MHT CET · Maths · Mathematical Reasoning

Which of the following is the negation of the statement " For all \(\mathrm{M}>0\), there exist \(x \in s\) such that \(x \geqslant \mathrm{M}\) "

A \(\exists \mathrm{M}>0\) such that \(x \geqslant \mathrm{M}\) for all \(x \in \mathrm{~s}\)
B \(\exists \mathrm{M}>0, \exists x \in \mathrm{~s}\) such that \(x \geqslant \mathrm{M}\)
C \(\exists \mathrm{M}>0\) such that \(x < \mathrm{M}\) for all \(x \in \mathrm{~s}\)
D \(\exists \mathrm{M}>0\), there exist \(x \in \mathrm{~s}\) such that \(x < \mathrm{M}\)

Verified Solution

Answer & Solution

Correct Answer

(C) \(\exists \mathrm{M}>0\) such that \(x < \mathrm{M}\) for all \(x \in \mathrm{~s}\)

Step-by-step Solution

Detailed explanation

The original statement is: \(\forall \mathrm{M}>0, \exists x \in s \text{ such that } x \geqslant \mathrm{M}\) To negate this statement, we change quantifiers and negate the predicate:

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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