k-4)$.

Now according to the condition stated above $(k-2)(k-4) < 0$ and one option i.e 3  (Option C) is satisfying the inequality.
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