Define a relation R on the interval [0 pi/2) by xRy iff | Sets & Relations PYQ for JEE Main 2026

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Q) Define a relation 𝑅 on the interval [0,πœ‹/2) by π‘₯𝑅𝑦 if and only if sec^2 π‘₯βˆ’tan^2 𝑦=1. Then 𝑅 is :Q) Define a relation 𝑅 on the interval [0,πœ‹/2) by π‘₯𝑅𝑦 if and only if sec^2 π‘₯βˆ’tan^2 𝑦=1. Then 𝑅 is :
Define a relation R on the interval [0,─ β”œ Ο€/2) by xRy if and only if γ€–secγ€—^2 x-γ€–tanγ€—^2 y=1. ThenDefine a relation R on the interval [0,─ β”œ Ο€/2) by xRy if and only if γ€–secγ€—^2 x-γ€–tanγ€—^2 y=1. Then
Define a relation R on the interval [0, Ο€/2) by xRy if and only if (sec)^2 x- (tan)^2 y=1.Then R isDefine a relation R on the interval [0, Ο€/2) by xRy if and only if (sec)^2 x- (tan)^2 y=1.Then R is
A = {x ∈ (0 pi) - {pi/2}: log|sinx| + log|cosx| = 2} is | Sets & Relations PYQs for JEE Main 2026A = {x ∈ (0 pi) - {pi/2}: log|sinx| + log|cosx| = 2} is | Sets & Relations PYQs for JEE Main 2026
Let A = {-3 -2 -1 0 1 2 3} and R be a relation on A | Sets & Relations PYQs for JEE Main 2026Let A = {-3 -2 -1 0 1 2 3} and R be a relation on A | Sets & Relations PYQs for JEE Main 2026
Let S = N U {0} Define a relation R from S to R by | Sets & Relations PYQ for JEE Main 2026Let S = N U {0} Define a relation R from S to R by | Sets & Relations PYQ for JEE Main 2026
Kitne PYQ kare ? πŸ†| IIT JEE | Snehit Mishra | JEE 2025 | JEE 2026 πŸ”₯ | #iit #jee2025Kitne PYQ kare ? πŸ†| IIT JEE | Snehit Mishra | JEE 2025 | JEE 2026 πŸ”₯ | #iit #jee2025
Let A = {0, 1, 2, 3, 4, 5}. Let R be a relation on A | Sets & Relations PYQs for JEE Main 2026Let A = {0, 1, 2, 3, 4, 5}. Let R be a relation on A | Sets & Relations PYQs for JEE Main 2026