Let \(X\) be a non-void set. If \(\rho_1\) and \(\rho_2\) be the transitive
relations on \(X\), then [ \( \circ\) denotes composition ]
A
\(\rho_1 \circ \rho_2\) is transitive relation
B
\(\rho_1 \circ \rho_2\) is not transitive relation
C
\(\rho_1 \circ \rho_2\) is equivalence relation
D
\(\rho_1 \circ \rho_2\) is not any relation on \(X\)
Hint
Let's define a set \(X = \{1, 2, 3, 4, 5\}\). Now, let's create two relations on \(X\),
\(\rho_1\) and \(\rho_2\), both of which are transitive:Let \(\rho_1 = \{(1, 2), (3, 4)\}\)
Let \(\rho_2 = \{(2, 3), (4, 5)\}\) (Note: These are trivially transitive).
Now, let's find the composition \(\rho_1 \circ \rho_2\). By definition of composition,
\((x, z) \in \rho_1 \circ \rho_2\) if there is some intermediate element \(y\) such
that \((x, y) \in \rho_1\) and \((y, z) \in \rho_2\).
So, the resulting composition is:\(\rho_1 \circ \rho_2 = \{(1, 3), (3, 5)\}\) which is clearly not transitive.
If \(R\) and \(Q\) are equivalence relations on set \(A\),
then which of the following is not an equivalence relation
A \(R^{-1} \cap Q^{-1}\)
B \(Q^{-1}\)
C \(R \cup Q\)
D \(R \cap Q\)
Hint
Let \( A=\{1,2,3\}, P =\{(1,1),(2,2),(3,3),(2,3),(3,2)\}\)
and \(Q =\{(1,1),(2,2),(3,3),(1,2),(2,1)\}\). It is easy to see that both \(P \text{ and } Q\)
are equivalence relation on \(A\) but \(P \cup Q\) is not an equivalence relation.
Let \( \rho \) be a relation defined on set of natural numbers \( \mathbb{N} \), as \( \rho = \{(x, y) \in \mathbb{N} \times \mathbb{N} : 2x + y = 41\} \).
Then domain A and range B are
A \( A \subset \{x \in \mathbb{N} : 1 \le x \le 20\} \) and \( B \subset \{y \in \mathbb{N} : 1 \le y \le 39\} \)
B \( A = \{x \in \mathbb{N} : 1 \le x \le 15\} \) and \( B = \{y \in \mathbb{N} : 2 \le y \le 30\} \)
C \( A \equiv \mathbb{N}, B \equiv \mathbb{Q} \)
D \( A \equiv \mathbb{Q}, B \equiv \mathbb{Q} \)
Hint
See that \(y= 41 - 2x\) and for \(x \in \{1,2,...,20 \}\), \( y \in \{1,3,5,...,39\}\)
Let \( f : \mathbb{R} \to \mathbb{R} \) be a function defined by \( f(x) = \frac{e^{|x|} - e^{-x}}{e^x + e^{-x}} \), then
A
\( f \) is both one-one and onto
B
\( f \) is one-one but not onto
C
\( f \) is onto but not one-one
D
\( f \) is neither one-one nor onto
Hint
\Injectivity (One-One):
For \( x < 0 \), \( |x| = -x \).
$$ f(x) = \frac{e^{-x} - e^{-x}}{e^x + e^{-x}} = 0 $$
Since \( f(-1) = f(-2) = 0 \), multiple inputs yield the same output. Therefore, \( f \) is not one-one.
2. Surjectivity (Onto):
For \( x \ge 0 \), \( |x| = x \).
$$ f(x) = \frac{e^x - e^{-x}}{e^x + e^{-x}} $$
The range for \( x \ge 0 \) is \( [0, 1) \) ( use limits), the total range of \( f \) is \( [0, 1) \).
Since the range \( [0, 1) \neq \) codomain \( \mathbb{R} \), \( f \) is not onto.
Let \(X=\{v,i,n,o,d\}\) and \(Y=\{p,m\}\). The number of
onto ( surjective) functions from \(X\) to \(Y\) is
A \(32\)
B \(28 \)
C \(30\)
D \(36\)
Hint
To find the number of onto functions,
first calculate the total number of unrestricted mappings
from \(\{v, i, n, o, d\}\) to \(\{p, m\}\), which is \(2^5\).
Then, simply subtract the cases where the mapping is not
onto—specifically, the exact number of scenarios where all \(5\)
elements map exclusively to just a single output in the codomain.
The domain of the definition of the function \(f(x)=\frac{1}{4-x^2} + log_{10} (x^3-x)\) is
A
\( (-1,0) \cup (1,2) \cup (3,\infty) \)
B
\( (-2,-1) \cup (-1,0) \cup (2,\infty) \)
C
\( (-1,0) \cup (1,2) \cup (2,\infty) \)
D
\( (1,2) \cup (2,\infty)\)
Hint
Critical points of \(x^3-x\) are \(-1,0\) and \(1\)
Let \( f : R \rightarrow R \) be such that \( f \) is injective and \( f(x)f(y) = f(x+y) \) for \( \forall x, y \in R \). If \( f(x) \), \( f(y) \),
\( f(z) \) are in G.P., then \( x, y, z \) are in
A AP always and \(f(0)=1\)
B GP always and \(f(0)=0\)
C AP depending on the value of \( x, y, z \)
D GP depending on the value of \( x, y, z \)
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