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જો $d \in R$, અને $A = \left[ {\begin{array}{*{20}{c}} { - 2}&{4 + d}&{\left( {\sin \,\theta } \right) - 2}\\ 1&{\left( {\sin \,\theta } \right) + 2}&d\\ 5&{\left( {2\sin \,\theta } \right) - d}&{\left( { - \sin \,\theta } \right) + 2 + 2d} \end{array}} \right]$, $\theta \in \left[ {0,2\pi } \right]$. જો $det (A)$ ની ન્યૂનતમ કિમંત $8$, હોય તો $d$ મેળવો.
$-5$
$-7$
$2\left( {\sqrt 2 + 1} \right)$
$2\left( {\sqrt 2 + 2} \right)$
Solution
$\left| A \right| = \left| {\begin{array}{*{20}{c}}
{ – 2}&{4 + d}&{\left( {\sin \theta – 2} \right)}\\
1&{\left( {\sin \theta } \right) + 2}&d\\
5&{\left( {2\sin \theta } \right) – d}&{\left( { – \sin \theta } \right) + 2 + 2d}
\end{array}} \right|$
$ = \left| {\begin{array}{*{20}{c}}
{ – 2}&{4 + d}&{\left( {\sin \theta – 2} \right)}\\
1&{\left( {\sin \theta } \right)}&d\\
1&0&0
\end{array}} \right|$ (New ${R_3} = {R_3} – 2{R_2} + {R_1}$)
$ = \left( {4 + d} \right)d – {\sin ^2}\theta + 4 = {\left( {d + 2} \right)^2} – {\sin ^2}\theta $
Because minimum value of $\left| A \right| = 8 \Rightarrow {\left( {d + 2} \right)^2} = 9 \Rightarrow d = 1$ or $-5$
