If x is a positive integer- then x- x-1 - x-2 - x-1 - x-2 - x-3 - x-2 - x-3 - x-4 -

The correct option is B 2x!(x+1)!(x+2) ! x! amp; ( x+1 ) ! amp; ( x+2 ) ! ( x+1 ) ! amp; ( x+2 ) ! amp; ( x+3 ) ! ( x+2 ) ! ...

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Definition of a Determinant

If x is a pos...

Question

If x is a positive integer, then $∣ ∣ ∣ ∣ ∣ \begin{matrix} x! & (x + 1)! & (x + 2)! (x + 1)! & (x + 2)! & (x + 3)! (x + 2)! & (x + 3)! & (x + 4)! \end{matrix} ∣ ∣ ∣ ∣ ∣$ =

2 x! (x + 1)!

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

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2 x! (x + 3)

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(x + 1)! (x + 2)! (x + 3)!

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Solution

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∣ ∣ ∣ ∣ ∣ \begin{matrix} x^{2} + x & x + 1 & x - 2 2 x^{2} + 3 x - 1 & 3 x & 3 x - 3 x^{2} + 2 x + 3 & 2 x - 1 & 2 x - 1 \end{matrix} ∣ ∣ ∣ ∣ ∣

3 s i n^{- 1} \frac{2 x}{1 + x^{2}} - 4 c o s^{- 1} \frac{1 - x^{2}}{1 + x^{2}} + 2 t a n^{- 1} \frac{2 x}{1 + x^{2}} = \frac{π}{3}

Solve:

(i) $x (x + 1) + (x + 2) (x + 3) = 42$

(ii) $\frac{1}{x + 1} - \frac{2}{x + 2} = \frac{3}{x + 3} - \frac{4}{x + 4}$

(x^{3} - 2 x^{2} + 3 x - 4) (x - 1) - (2 x - 3) (x^{2} - x + 1)

x^{2} (x - y) y^{2} (x + 2 y)

∣ ∣ ∣ ∣ ∣ \begin{matrix} x^{2} + x & x + 1 & x - 2 2 x^{2} + 3 x - 1 & 3 x & 3 x - 3 x^{2} + 2 x + 3 & 2 x - 1 & 2 x - 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ = a x - 12

a

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