Karnataka PUC 12th Maths Question Paper 2025

A relation $R$ is reflexive if $(a, a) \in R$ for all $a \in \{1, 2, 3\}$. Since $(1, 1), (2, 2), (3, 3) \notin R$, it is not reflexive. It is symmetric if $(a, b) \in R \implies (b, a) \in R$. Since $(2, 3) \in R$ but $(3, 2) \notin R$, it is not symmetric. It is transitive if $(a, b) \in R$ and $(b, c) \in R \implies (a, c) \in R$. Here, the condition 'if $(a, b) \in R$ and $(b, c) \in R$' is never met for any distinct elements to form a violation, hence it is vacuously transitive. Therefore, the relation is transitive.

The principal value branch of $\cot^{-1} x$ is defined as the open interval $(0, \pi)$.

Statement 1: The function $\sin^{-1}(\sin x) = x$ is valid if $x$ lies within the principal value branch of $\sin^{-1}$, which is $[-\frac{\pi}{2}, \frac{\pi}{2}]$. Statement 2: The function $\cos(\cos^{-1}x) = x$ is valid if $x$ lies within the domain of $\cos^{-1}$, which is $[-1, 1]$. However, for the given $x$ in $[0, \pi]$, since the input to $\cos^{-1}$ must be in $[-1, 1]$, if $x$ is interpreted as the domain constraint, both identities hold true. Since both are standard trigonometric identities within their defined intervals, both are true.

If a matrix has 8 elements, its possible orders $(m \times n)$ are factors of 8. The pairs $(m, n)$ such that $mn=8$ are $(1, 8), (8, 1), (2, 4), (4, 2)$. Thus, there are 4 possible orders.

Given $|A| = 8$. We know $|A'| = |A| = 8$. Also, $|AA'| = |A| \cdot |A'|$. Therefore, $|AA'| = 8 \times 8 = 64$.