If t - 1 and -t -1, t ∈ R, are the roots of (a + 2) $x^{2}$ + 2ax - 1 = 0 then complete set of values of 'a' is :

( 0, ∞)

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( - ∞, 0)

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( -∞, ∞)

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∅

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Solution

The correct option is D
∅

$\frac{- 2 a}{a + 2}$ = -2 => a+2 = a

Therefore set of values of a is null , i.e ∅

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If t - 1 and -t -1, t ∈ R, are the roots of (a + 2) $x^{2}$ + 2ax - 1 = 0 then complete set of values of 'a' is :

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If t - 1 and -t -1, t ∈ R, are the roots of (a + 2) x2 + 2ax - 1 = 0 then complete set of values of 'a' is :

The correct option is D ∅ −2aa+2 = -2 => a+2 = a Therefore set of values of a is null , i.e ∅

If t - 1 and -t -1, t ∈ R, are the roots of (a + 2) $x^{2}$ + 2ax - 1 = 0 then complete set of values of 'a' is :

The correct option is D
∅

$\frac{- 2 a}{a + 2}$ = -2 => a+2 = a

Therefore set of values of a is null , i.e ∅