$∣ ∣ ∣ ∣ \begin{matrix} x + 2 & x + 3 & x + 5 x + 4 & x + 6 & x + 9 x + 8 & x + 11 & x + 15 \end{matrix} ∣ ∣ ∣ ∣$ =

3 x^{2} + 4 x + 5

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x^{3} + 8 x + 5

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Solution

The correct option is B $- 2$
$∣ ∣ ∣ ∣ \begin{matrix} x + 2 & x + 3 & x + 5 x + 4 & x + 6 & x + 9 x + 8 & x + 11 & x + 15 \end{matrix} ∣ ∣ ∣ ∣$ $= ∣ ∣ ∣ ∣ \begin{matrix} x + 2 & x + 3 & x + 5 2 & 3 & 4 4 & 5 & 6 \end{matrix} ∣ ∣ ∣ ∣, R_{3} \to R_{3} - R_{2}, R_{2} \to R_{2} - R_{1}$
$= (x + 2) (18 - 20) - (x + 3) (12 - 16) + (x + 5) (10 - 12)$ , Expand along first row
$= - 2 x - 4 + 4 x + 12 - 2 x - 10 = - 2$

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Similar questions

Δ = ∣ ∣ ∣ ∣ \begin{matrix} x + 2 & x + 3 & x + 5 x + 4 & x + 6 & x + 9 x + 8 & x + 11 & x + 15 \end{matrix} ∣ ∣ ∣ ∣

\frac{cos 3 x}{sin 2 x sin 4 x} + \frac{cos 5 x}{sin 4 x sin 6 x} + \frac{cos 7 x}{sin 6 x sin 8 x} + \frac{cos 9 x}{sin 8 x sin 10 x} = \frac{1}{2} (c o s e c x) [c o s e c 2 x - c o s e c a x]

(6 x^{5} - 4 x^{4} + 8 x^{3} + 2 x^{2}) \div 2 x^{2}

\frac{(cos x - cos 3 x) (sin 8 x + sin 2 x)}{(sin 5 x - sin x) (cos 4 x - cos 6 x)}

(6 x^{5} - 4 x^{4} + 8 x^{3} + 2 x^{3}) \div 2 x^{2}

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