Solution of - x - x -1-2 isA- -1-1B- -2-1C- -3-2D- -4-2

The correct option is B { 2, 1 }|x||x+1|=2 ∵ [|a||b|=|ab|] ⇒ |x(x+1)|=2 ⇒ x^2+x=±2 ⇒ x^2+x 2=0 or x^2+x+2=0 ⇒ (x 1)(x+2)=0 or x = 1 ±√(1 4(1)2)2 ⇒ x=1, 2 or ...

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Standard XII

Mathematics

Absolute Value Function

Solution of |...

Question

Solution of $| x | | x + 1 | = 2$ is

{- 1, 1}

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{- 2, 1}

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{- 3, 2}

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{- 4, - 2}

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Solution

The correct option is B ${- 2, 1}$
$| x | | x + 1 | = 2$
$∵ [| a | | b | = | a b |]$
$\Rightarrow | x (x + 1) | = 2$
$\Rightarrow x^{2} + x = \pm 2$
$\Rightarrow x^{2} + x - 2 = 0$ or $x^{2} + x + 2 = 0$
$\Rightarrow (x - 1) (x + 2) = 0$ or $x = \frac{- 1 \pm \sqrt{1 - 4 (1) 2}}{2}$
$\Rightarrow x = 1, - 2$ or
no real value of $x$ $($ as $D < 0)$
Hence, required solution is $x = {1, - 2}$

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Absolute Value Function

Standard XII Mathematics

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Solution of |x||x+1|=2 is

The correct option is B {−2,1}|x||x+1|=2 ∵[|a||b|=|ab|] ⇒|x(x+1)|=2 ⇒x2+x=±2 ⇒x2+x−2=0 or x2+x+2=0 ⇒(x−1)(x+2)=0 or x=−1±√1−4(1)22 ⇒x=1,−2 or no real value of x (as D<0) Hence, required solution is x={1,−2}

Solution of $| x | | x + 1 | = 2$ is