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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. – Akshat Jain · 3 years, 4 months agoLog 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 . – Priyansh Sangule · 3 years, 4 months agoLog in to reply