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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