Prove that value of - for which 2x2 - 22 - - 1 x - - - - 1 - 0 may have one root less than - and other root greater than - are given by - - 0 or - - -1-

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Standard X

Mathematics

Area of a Triangle Given Its Vertices

Prove that va...

Question

Prove that value of $λ$ for which $2 x^{2}$ - 2(2 $λ$ + 1) x + $λ$ ( $λ$ + 1) = 0 may have one root less than $λ$ and other root greater than $λ$ are given by $λ$ > 0 or $λ$ < -1.

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Solution

If f (x) = ( - $α$ ) (x - $β$ ),
then f ( $λ$ ) = ( $λ$ - $α$ ) ( $λ$ - $β$ ) = ive as the two factors are of opposite signs by given conditions.
Also roots must be real
$∴ Δ \geq 0 \Rightarrow 8 (λ^{2} + λ + \frac{1}{2})$
$= 8 [{(λ + \frac{1}{2})}^{2} + \frac{1}{4}]$ is +ive, which is true.
Now f ( $λ$ ) = -ive
$\Rightarrow 2 λ^{2} - 2 (2 λ + 1) λ + λ (λ + 1) < 0$
or $- λ^{2} - λ < 0 o r λ (λ + 1) > 0$
or $λ < - 1 o r λ > 0$

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Prove that value of λ for which 2x2 - 2(2λ + 1) x + λ (λ + 1) = 0 may have one root less than λ and other root greater than λ are given by λ > 0 or λ < -1.

Prove that value of $λ$ for which $2 x^{2}$ - 2(2 $λ$ + 1) x + $λ$ ( $λ$ + 1) = 0 may have one root less than $λ$ and other root greater than $λ$ are given by $λ$ > 0 or $λ$ < -1.