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Determinant

The determina...

Question

The determinant $∣ ∣ ∣ ∣ ∣ \begin{matrix} s i n (x - α) & c o s (x - α) & a + x s i n α s i n (x - β) & c o s (x - β) & b + x s i n β s i n (x - γ) & c o s (x - γ) & c + x s i n γ \end{matrix} ∣ ∣ ∣ ∣ ∣$ is independent of

A

a

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B

x

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C

α

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D

β

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Solution

The correct option is A $x$
$\frac{d}{d x} | D | = 0$ so independent of $x$

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Similar questions

\frac{s i n (x - α)}{s i n (x - β)} = \frac{a}{b}, \frac{c o s (x - α)}{c o s (x - β)} = \frac{A}{B}

\neq

0, then prove that -

c o s (α - β) = \frac{a A + b B}{a B + b A} .

β

α

are +ive angles less than two right angles and

c o s α + c o s (α + β) + c o s (α + β + γ) = 0

s i n α + s i n (α + β) + s i n (α + β + γ) = 0

β = γ = 2 π / 3

α, β

are the solutions of

sin x = - \frac{1}{2}

[0, 2 π]

α, γ

are the solutions of

cos x = - \frac{\sqrt{3}}{2}

[0, 2 π],

\int [1 + t a n x t a n (x + α)] d x

α

β

, so that the function

f (x)

f (x) = - 2 sin x,

- π \leq x \leq - \frac{π}{2}

= α sin x + β,

- \frac{π}{2} < x < \frac{π}{2}

= cos x,

\frac{π}{2} \leq x \leq π

is continuous on

[- π, π]

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