Which of the following is incorrect for the solution set of the inequation | | x | - 1 | | 1 - x | , x R , is

Question

Which of the following is incorrect for the solution set of the inequation $| | x | - 1 |$ $<$ $| 1 - x |$ $, x \in R$ , is

Smart Achievers · Accepted Answer

(1, 2, 3)

Let us consider the following cases:

Case I: When $x \geq 0$

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

$∴$ $| | x | - 1 |$ $<$ $| 1 - x |$

$\Rightarrow$ $| x - 1 |$ $<$ $| x - 1 |$ , which is not true for $x \geq 0$

Case II: when $x < 0$

$| x |$ $= - x$

$∴$ $| | x | - 1 |$ $<$ $| 1 - x |$

$=$ $| - x - 1 |$ $<$ $| 1 - x |$ $=$ $| x + 1 |$ $<$ $| x - 1 |$

If $x < - 1$ , then $| x + 1 |$ $= - (x + 1)$ and $| x - 1 |$ $= - (x - 1)$

$∴$ $| x + 1 |$ $<$ $| x - 1 |$ $= - (x + 1) < - (x - 1)$

$\Rightarrow$ $- 1 < 1$ , which is always true.

Thus, $(- \infty, - 1)$ is the solution set.

If $- 1 < x < 0,$ then $| x + 1 |$ $= x + 1,$ $| x - 1 |$ $= - (x - 1)$

$∴$ $| x + 1 |$ $<$ $| x - 1 |$ $= x + 1 < - (x - 1) \Rightarrow 2 x < 0 \Rightarrow x < 0$

But, $- 1 < x < 0 .$

Therefore, $x \in (- 1, 0)$ .

Hence, $(- \infty, 0)$ is the solution set.

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