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3 and 4 .Determinants and Matrices
hard
$xyz$ ના ગુણાકારની ન્યૂનતમ કિમત મેળવો કે જેથી $\left| {\begin{array}{*{20}{c}}
x&1&1 \\
1&y&1 \\
1&1&z
\end{array}} \right|$ ની કિમંત અનૃણ મળે.
A
$-2\sqrt 2$
B
$-1$
C
$-16\sqrt 2$
D
$-8$
(JEE MAIN-2015)
Solution
$\begin{array}{*{20}{c}}
x&1&1\\
1&y&1\\
1&1&z
\end{array} \ge 0$
$xyz – x – y – z + 2 \ge 0$
$xyz + 2 \ge x + y + z \ge 3{\left( {xyz} \right)^{1/3}}$
$xyz + 2 – 3{\left( {xyz} \right)^{1/3}} \ge 0$
at $\left( {xyz} \right) = {t^3}$
${t^3} – 3t + 2 \ge 0$
$\left( {t + 2} \right){\left( {t – 1} \right)^2} \ge 0$
$\left[ {t = – 2} \right]{t^3} = – 8$
Standard 12
Mathematics
