JEE Main 2015 · 10 Apr · Shift 1 · Mathematics Q64

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The least value of the product $xyz$ for which the determinant $\begin{vmatrix}x&1&1\\1&y&1\\1&1&z\end{vmatrix}$ is non-negative, is :

-2\sqrt2

-16\sqrt2

−8

−1

Answer and worked solution

Official answer: C

$\det=xyz-(x+y+z)+2\ge0$ , so $xyz+2\ge x+y+z$ . For the least (negative) value of $xyz$ take $x=y=z=t<0$ : $t^3-3t+2=(t-1)^2(t+2)\ge0$ needs $t\ge-2$ .

So $xyz\ge-8$ , and the least value is $-8$ .

Topics:Determinants and properties AM-GM-HM inequalities

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JEE Main 2015 · 10 Apr · Shift 1, Mathematics question 64