9 public questions tagged with this topic.
Notice that after simplifying, Symmetric relation is left. The method is simple: formula → substitution → calculation → answer.
Ref: NCERT Mathematics Class XI — Sets, Relations and Functions
Let's break it down: solving fully: apply relevant formula, put numbers, compute. Result comes as R = (7, 7), (8, 8), (9, 9). Easy verification shows it satisfies condition.
Ref: NCERT Mathematics Class XI — Sets, Relations and Functions
Applying the basic concept, complete solution: write formula, plug data, do arithmetic. Final value matches Reflexive relation. Simple and accurate.
Ref: NCERT Mathematics Class XI — Sets, Relations and Functions
Think of it like this: using formula we get R = (1, 2), (2, 1). Calculation gives this directly, check steps: apply rule, substitute values, simplify to get R = (1, 2), (2, 1).
Ref: NCERT Mathematics Class XII — Matrices and Determinants
Understanding this is easy: we verify a * b = b * a by full calculation: use standard identity, substitute, simplify step by step to reach a * b = b * a.
Ref: NCERT Mathematics Class XI — Sets, Relations and Functions
In easy words, using formula we get P(A) + P(B). Calculation gives this directly, check steps: apply rule, substitute values, simplify to get P(A) + P(B).
Ref: NCERT Mathematics Class XII — Probability
The logic is straightforward: after simplifying, The relation is an equivalence relation is left. The method is simple: formula → substitution → calculation → answer.
Ref: NCERT Mathematics Class XI — Sets, Relations and Functions
From the definition, after simplifying, (5, 2), (9, 4), (11, 5) is left. The method is simple: formula → substitution → calculation → answer.
Ref: NCERT Mathematics Class XI — Sets, Relations and Functions
We know that using formula we get Ⴔ. Calculation gives this directly, check steps: apply rule, substitute values, simplify to get Ⴔ.
Ref: NCERT Mathematics Class XI — Sets, Relations and Functions
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