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જો સંખ્યાઓ $2, b, c$ એ સમાંતર શ્રેણીમાં હોય અને $A = \left[ {\begin{array}{*{20}{c}}
1&1&1 \\
2&b&c \\
4&{{b^2}}&{{c^2}}
\end{array}} \right]$ છે જો $det(A) \in [2,16]$ તો $c$ ની કિમંત .. . . અંતરાલ માં આવેલી છે .
$[3,2 + 2^{2/4} ]$
$(2 + 2^{3/4},4)$
$( 2,3)$
$[4, 6]$
Solution
$\left| {\begin{array}{*{20}{c}}
1&1&1\\
2&b&c\\
4&{{b^2}}&{{c^2}}
\end{array}} \right|$
${C_2} \to {C_2} – {C_1},{C_3} \to {C_3} – {C_1}$
$ \Rightarrow \left| {\begin{array}{*{20}{c}}
1&0&0\\
2&{b – 2}&{c – 2}\\
4&{{b^2} – 4}&{{c^2} – 4}
\end{array}} \right|$
$ = \left( {b – 2} \right)\left( {c – 2} \right)\left| {\begin{array}{*{20}{c}}
1&1\\
{b + 2}&{c + 2}
\end{array}} \right|$
$\left| A \right| = \left( {b – 2} \right)\left( {c – 2} \right)\left( {c – b} \right)$
$2,b,c$ are in $AP \Rightarrow 2,2 + d,2 + 2d$
$ \Rightarrow \left| A \right| = \left( d \right)\left( {2d} \right)\left( d \right) = 2{d^3} \in \left[ {2,16} \right]$
$ \Rightarrow {d^3} \in \left[ {1,8} \right]$
$ \Rightarrow 2d \in \left[ {2,4} \right]$
$ \Rightarrow 2 + 2d \in \left[ {4,6} \right]$
