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solve: |4-x|+1<1

Note by Akash Sinha 4 years, 9 months ago

Easy Math Editor

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2 \times 3

2^{34}

a_{i-1}

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\(|4-x| + 1 < 1\)

Simplifying, we get-

\(|4-x| < 0\)

But since the LHS is in Modulus, it has to be positive or equal to zero. Hence, the above inequality holds for no real numbers.

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From the given inequality ,

\(| 4-x | < 0\)

But since we know , for any \(a \in \mathbb{R} \)

\[ | a | = \begin{cases} -a & \text{if } x < 0 \\ a & \text{if } x \geq 0 \end{cases} \]

also , \( | a | \geq 0 \)

But here , \(| 4-x | < 0\) , which is not possible .

Thus no such Real values are there for x .

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Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

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TopNewest\(|4-x| + 1 < 1\)

Simplifying, we get-

\(|4-x| < 0\)

But since the LHS is in Modulus, it

hasto be positive or equal to zero. Hence, the above inequality holds for no real numbers.Log in to reply

From the given inequality ,

\(| 4-x | < 0\)

But since we know , for any \(a \in \mathbb{R} \)

\[ | a | = \begin{cases} -a & \text{if } x < 0 \\ a & \text{if } x \geq 0 \end{cases} \]

also , \( | a | \geq 0 \)

But here , \(| 4-x | < 0\) , which is not possible .

Thus no such

Realvalues are there for x .Log in to reply