Relations and Functions — JEE MCQs with answers

Answer: D — If $b$ is related to $a$ then, by symmetry and transitivity, an element is related to $a$ exactly when it is related to $b$, so the two classes are the same subset. p.4 illustrates this with $[0] = [2r]$ for every $r$. The correct option is D. Option A is wrong: both classes contain $b$, so their intersection is not empty. Option B is wrong: in Example 6, $3 \in [1]$ although $3 \ne 1$. Option C is wrong: the two classes are equal, so each is contained in the other.

Answer: A — Option A is correct: $f$ applied to $\tfrac{1}{y}$ returns $y$, and $\tfrac{1}{y}$ is a legitimate member of the domain because it is never zero. Option B is false, since $f(2) = \tfrac{1}{2}$ and not $2$. Option C yields $-y$ rather than $y$. Option D yields $\tfrac{1}{1-y}$, which is not $y$ in general and is undefined at $y = 1$.

Answer: C — C is correct: with $f$ landing in $B$ and $g$ defined on $B$, the definition builds $gof : A \to C$. A and B propose $fog$, which would need $g$ to land in the domain of $f$; here $g$ lands in $C$ and $f$ is defined on $A$, and these are different sets. D keeps the correct composite but adds the impossible one alongside it.

Answer: D — Option D is correct: Section 1.1 spells out $(a, b) \in R$ in exactly this way before shortening it to $a\,R\,b$. Option A is wrong: the order of the entries settles which object is related to which; the first entry a is related to the second entry b. Option B is wrong: relatedness holds between a and b, not between them and R. Option C is wrong: R is the relation doing the relating, not one of the related objects.

Answer: A — Testing each value in $a \le a^2$: for $a = \frac{1}{2}$, $a^2 = \frac{1}{4}$ and $\frac{1}{2} \le \frac{1}{4}$ is false, so $\left(\frac{1}{2}, \frac{1}{2}\right) \notin R$ and R is not reflexive. Option B is wrong because $(-1)^2 = 1$ and $-1 \le 1$, so $(-1, -1) \in R$. Option C is wrong because $2^2 = 4$ and $2 \le 4$, so $(2, 2) \in R$. Option D is wrong because $3^2 = 9$ and $3 \le 9$, so $(3, 3) \in R$. Option A is therefore correct.

Answer: D — The four images are $x$, $v$, $z$ and $y$. The only element of $\{v, w, x, y, z\}$ that does not appear among them is $w$, so $w$ is the element Definition 6 stumbles on and $f$ is not onto. Each of the other three options names a letter that an arrow does reach: $v = f(2)$, $x = f(1)$, $z = f(3)$.

Answer: C — C is correct: $fog(x) = f(3x) = (3x)^2 + 1 = 9x^2 + 1$. A is $gof(x) = 3(x^2+1) = 3x^2 + 3$, the composite in the other order. B substitutes $3x$ but fails to square the $3$. D doubles the $3$ instead of squaring it, reading $(3x)^2$ as $2 \cdot 3x^2$.

Answer: D — Option D is correct: testing all nine pairs of $A \times B$: 2 divides 4 and 6 but not 9; 3 divides 6 and 9 but not 4; 4 divides 4 but neither 6 nor 9. Option A is wrong: this is the whole of $A \times B$, which is what would be got by admitting every pair instead of testing the condition. Option B is wrong: this is what the reversed test, b divides a, yields: among the nine pairs only 4 divides 4. Option C is wrong: this comes from taking $b = 2a$ for each a and not checking that the result lies in B; $8 \notin B$, so $(4, 8)$ is not even a pair of $A \times B$.

Answer: C — $(0, 0) \notin R$, since $0 \times 0 = 0$ is not positive, so R is not reflexive. $ab > 0$ is unchanged if $a$ and $b$ are exchanged, so R is symmetric. If $ab > 0$ and $bc > 0$ then $a$ and $b$ have the same sign and so do $b$ and $c$, hence $a$ and $c$ do, giving $ac > 0$; so R is transitive. Option A is wrong because signs do pass along a chain, so transitivity holds; it is reflexivity that fails. Option B is wrong because symmetry and transitivity both hold; only reflexivity fails, and it fails at $a = 0$ alone. Option D is wrong because reflexivity fails at the single number 0, since $0 \times 0$ is not positive. Option C is therefore correct.

Answer: D — D is correct: the sentence gives both directions, and the converse direction runs from one-one and onto to invertible. A and B state halves of the other direction, the one that starts from invertibility. C weakens the hypothesis to one-one alone, which is not what is claimed.