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$\left| {\,\begin{array}{*{20}{c}}{x + 1}&{x + 2}&{x + 4}\\{x + 3}&{x + 5}&{x + 8}\\{x + 7}&{x + 10}&{x + 14}\end{array}\,} \right| = $
$2$
$-2$
${x^2} - 2$
એકપણ નહી.
Solution
(b) $\Delta = \left| {\,\begin{array}{*{20}{c}}{ – 1}&{ – 2}&{x + 4}\\{ – 2}&{ – 3}&{x + 8}\\{ – 3}&{ – 4}&{x + 14}\end{array}\,} \right|,$ by $\begin{array}{l}{C_1} \to {C_1} – {C_2}\\{C_2} \to {C_2} – {C_3}\end{array}$
= $\left| {\,\begin{array}{*{20}{c}}
{ – 1}&{ – 1}&x\\
{ – 2}&{ – 1}&x\\
{ – 3}&{ – 1}&{x + 2}
\end{array}\,} \right|,$ , by $\begin{array}{l}{C_2} \to {C_2} – {C_1}\\{C_3} \to {C_3} + 4{C_1}\end{array}$
$ = – ( – x – 2 + x) + 1\,.\,( – 2x – 4 + 3x) + x(2 – 3)$
= $2 + x – 4 – x = – 2$.
Trick : Put $x=1$. Then $\left| {\,\begin{array}{*{20}{c}}2&3&5\\4&6&9\\8&{11}&{15}\end{array}\,} \right| = – 2$
Note : Since there is a option “None of these”, therefore we should check for one more different value of $x$. Put $x = – 1$.
$\left| {\,\begin{array}{*{20}{c}}0&1&3\\2&4&7\\6&9&{13}\end{array}\,} \right| = – 1(26 – 42) + 3(18 – 24) = – 2$
Therefore answer is $ (b).$
