The set of real values of x satisfying ||x−1|−1|≤1, is

Since x -1 changes its sign as x passes through 1.So, following cases arise.CASE I When x< 1In this case, we have |x−1|=−(x−1)∴ ||x−1|−1|≤1⇒ |−(x−1)−1|≤1⇒ |−x|≤1⇒ |x|≤1⇒ −1≤x≤1 [∵|x|≤a⇔−a≤x≤a] ⇒ −1≤x<1 [∵x<1]∴ x∈[−1, 1)CASE II When x ≥1In this case, we have |x−1|=x−1∴||x−1|−1|≤1⇒|x−1−1|≤1⇒|x−2|≤1⇒2−1≤x≤2+1⇒x∈[1, 3]Hence, x∈[−1, 3]

The set of real values of x satisfying ||x−1|−1|≤1, is

Since x -1 changes its sign as x passes through 1.So, following cases arise.CASE I When x< 1In this case, we have |x−1|=−(x−1)∴ ||x−1|−1|≤1⇒ |−(x−1)−1|≤1⇒ |−x|≤1⇒ |x|≤1⇒ −1≤x≤1 [∵|x|≤a⇔−a≤x≤a] ⇒ −1≤x<1 [∵x<1]∴ x∈[−1, 1)CASE II When x ≥1In this case, we have |x−1|=x−1∴||x−1|−1|≤1⇒|x−1−1|≤1⇒|x−2|≤1⇒2−1≤x≤2+1⇒x∈[1, 3]Hence, x∈[−1, 3]

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The set of real values of $x$ satisfying $| | x - 1 | - 1 | \leq 1,$ is

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[-1, 1 l

[0, 2]

none of these

[-1, 3]

answer is A.

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Detailed Solution

Since $x - 1$ changes its sign as $x$ passes through 1.

So, following cases arise.

CASE I When $x < 1$

In this case, we have $| x - 1 | = - (x - 1)$

$\begin{array}{l} ∴ | | x - 1 | - 1 | \leq 1 \\ \Rightarrow | - (x - 1) - 1 | \leq 1 \\ \Rightarrow | - x | \leq 1 \\ \Rightarrow | x | \leq 1 \\ \Rightarrow - 1 \leq x \leq 1 [∵ | x | \leq a \Leftrightarrow - a \leq x \leq a] \\ \Rightarrow - 1 \leq x < 1 [∵ x < 1] \\ ∴ x \in [- 1, 1) \end{array}$

CASE II When $x \geq 1$

In this case, we have $| x - 1 | = x - 1$

$\begin{array}{l} ∴ | | x - 1 | - 1 | \leq 1 \\ \Rightarrow | x - 1 - 1 | \leq 1 \\ \Rightarrow | x - 2 | \leq 1 \Rightarrow 2 - 1 \leq x \leq 2 + 1 \Rightarrow x \in [1, 3] \end{array}$