Sets, Relations & Functions
Properties of Relations
nta_pyq_2025_apr
Grade 11
Question:
Define a relation $R$ on the interval $\left[0, \frac{\pi}{2}\right)$ by $x\,R\,y$ if and only if $\sec^2 x - \tan^2 y = 1$. Then $R$ is:
both reflexive and transitive but not symmetric
an equivalence relation
reflexive but neither symmetric nor transitive
both reflexive and symmetric but not transitive
Step-by-Step Solution
Key Concept: Check reflexivity: $\sec^2 x - \tan^2 x = 1$ (always true). For symmetry, use $1+\tan^2 x - \tan^2 y = 1 \Rightarrow \tan^2 x = \tan^2 y \Leftrightarrow$ same holds with $x,y$ swapped. For transitivity, add the two equalities.
Reflexive: $\sec^2 x - \tan^2 x = 1$ ✓. Symmetric: $\sec^2 x - \tan^2 y = 1 \Rightarrow 1+\tan^2 x = 1+\tan^2 y \Rightarrow \sec^2 y - \tan^2 x = 1$ ✓. Transitive: if $\sec^2 x - \tan^2 y = 1$ and $\sec^2 y - \tan^2 z = 1$, adding gives $\sec^2 x + 1 - \tan^2 z = 2 \Rightarrow \sec^2 x - \tan^2 z = 1$ ✓. R is an equivalence relation.
Correct Answer: an equivalence relation