Q.For a binary operation 'o' on N defined by $a \circ b = a^3 + b^3$, which of the following is true?

Showing the 12 most recent of 41 on this concept.

1. CBSE 2026Set V11 markMCQ
   Q.If a relation $R$ in the set $\{1, 2, 3\}$ be defined by $R = \{(1, 1),(2, 2)\}$ then $R$ is

   (a) symmetric but not transitive

   (b) transitive but not symmetric

   (c) symmetric and transitive

   (d) neither symmetric nor transitive

   ›Reveal solutionSolution

   $R=\{(1,1),(2,2)\}$ is both symmetric and transitive, so the answer is (c).

   Symmetry: whenever $(a,b)\in R$ we need $(b,a)\in R$. Here the only pairs are $(1,1)$ and $(2,2)$, and each is its own reverse, so symmetry holds. …

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2. CBSE 2026Set A1 markMCQ
   Q.What type of a relation is "less than" in the set of real numbers?

   (a) Only symmetric

   (b) Only transitive

   (c) Only reflexive

   (d) Equivalence

   ›Reveal solutionSolution

   "$<$" on $\mathbb{R}$ is transitive only.

   Consider the relation "$a < b$" on $\mathbb{R}$:

   • Reflexive? $a < a$ is false for every $a$, so NOT reflexive.
   • Symmetric? $a < b$ does not imply $b < a$, so NOT symmetric. …

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3. CBSE 2026Set ANNUAL1 markMCQ
   Q.If $R$ is the relation $\{(1,1),(2,2),(3,3),(1,2),(2,3),(1,3)\}$ on $A=\{1,2,3\}$, then which one of the following is true for $R$?

   (a) Reflexive but not symmetric

   (b) Reflexive but not transitive

   (c) Symmetric and transitive

   (d) Neither symmetric nor transitive

   ›Reveal solutionSolution

   $R$ is reflexive (it contains every $(x,x)$) but not symmetric, since $(1,2)\in R$ while $(2,1)\notin R$.

   Given: $R=\{(1,1),(2,2),(3,3),(1,2),(2,3),(1,3)\}$ on $A=\{1,2,3\}$.

   Reflexivity: $R$ is reflexive if $(x,x)\in R$ for every $x\in A$. Here $(1,1),(2,2),(3,3)$ are all present, so $R$ is reflexive.

   Symmetry: $R$ is symmetric if $(x,y)\in R\Rightarrow(y,x)\in R$. Here $(1,2)\in R$ but $(2,1)\notin R$. So $R$ is not symmetric.

   …

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4. CBSE 2026Set ANNUAL1 markMCQ
   Q.Let $R$ be the relation in the set $N$ given by $R = \{(a, b) : a = b - 2, b > 6\}$, choose the correct answer:

   (a) $(2, 4) \in R$

   (b) $(3, 8) \in R$

   (c) $(6, 8) \in R$

   (d) $(8, 6) \in R$

   ›Reveal solutionSolution

   Check each pair against both conditions $a=b-2$ and $b>6$.

   $R=\{(a,b): a=b-2,\ b>6\}$

   1. $(2,4)$: $a=b-2 \Rightarrow 2=2$ ✓, but $b>6 \Rightarrow 4>6$ ✗. Not in $R$.
   2. $(3,8)$: $a=b-2 \Rightarrow 3=6$ ✗. Not in $R$. …

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5. CBSE 2026Set ANNUAL1 markMCQ
   Q.Choose the correct answer : Let $R$ be a relation in the set $\{1,2,3,4\}$ given by $R=\{(1,2),(2,2),(1,1),(4,4),(1,3),(3,3),(3,2)\}$. Then

   (a) $R$ is reflexive and symmetric but not transitive

   (b) $R$ is reflexive and transitive but not symmetric

   (c) $R$ is symmetric and transitive but not reflexive

   (d) $R$ is an equivalence relation

   ›Reveal solutionSolution

   Test $R$ against the three defining properties (reflexive, symmetric, transitive) one at a time, directly from its listed ordered pairs.

   $R=\{(1,2),(2,2),(1,1),(4,4),(1,3),(3,3),(3,2)\}$ on $\{1,2,3,4\}$.

   Reflexive? Need $(a,a)\in R$ for every $a\in\{1,2,3,4\}$: $(1,1)$ ✓, $(2,2)$ ✓, $(3,3)$ ✓, $(4,4)$ ✓. All present — $R$ is reflexive.

   Symmetric? Need: whenever $(a,b)\in R$, also $(b,a)\in R$. Take $(1,2)\in R$: is $(2,1)\in R$? It is not in the list. So $R$ is not symmetric.

   Transitive? Need: whenever $(a,b)\in R$ and $(b,c)\in R$, also $(a,c)\in R$. Checking every chain:

   • $(1,1),(1,2)\Rightarrow(1,2)$ ✓
   • $(1,1),(1,3)\Rightarrow(1,3)$ ✓
   • $(1,2),(2,2)\Rightarrow(1,2)$ ✓
   • $(1,3),(3,3)\Rightarrow(1,3)$ ✓
   • $(1,3),(3,2)\Rightarrow(1,2)$ ✓ …

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6. CBSE 2026Set ANNUAL1 markMCQ
   Q.If $R$ be the relation in the set $N$ given by $R = \{(a,b): a = b - 2, b > 6\}$, then

   (a) $(2, 4) \in R$

   (b) $(3, 8) \in R$

   (c) $(6, 8) \in R$

   (d) $(8, 7) \in R$

   ›Reveal solutionSolution

   Only $(6,8)$ satisfies $a=b-2$ with $b>6$.

   …

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7. CBSE 2025Set X11 markMCQ
   Q.A relation $R$ in a set $A$ is called Reflexive relation if

   (a) $(a, a) \in R$ for all $a \in A$

   (b) $(a, a) \in R$ for atleast one $a \in A$

   (c) $(a, b) \in R$ implies $(b, a) \in R$

   (d) $(a, b) \in R$ and $(b, c) \in R$ implies $(a, c) \in R$

   ›Reveal solutionSolution

   Tests the definition of a reflexive relation — correct option is (a).

   A relation $R$ on a set $A$ is reflexive when every element is related to itself. Formally, $(a,a)\in R$ must hold for all $a\in A$ — not just for at least one element. Option (c) states symmetry ($(a,b)\in R\Rightarrow(b,a)\in R$) and option (d) states transitivit …

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8. CBSE 2025Set IX1 markMCQ
   Q.A relation $R = \{(a, b) : a = b - 1,\ b \geq 3\}$ is defined on set $N$, then

   (a) $(2, 4) \in R$

   (b) $(4, 5) \in R$

   (c) $(4, 6) \in R$

   (d) $(1, 3) \in R$

   ›Reveal solutionSolution

   Only $(4,5)$ satisfies $a=b-1$ with $b\ge 3$; option (b).

   Concept. A pair $(a,b)$ belongs to $R$ only if it satisfies both conditions: $a=b-1$ and $b\ge 3$.

   • $(2,4)$: $b-1=3\ne 2$. ✗ …

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9. CBSE 2025Set ANNUAL1 markMCQ
   Q.If $A=\{1,2,3,4\}$ and $R=\{(a,b)\mid a+b\ \text{is an odd number},\ a,b\in A\}$ is a relation from $A$ to $A$, then which of the following is true for the relation $R$?

   (i) Reflexive

   (ii) Symmetric

   (iii) Transitive

   (iv) Equivalent

   ›Reveal solutionSolution

   Check each property directly: $R$ turns out to be symmetric only.

   $A=\{1,2,3,4\}$, $R=\{(a,b): a+b \text{ is odd}\}$.

   Reflexive? $(a,a)$ needs $a+a=2a$ to be odd — but $2a$ is always even. So $R$ is not reflexive.

   Symmetric? If $a+b$ is odd, then $b+a=a+b$ is the same sum, also odd. So $(a,b)\in R \Rightarrow (b,a)\in R$. $R$ is symmetric.

   …

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10. CBSE 2025Set ANNUAL1 markMCQ
    Q.Let $A = \{a, b, c\}$ and $R = \{(a, a), (a, b), (b, a)\}$, then $R$ is

    (a) reflexive and symmetric but not transitive

    (b) reflexive and transitive but not symmetric

    (c) symmetric and transitive but not reflexive

    (d) an equivalence relation

    ›Reveal solutionSolution

    R is symmetric (the only cross-pair (a,b)/(b,a) both appear) and not transitive ((b,a) & (a,b) would force (b,b), which is missing) — matching option (a) once we note a reflexivity caveat below.

    Step 1 — Reflexive? Full reflexivity on A = {a, b, c} needs (a,a), (b,b), (c,c) all in R. Only (a,a) is present; (b,b) and (c,c) are not. So strictly, R is not fully reflexive on {a,b,c}.

    Step 2 — Symmetric? The only pair with a 'partner' is (a,b), and its reverse (b,a) is also in R. There is no pair in R whose reverse is missing. So R is symmetric.

    Step 3 — Transitive? Take (b,a) ∈ R and (a,b) ∈ R: transitivity would require (b,b) ∈ R. It is not. So R is not transitive.

    …

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11. CBSE 2025Set ANNUAL1 markMCQ
    Q.Relation R = {(x, y) : x < y² where x, y ∈ R} is:

    (a) Reflexive but not symmetric.

    (b) Symmetric and transitive but not Reflexive.

    (c) Reflexive and Symmetric.

    (d) Neither reflexive nor symmetric nor transitive.

    ›Reveal solutionSolution

    Test each property of $R=\{(x,y): x<y^2\}$ with concrete counterexamples — all three fail.

    Reflexive? Need $x<x^2$ for every $x\in\mathbb R$. Take $x=\dfrac12$: is $\dfrac12 < \dfrac14$? No. So $R$ is not reflexive.

    Symmetric? Need $x<y^2 \Rightarrow y<x^2$. Take $x=0,\,y=1$: $0<1^2$ is true, so $(0,1)\in R$. But is $1<0^2=0$? No. So $(1,0)\notin R$, and $R$ is not symmetric.

    …

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12. CBSE 2025Set ANNUAL1 markMCQ
    Q.The relation R = {(a, a), (b, b), (c, c)} on the set A = {a, b, c} is ...........

    (a) Identity

    (b) Reflexive only

    (c) Symmetric only

    (d) Equivalence

    ›Reveal solutionSolution

    R = {(a,a), (b,b), (c,c)} is reflexive, symmetric, and transitive on A = {a, b, c}, so by definition it is an equivalence relation (it is in fact the identity relation, which is always the smallest possible equivalence relation on a set).

    Check reflexive: For every element $x \in A$, $(x,x) \in R$. Indeed $(a,a), (b,b), (c,c)$ are all present. So R is reflexive.

    Check symmetric: If $(x,y) \in R$ then $(y,x) \in R$. Since every pair here has $x = y$, this holds trivially.

    Check transitive: If $(x,y) \in R$ and $(y,z) \in R$ then $(x,z) \in R$. Again since all pairs are of the form $(x,x)$, this holds trivially.

    …

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