Solve - x -1-1- x -2A- -1-1-B- -1- -1- -1- -D- -1- -

The correct option is D [1,∞)|x + 1| − |1 − x| = 2 If x lt;−1, −x − 1 − 1 + x = 2 ⇒−2 = 2 (Not possible) If −1≤ x lt; 1, x + 1 1 + x = 2 ⇒ x = 1 If x ...

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Standard XI

Mathematics

Properties of Inequalities

Solve | x +1|...

Question

Solve $| x + 1 | - | 1 - x | = 2$

[- 1, 1]

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(- \infty, - 1]

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(- \infty, - 1] \cup [1, \infty)

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[1, \infty)

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Solution

The correct option is D $[1, \infty)$
$| x + 1 | - | 1 - x | = 2$
If $x < - 1$ ,
$- x - 1 - 1 + x = 2$
$\Rightarrow - 2 = 2$ (Not possible)

If $- 1 \leq x < 1$ ,
$x + 1 - 1 + x = 2$
$\Rightarrow x = 1$

If $x \geq 1$ ,
$x + 1 + 1 - x = 2$
$\Rightarrow 2 = 2$ Always true
Hence, $x \in [1, \infty)$

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Q. Solve for x.

\frac{1}{2 a + b + 2 x} = \frac{1}{2 a} + \frac{1}{b} + \frac{1}{2 x}

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Solve |x+1|−|1−x|=2

The correct option is D [1,∞)|x+1|−|1−x|=2 If x<−1, −x−1−1+x=2 ⇒−2=2 (Not possible) If −1≤x<1, x+1−1+x=2 ⇒x=1 If x≥1, x+1+1−x=2 ⇒2=2 Always true Hence, x∈[1,∞)

Solve $| x + 1 | - | 1 - x | = 2$

The correct option is D $[1, \infty)$
$| x + 1 | - | 1 - x | = 2$
If $x < - 1$ ,
$- x - 1 - 1 + x = 2$
$\Rightarrow - 2 = 2$ (Not possible)

If $- 1 \leq x < 1$ ,
$x + 1 - 1 + x = 2$
$\Rightarrow x = 1$

If $x \geq 1$ ,
$x + 1 + 1 - x = 2$
$\Rightarrow 2 = 2$ Always true
Hence, $x \in [1, \infty)$