Q.Check whether the relation R in the set N of natural numbers given by $R = \{(a, b) : a \text{ is divisor of } b\}$ is reflexive, symmetric or transitive. Also determine whether R is an equivalence relation.

Showing the 12 most recent of 44 on this concept.

1. CBSE 2026Set 65/2/11 markMCQ
   Q.A relation $R$ on set $A = \{1, 2, 3\}$ defined as $R = \{(1, 1), (2, 2), (1, 2)\}$ is (A) Reflexive only (B) Reflexive and Transitive (C) Symmetric and Transitive (D) Transitive only

   ›Reveal solutionSolution

   On $A=\{1,2,3\}$, $R$ is not reflexive (missing $(3,3)$), not symmetric (has $(1,2)$ but not $(2,1)$), and is transitive. So $R$ is transitive only — option (D).

   Check each property of $R=\{(1,1),(2,2),(1,2)\}$ on $A=\{1,2,3\}$.

   Reflexive? Requires $(a,a)\in R$ for every $a\in A$, i.e. $(1,1),(2,2),(3,3)$. Since $3\in A$ but $(3,3)\notin R$, $R$ is not reflexive.

   Symmetric? Requires $(b,a)\in R$ whenever $(a,b)\in R$. Here $(1,2)\in R$ but $(2,1)\notin R$, so $R$ is not symmetric.

   Transitive? Requires $(a,c)\in R$ whenever $(a,b),(b,c)\in R$. The only linking pairs are: …

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2. CBSE 2026Set A1 markMCQ
   Q.$\tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) =$

   (a) $\frac{\pi}{3}$

   (b) $\frac{\pi}{6}$

   (c) $-\frac{\pi}{3}$

   (d) $-\frac{\pi}{6}$

   ›Reveal solutionSolution

   $\tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) = -\frac{\pi}{6}$.

   The principal value of $\tan^{-1}$ lies in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.

   We need the angle $\theta$ in this range with $\tan\theta = -\frac{1}{\sqrt{3}}$.

   …

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3. CBSE 2026Set A1 markMCQ
   Q.$2\tan^{-1}\frac{1}{3} =$

   (a) $\tan^{-1}\frac{3}{2}$

   (b) $\tan^{-1}\frac{3}{4}$

   (c) $\tan^{-1}\frac{4}{3}$

   (d) $\tan^{-1}\frac{2}{3}$

   ›Reveal solutionSolution

   $2\tan^{-1}\frac{1}{3} = \tan^{-1}\frac{3}{4}$.

   Use the double-angle identity valid for $|x|<1$:

   $2\tan^{-1}x = \tan^{-1}\frac{2x}{1-x^2}.$

   Here $x = \frac{1}{3}$: …

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4. CBSE 2026Set A1 markMCQ
   Q.$x \in R,\ \cot(\tan^{-1}x + \cot^{-1}x) =$

   (a) $1$

   (b) $\frac{1}{2}$

   (c) $0$

   (d) $\frac{1}{3}$

   ›Reveal solutionSolution

   $\cot\left(\tan^{-1}x + \cot^{-1}x\right) = \cot\frac{\pi}{2} = 0$.

   For all real $x$, the complementary identity gives

   $\tan^{-1}x + \cot^{-1}x = \frac{\pi}{2}.$

   Therefore …

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5. CBSE 2026Set A1 markMCQ
   Q.$\tan^{-1}2 + \tan^{-1}3 =$

   (a) $-\frac{\pi}{4}$

   (b) $\frac{\pi}{4}$

   (c) $\frac{3\pi}{4}$

   (d) $\pi$

   ›Reveal solutionSolution

   $\tan^{-1}2 + \tan^{-1}3 = \frac{3\pi}{4}$.

   Use the addition formula. With $a=2,\ b=3$ we have $ab=6>1$, so

   $\tan^{-1}a + \tan^{-1}b = \pi + \tan^{-1}\frac{a+b}{1-ab}.$

   Compute:

   $\frac{a+b}{1-ab} = \frac{5}{1-6} = \frac{5}{-5} = -1.$

   So …

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6. CBSE 2026Set A1 markMCQ
   Q.$\tan^{-1}\frac{x}{y} - \tan^{-1}\frac{x-y}{x+y} =$

   (a) $-\frac{3\pi}{4}$

   (b) $\frac{\pi}{2}$

   (c) $\frac{\pi}{4}$

   (d) $\frac{\pi}{3}$

   ›Reveal solutionSolution

   $\tan^{-1}\frac{x}{y} - \tan^{-1}\frac{x-y}{x+y} = \frac{\pi}{4}$.

   Use $\tan^{-1}a - \tan^{-1}b = \tan^{-1}\frac{a-b}{1+ab}$ with $a=\frac{x}{y}$, $b=\frac{x-y}{x+y}$.

   Numerator:

   $a-b = \frac{x}{y} - \frac{x-y}{x+y} = \frac{x(x+y) - y(x-y)}{y(x+y)} = \frac{x^2+xy-xy+y^2}{y(x+y)} = \frac{x^2+y^2}{y(x+y)}.$

   Denominator: …

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7. CBSE 2026Set A1 markMCQ
   Q.$|x| \le 1,\ \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) =$

   (a) $2\cos^{-1}x$

   (b) $2\sin^{-1}x$

   (c) $2\tan^{-1}x$

   (d) $\tan^{-1}2x$

   ›Reveal solutionSolution

   $\cos^{-1}\frac{1-x^2}{1+x^2} = 2\tan^{-1}x$ (for $0\le x\le 1$).

   Put $x = \tan\theta$, so $\theta = \tan^{-1}x$. Then

   $\frac{1-x^2}{1+x^2} = \frac{1-\tan^2\theta}{1+\tan^2\theta} = \cos 2\theta.$

   Therefore …

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8. CBSE 2026Set ANNUAL1 markMCQ
   Q.Let R = {(4, 4), (6, 6), (7, 7), (4, 6), (6, 4), (4, 7), (6, 7)} be a relation defined on A = {4, 6, 7}, then this relation R is ................. .

   (a) Reflexive, not symmetric and not transitive

   (b) Reflexive, symmetric and transitive

   (c) Reflexive and transitive but not symmetric

   (d) Neither Reflexive, nor symmetric and nor transitive

   ›Reveal solutionSolution

   Check reflexivity, symmetry and transitivity directly against the listed ordered pairs.

   Reflexive: $(4,4),(6,6),(7,7)$ are all present in $R$, so $R$ is reflexive.

   Symmetric: $(4,6)\in R$ and $(6,4)\in R$ is fine, but $(4,7)\in R$ while $(7,4)\notin R$. So $R$ is NOT symmetric.

   …

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9. CBSE 2025Set E1 markMCQ
   Q.$\cot^{-1}\left(\tan\frac{\pi}{7}\right) =$

   (a) $\frac{\pi}{7}$

   (b) $\frac{5\pi}{14}$

   (c) $\frac{9\pi}{14}$

   (d) $\frac{3\pi}{14}$

   ›Reveal solutionSolution

   Convert the tangent to a cotangent using complementary angles; the answer is $\frac{5\pi}{14}$.

   Use $\tan\theta=\cot\left(\frac{\pi}{2}-\theta\right)$:

   $\tan\frac{\pi}{7}=\cot\left(\frac{\pi}{2}-\frac{\pi}{7}\right)=\cot\frac{7\pi-2\pi}{14}=\cot\frac{5\pi}{14}.$

   …

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10. CBSE 2025Set E1 markMCQ
    Q.$\tan^{-1}(-\sqrt{3}) =$

    (a) $\frac{\pi}{6}$

    (b) $\frac{\pi}{3}$

    (c) $\frac{2\pi}{3}$

    (d) $-\frac{\pi}{3}$

    ›Reveal solutionSolution

    $\tan^{-1}$ is an odd function with principal range $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$; the value is $-\frac{\pi}{3}$.

    Since $\tan\frac{\pi}{3}=\sqrt{3}$ and $\tan^{-1}(-x)=-\tan^{-1}x$, …

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11. CBSE 2025Set E1 markMCQ
    Q.$\tan^{-1}(\sqrt{3}) - \cot^{-1}(-\sqrt{3}) =$

    (a) $0$

    (b) $-\frac{\pi}{2}$

    (c) $\pi$

    (d) $\frac{\pi}{2}$

    ›Reveal solutionSolution

    Evaluate each inverse function in its principal range and subtract; result $-\frac{\pi}{2}$.

    First, $\tan^{-1}(\sqrt{3})=\frac{\pi}{3}$.

    For $\cot^{-1}(-\sqrt{3})$, the principal range of $\cot^{-1}$ is $(0,\pi)$. We need $\cot\theta=-\sqrt{3}$ with $\theta\in(0,\pi)$. Since $\cot\frac{\pi}{6}=\sqrt{3}$,

    $\cot^{-1}(-\sqrt{3})=\pi-\frac{\pi}{6}=\frac{5\pi}{6}.$

    …

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12. CBSE 2025Set E1 markMCQ
    Q.$\tan^{-1}\frac{1}{2} + \tan^{-1}\frac{1}{3} =$

    (a) $\pi$

    (b) $\frac{\pi}{4}$

    (c) $\frac{\pi}{2}$

    (d) $\frac{\pi}{3}$

    ›Reveal solutionSolution

    Use the sum formula for inverse tangents; the sum is $\frac{\pi}{4}$.

    When $xy<1$, $\tan^{-1}x+\tan^{-1}y=\tan^{-1}\dfrac{x+y}{1-xy}$. Here $xy=\frac16<1$, so …

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