Values of x for - x 2-2- x -1- x -1-2 is-reA- - 2-5B- -5C- -2 -5D- - 2

The correct options are A (±2+√(5))|x^2 nbsp; 2|x|+1||x|+1= 2 ⇒ (|x| 1)^2=2(|x|+1), x∈ℝ ⇒ (|x| 1)^2=2|x|+2 ⇒ |x|^2 2|x|+1 2|x| 2=0 nbsp; ⇒ |x|^2 4|x| 1=0 ...

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Byju's Answer

Standard XII

Mathematics

Absolute Value Function

Values of x f...

Question

Value(s) of $x$ for $\frac{| x^{2} - 2 | x | + 1 |}{| x | + 1} = 2$ is/are

(\pm 2 + \sqrt{5})

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\pm \sqrt{5}

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\pm (2 \pm \sqrt{5})

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\pm 2

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Solution

The correct option is A $(\pm 2 + \sqrt{5})$
$\frac{| x^{2} - 2 | x | + 1 |}{| x | + 1} = 2$
$\Rightarrow (| x | - 1)^{2} = 2 (| x | + 1), x \in R$
$\Rightarrow (| x | - 1)^{2} = 2 | x | + 2$
$\Rightarrow | x |^{2} - 2 | x | + 1 - 2 | x | - 2 = 0 \Rightarrow | x |^{2} - 4 | x | - 1 = 0 \Rightarrow | x | = 2 \pm \sqrt{5}$
But $| x | = 2 - \sqrt{5} < 0$
$∴ | x | = 2 + \sqrt{5} \Rightarrow x = \pm (2 + \sqrt{5})$

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Similar questions

Q. If

x = \sqrt{\frac{\sqrt{5} + 1}{\sqrt{5} - 1}}

, value of

x^{2} + x - 1

If 2 $^{x}$ x 8 $^{1 / 5}$ = 2 $^{1 / 5}$ , then x is equal to

Q. The distance between the origin and the tangent to the curve

y = e^{2 x} + x^{2}

drawn at the point

x = 0

Find $x$ :
$\frac{2}{5} + x = \frac{3}{5}$

Find x:
$\frac{2}{5} + x = \frac{3}{5}$

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Value(s) of x for |x2−2|x|+1||x|+1=2 is/are

The correct option is A (±2+√5)|x2−2|x|+1||x|+1=2 ⇒(|x|−1)2=2(|x|+1), x∈R ⇒(|x|−1)2=2|x|+2 ⇒|x|2−2|x|+1−2|x|−2=0⇒|x|2−4|x|−1=0⇒|x|=2±√5 But |x|=2−√5<0 ∴|x|=2+√5⇒x=±(2+√5)

Value(s) of $x$ for $\frac{| x^{2} - 2 | x | + 1 |}{| x | + 1} = 2$ is/are