If $a, b, c$ are distinct numbers in arithmetic progression and roots of the quadratic equation $(a + 2 b - 3 c) x^{2} + (b + 2 c - 3 a) x + (c + 2 a - 3 b) = 0$ are $α$ and $β$ then both $α$ and $β$ belong to domain of

s i n^{- 1} x

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s e c^{- 1} (s e c x)

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s e c (s e c^{- 1} x)

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t a n^{- 1} x

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Solution

The correct options are
A $s i n^{- 1} x$
B $s e c^{- 1} (s e c x)$
D $t a n^{- 1} x$
Let $A = a + 2 b - 3 c, B = b + 2 c - 3 a, C = c + 2 a - 3 b$
Since a,b,c are in AP
$b = a + d$ (d is common difference)
$c = a + 2 d$
$α β = \frac{c + 2 a - 3 b}{a + 2 b - 3 c} = \frac{a + 2 d + 2 a - 3 a - 3 d}{a + 2 a + 2 d - 3 a - 6 d} = \frac{1}{4}$
$α + β = \frac{- (b + 2 c - 3 a)}{a + 2 b - 3 c} = \frac{- (a + d + 2 a + 4 d - 3 a)}{a + 2 a + 2 d - 3 a - 6 d} = \frac{5}{4}$
$α = \frac{1}{4}, β = 1$
$∴ α, β = 1, \frac{1}{4}$
Domain of ${sin}^{- 1} x$ is $[- 1, 1]$
Domain of ${sec}^{- 1} (s e c x)$ is $(0, π) - \frac{π}{2}$
Domain of $sec (s e c^{- 1} x)$ is $R - (- 1, 1)$
Domain of ${tan}^{- 1} x$ is $(- \infty, \infty)$
These values of $α$ , $β$ belong to the domain of A, B and D.

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