If x is a positive integer, then = | , * 20 c x! & (x + 1)! & (x … - Vidyadip

MCQ

If $x$ is a positive integer, then $\Delta = \left| {\,\begin{array}{*{20}{c}}{x!}&{(x + 1)!}&{(x + 2)!}\\{(x + 1)!}&{(x + 2)!}&{(x + 3)!}\\{(x + 2)!}&{(x + 3)!}&{(x + 4)!}\end{array}\,} \right|$ is equal to

A
$2(x!)(x + 1)!$
✓
$2(x!)(x + 1)!(x + 2)!$
C
$2(x!)(x + 3)!$
D
None of these

✓

Answer

Correct option: B.

$2(x!)(x + 1)!(x + 2)!$

b
(b) $\Delta = \left| {\begin{array}{*{20}{c}}{x!}&{(x + 1)!}&{(x + 2)\,!}\\{(x + 1)!}&{(x + 2)!}&{(x + 3)\,!}\\{(x + 2)!}&{(x + 3)!}&{(x + 4)\,!}\end{array}} \right|$

= $x\,!\,(x + 1)\,!\,(x + 2)\,!\,\left| {\,\begin{array}{*{20}{c}}1&{(x + 1)}&{(x + 2)\,(x + 1)}\\1&{(x + 2)}&{(x + 3)\,(x + 2)}\\1&{(x + 3)}&{(x + 4)\,(x + 3)}\end{array}\,} \right|$

Applying ${R_1} \to {R_2} - {R_1},{R_2} \to ({R_3} - {R_2})$ we get

$ = x\,!(x + 1)\,!(x + 2)\,!$ $\,\left| {\,\begin{array}{*{20}{c}}0&1&{2(x + 2)}\\0&1&{2(x + 3)}\\1&{(x + 3)}&{(x + 4)\,(x + 3)}\end{array}\,} \right|$

$ = 2\,x!(x + 1)!(x + 2)!$,(on simplification).

Trick: Put $x = 1$ and then match the alternate.

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