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.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

JEE STD 11 Science question papers→Gujarat Board STD 11 Science question papers→Gujarat Board question paper generator→

Similar questions

If the orthocentre of the triangle formed by the lines $\mathrm{y}=\mathrm{x}+1, \mathrm{y}=4 \mathrm{x}-8$ and $\mathrm{y}=\mathrm{mx}+\mathrm{c}$ is at $(3,-1)$, then $\mathrm{m}-\mathrm{c}$ is :

Let three vectors $\overrightarrow{ a }=\alpha \hat{ i }+4 \hat{ j }+2 \hat{ k }$, $\overrightarrow{ b }=5 \hat{ i }+3 \hat{ j }+4 \hat{ k }, \overrightarrow{ c }=x \hat{ i }+y \hat{ j }+z \hat{ k }$ from a triangle such that $\overrightarrow{ c }=\overrightarrow{ a }-\overrightarrow{ b }$ and the area of the triangle is $5 \sqrt{6}$. if $\alpha$ is a positive real number, then $|\overrightarrow{ c }|^2$ is $:$

If a variable takes the discrete values $\alpha - 4,\,\alpha - \frac{7}{2},\,\alpha - \frac{5}{2},\,\alpha - 3,\,\alpha - 2,\,\alpha + \frac{1}{2},\,\alpha - \frac{1}{2},\,\alpha + 5\,(\alpha > 0)$, then the median is

If $x$ takes non-positive permissible value, then ${\sin ^{ - 1}}x =$

Let $y$ be the function which passes through $(1, 2)$ having slope $(2x + 1)$. The area bounded between the curve and $x -$ axis is

If ${G_1}$ and ${G_2}$ are two geometric means and $A$ the arithmetic mean inserted between two numbers, then the value of $\frac{{G_1^2}}{{{G_2}}} + \frac{{G_2^2}}{{{G_1}}}$is

If $\int {\,\,\frac{{{x^4} + 1}}{{x{{\left( {{x^2} + 1} \right)}^2}}}} $ $dx$ =$ A $ $ln$ $ |x| $ $+ $$\frac{B}{{1 + {x^2}}}$$ +$ $ c$ , where $c$ is the constant of integration then :

Let $f (x)$ be integrable over $(a, b) , b > a > 0$.

If $I_1 = \int\limits_{\frac{\pi }{6}}^{\frac{\pi }{3}} \, f (\tan\, \theta + \cot\, \theta )\cdot sec^2\, \theta\, d\, \theta$ &

$I_2 = \int\limits_{\frac{\pi }{6}}^{\frac{\pi }{3}} \, f (\tan\, \theta + \cot\, \theta )\cdot cosec^2\, \theta\, d \, \theta$ ,

then the ratio $\frac{{{I_1}}}{{{I_2}}}$ :

Two parabolas have the same focus. If their directrices are the $x -$ axis $\&$ the $ y -$ axis respectively, then the slope of their common chord is :

Let $A = \{1, 2, 3\}$. The total number of distinct relations that can be defined over $A$ is