Exercise 1.1 — Relations and Functions

1. 1.1.16 marksNCERT Class 12 Mathematics, Relations and Functions, Reprint 2026-27

   Determine whether each of the following relations are reflexive, symmetric and transitive: (i) Relation R in the set A={1,2,...,13,14} defined as R={(x,y): 3x-y=0}. (ii) Relation R in the set N of natural numbers defined as R={(x,y): y=x+5 and x<4}. (iii) Relation R in the set A={1,2,3,4,5,6} as R={(x,y): y is divisible by x}. (iv) Relation R in the set Z of all integers defined as R={(x,y): x-y is an integer}. (v) Relation R in the set A of human beings in a town at a particular time given by (a) R={(x,y): x and y work at the same place} (b) R={(x,y): x and y live in the same locality} (c) R={(x,y): x is exactly 7 cm taller than y} (d) R={(x,y): x is wife of y} (e) R={(x,y): x is father of y}

   Hint. For each relation, test reflexivity with (a,a), test symmetry by finding one pair whose reverse fails, and test transitivity by checking whether every chain closes up.

   (i) R={(x,y): y=3x} within A={1,...,14} gives pairs (1,3),(2,6),(3,9),(4,12). (1,1) is not in R (reflexive fails). (1,3) is in R but (3,1) is not (symmetric fails). (1,3) and (3,9) are in R but (1,9) is not, since 3(1)-9=-6, not 0 (transitive fails).

   (ii) R={(1,6),(2,7),(3,8)} on N. (1,1) is not in R (not reflexive). (1,6) is in R but (6,1) is not (not symmetric). No pair's second coordinate (6,7,8) ever appears as a first coordinate (only 1,2,3 do), so there is no chain to violate — transitive holds vacuously.

   (iii) 'y divisible by x' on {1,...,6}. x divides x always, so reflexive holds. (1,2) is in R (2 divisible by 1) but (2,1) is not (1 not divisible by 2), so not symmetric. If x divides y and y divides z, then x divides z (standard divisibility property), so transitive holds.

   (iv) 'x-y is an integer' on Z. Since x,y are always integers, x-y is always an integer, so R=Z x Z, the universal relation. This is trivially reflexive, symmetric, and transitive — an equivalence relation.

   (v)(a) Same workplace: reflexive, symmetric, transitive — an equivalence relation. (b) Same locality: likewise an equivalence relation. (c) 'exactly 7 cm taller': not reflexive (x is not taller than itself), not symmetric (if x is 7cm taller than y, y is 7cm shorter, not taller, than x), not transitive (7cm taller than 7cm taller is 14cm taller, not 7cm). (d) 'x is wife of y': not reflexive, not symmetric (x wife of y means y is husband, not wife, of x), not transitive. (e) 'x is father of y': not reflexive, not symmetric, not transitive (a father's father is a grandfather, not a father).

   ✦ (i) Neither reflexive, symmetric, nor transitive. (ii) Neither reflexive nor symmetric, but transitive. (iii) Reflexive and transitive, not symmetric. (iv) Reflexive, symmetric, and transitive (equivalence relation). (v)(a),(b) equivalence relations; (c),(d),(e) neither reflexive, symmetric, nor transitive.

2. 1.1.24 marksNCERT Class 12 Mathematics, Relations and Functions, Reprint 2026-27

   Show that the relation R in the set R of real numbers, defined as R={(a,b): a <= b^2} is neither reflexive nor symmetric nor transitive.

   Hint. Look for a real number that is not less than or equal to its own square, then find pairs where the relation holds one way but not the reverse, and a chain that breaks.

   Not reflexive: take a=1/2. Since 1/2 is not <= (1/2)^2=1/4, the pair (1/2,1/2) is not in R, so reflexivity fails.

   Not symmetric: take a=1, b=2. Because 1<=2^2=4, the pair (1,2) is in R; but because 2 is not <= 1^2=1, the pair (2,1) is not in R, so symmetry fails.

   Not transitive: take a=100, b=-10, c=0. Because 100<=(-10)^2=100 (equality), (100,-10) is in R; because -10<=0^2=0, (-10,0) is in R; but because 100 is not <= 0^2=0, (100,0) is not in R — the chain breaks, so transitivity fails.

   ✦ R is neither reflexive, nor symmetric, nor transitive, confirmed by the counterexamples a=1/2 (reflexivity), (1,2) (symmetry), and (100,-10,0) (transitivity).

3. 1.1.33 marksNCERT Class 12 Mathematics, Relations and Functions, Reprint 2026-27

   Check whether the relation R defined in the set {1,2,3,4,5,6} as R={(a,b): b=a+1} is reflexive, symmetric or transitive.

   Hint. Compare (a,a) against the rule b=a+1, then check a single pair's reverse, then check a two-step chain.

   Reflexive: (a,a) needs a=a+1, which is never true. Not reflexive.

   Symmetric: (1,2) is in R since 2=1+1. Is (2,1) in R? That needs 1=2+1=3, false. Not symmetric.

   Transitive: (1,2) and (2,3) are both in R. Is (1,3) in R? That needs 3=1+1=2, false. Not transitive.

   ✦ R is neither reflexive, nor symmetric, nor transitive.

4. 1.1.43 marksNCERT Class 12 Mathematics, Relations and Functions, Reprint 2026-27

   Show that the relation R in R defined as R={(a,b): a <= b}, is reflexive and transitive but not symmetric.

   Hint. a<=a is always true; find one pair where a<=b holds but b<=a fails; chain two <= relations together.

   Reflexive: a<=a holds for every real a, since no real number is strictly greater than itself. Transitive: if a<=b and b<=c, then a<=c, because this is a standard order property of the real numbers. Not symmetric: since 1<=2 holds but 2<=1 does not, symmetry fails.

   ✦ R is reflexive and transitive, but not symmetric — confirmed by the counterexample 1<=2 without 2<=1.

5. 1.1.54 marksNCERT Class 12 Mathematics, Relations and Functions, Reprint 2026-27

   Check whether the relation R in R defined by R={(a,b): a <= b^3} is reflexive, symmetric or transitive.

   Hint. Test a fractional value against its own cube for reflexivity; find a pair that fails the reverse; build a three-term chain using large and small values.

   Not reflexive: a=1/2 gives 1/2 <= (1/2)^3=1/8? No.

   Not symmetric: a=1, b=2 gives 1<=2^3=8, true, so (1,2) is in R. Is 2<=1^3=1? No.

   Not transitive: take a=512, b=8, c=2. Is 512<=8^3=512? Yes (equality). Is 8<=2^3=8? Yes (equality). Is 512<=2^3=8? No — the chain breaks.

   ✦ R is neither reflexive, nor symmetric, nor transitive.

6. 1.1.63 marksNCERT Class 12 Mathematics, Relations and Functions, Reprint 2026-27

   Show that the relation R in the set {1,2,3} given by R={(1,2),(2,1)} is symmetric but neither reflexive nor transitive.

   Hint. Check the diagonal pairs for reflexivity, check the two given pairs for symmetry, then see what transitivity would force.

   Not reflexive: (1,1) is not in R, so R is not reflexive.

   Symmetric: (1,2) is in R and (2,1) is in R — both directions present, so R is symmetric.

   Not transitive: (1,2) and (2,1) are both in R, so transitivity would require (1,1) to be in R, but it is not.

   ✦ R is symmetric but neither reflexive nor transitive.

7. 1.1.73 marksNCERT Class 12 Mathematics, Relations and Functions, Reprint 2026-27

   Show that the relation R in the set A of all the books in a library of a college, given by R={(x,y): x and y have same number of pages} is an equivalence relation.

   Hint. 'Same number of pages' is really equality of a numeric attribute, so all three properties follow the same way equality does.

   Reflexive: every book has the same number of pages as itself, so (x,x) is in R for every book x. Symmetric: since 'x has the same page count as y' and 'y has the same page count as x' are the same statement, (x,y) in R implies (y,x) in R. Transitive: because equal page counts chain together — if x and y match, and y and z match, then x and z must match too — (x,y) and (y,z) in R implies (x,z) in R.

   ✦ R is an equivalence relation.

8. 1.1.85 marksNCERT Class 12 Mathematics, Relations and Functions, Reprint 2026-27

   Show that the relation R in the set A={1,2,3,4,5} given by R={(a,b): |a-b| is even}, is an equivalence relation. Show that all the elements of {1,3,5} are related to each other and all the elements of {2,4} are related to each other, but no element of {1,3,5} is related to any element of {2,4}.

   Hint. '|a-b| is even' means a and b have the same parity — reduce every check to a parity argument.

   Reflexive: |a-a|=0, which is even. Symmetric: |a-b|=|b-a|, so parity is unaffected by order. Transitive: |a-b| even means a,b have the same parity; |b-c| even means b,c have the same parity; so a,c share that same parity, making |a-c| even too.

   {1,3,5} are all odd, so any two of them differ by an even number — mutually related. {2,4} are both even, differing by 2 — related. Any element of {1,3,5} (odd) and any element of {2,4} (even) differ by an odd number, so they are not related.

   ✦ R is an equivalence relation. {1,3,5} forms one equivalence class, {2,4} forms another, and no element of one class relates to an element of the other.

9. 1.1.96 marksNCERT Class 12 Mathematics, Relations and Functions, Reprint 2026-27

   Show that each of the relation R in the set A={x in Z: 0<=x<=12}, given by (i) R={(a,b): |a-b| is a multiple of 4} (ii) R={(a,b): a=b} is an equivalence relation. Find the set of all elements related to 1 in each case.

   Hint. For (i), reduce to a mod-4 argument the same way Q8 reduced to a parity argument. For (ii), equality is the simplest possible equivalence relation.

   (i) Reflexive: |a-a|=0, a multiple of 4. Symmetric: |a-b|=|b-a|. Transitive: if 4 divides (a-b) and 4 divides (b-c), then 4 divides their sum (a-b)+(b-c)=(a-c). So R is an equivalence relation. Elements related to 1 within {0,...,12}: need |a-1| to be a multiple of 4, i.e. a=1,5,9 (a=-3 and a=13 fall outside the set).

   (ii) a=b is reflexive, symmetric, and transitive trivially (it is plain equality). The only element related to 1 is 1 itself, since a=b=1 forces a=1.

   ✦ (i) Equivalence relation; set related to 1 is {1,5,9}. (ii) Equivalence relation; set related to 1 is {1}.

10. 1.1.106 marksNCERT Class 12 Mathematics, Relations and Functions, Reprint 2026-27

    Give an example of a relation. Which is (i) Symmetric but neither reflexive nor transitive. (ii) Transitive but neither reflexive nor symmetric. (iii) Reflexive and symmetric but not transitive. (iv) Reflexive and transitive but not symmetric. (v) Symmetric and transitive but not reflexive.

    Hint. Build each example on the small set {1,2,3}, choosing pairs deliberately so exactly the required properties hold.

    (i) R={(a,b): a is not equal to b} on {1,2,3}. Symmetric (a!=b implies b!=a). Not reflexive ((a,a) never satisfies a!=a). Not transitive (1!=2 and 2!=1, but 1=1, so (1,1) would be needed and is excluded — concretely 1!=2, 2!=1 does not force anything, but using values 1,2,1: 1!=2 and 2!=1 both hold while the 'chain' 1 to 1 needs 1!=1, false, so a required conclusion fails).

    (ii) R={(1,2),(2,3),(1,3)} on {1,2,3}. Not reflexive (no (a,a) present). Not symmetric ((1,2) present, (2,1) absent). Transitive: the only chain (1,2),(2,3) requires (1,3), which is present.

    (iii) R={(1,1),(2,2),(3,3),(1,2),(2,1),(2,3),(3,2)} on {1,2,3}. Reflexive (all three diagonal pairs present). Symmetric ((1,2)/(2,1) and (2,3)/(3,2) both present as pairs). Not transitive: (1,2) and (2,3) are in R, but (1,3) is not.

    (iv) R={(1,1),(2,2),(3,3),(1,2),(1,3),(2,3)} on {1,2,3} (essentially 'less than or equal to'). Reflexive (all diagonal pairs present). Transitive (the only nontrivial chain (1,2),(2,3) requires and has (1,3)). Not symmetric ((1,2) present, (2,1) absent).

    (v) R={(1,1),(2,2)} on {1,2,3} (a proper subset missing (3,3)). Symmetric (trivially, only diagonal-type pairs). Transitive (trivially, no cross pairs to chain). Not reflexive ((3,3) is missing, since 3 is in the set).

    ✦ Five worked examples given above, one for each combination of properties, all built on the set {1,2,3}.

11. 1.1.115 marksNCERT Class 12 Mathematics, Relations and Functions, Reprint 2026-27

    Show that the relation R in the set A of points in a plane given by R={(P,Q): distance of the point P from the origin is same as the distance of the point Q from the origin}, is an equivalence relation. Further, show that the set of all points related to a point P != (0,0) is the circle passing through P with origin as centre.

    Hint. This reduces entirely to properties of equality of distances, and the geometric picture of 'same distance from a fixed point' is a circle.

    Reflexive: distance of P from O equals distance of P from O. Symmetric: if dist(P,O)=dist(Q,O) then dist(Q,O)=dist(P,O). Transitive: if dist(P,O)=dist(Q,O) and dist(Q,O)=dist(S,O), then dist(P,O)=dist(S,O). So R is an equivalence relation.

    The set of points related to a fixed P (P not the origin) is every point at the same distance from the origin as P, i.e. every point on the circle of radius |OP| centred at the origin — and this circle passes through P itself.

    ✦ R is an equivalence relation; the equivalence class of P (P != origin) is the circle centred at the origin passing through P.

12. 1.1.125 marksNCERT Class 12 Mathematics, Relations and Functions, Reprint 2026-27

    Show that the relation R defined in the set A of all triangles as R={(T1,T2): T1 is similar to T2}, is equivalence relation. Consider three right angle triangles T1 with sides 3,4,5, T2 with sides 5,12,13 and T3 with sides 6,8,10. Which triangles among T1, T2 and T3 are related?

    Hint. Similarity of triangles is reflexive, symmetric, and transitive by its standard geometric definition; check which side ratios actually match.

    Similarity is reflexive (a triangle is similar to itself), symmetric (if T1 is similar to T2, T2 is similar to T1), and transitive (if T1T2 and T2T3, then T1~T3) — these are standard properties of geometric similarity, so R is an equivalence relation.

    T1 has sides 3,4,5. T3 has sides 6,8,10 = 2x(3,4,5), so T1 and T3 are similar (all ratios equal 2). T2 has sides 5,12,13; comparing to T1's 3,4,5, the ratios 5/3, 12/4=3, 13/5=2.6 are not all equal, so T2 is not similar to T1 or T3.

    ✦ R is an equivalence relation. T1 and T3 are related (similar); T2 is related to neither.

13. 1.1.134 marksNCERT Class 12 Mathematics, Relations and Functions, Reprint 2026-27

    Show that the relation R defined in the set A of all polygons as R={(P1,P2): P1 and P2 have same number of sides}, is an equivalence relation. What is the set of all elements in A related to the right angle triangle T with sides 3,4 and 5?

    Hint. 'Same number of sides' is another equality-of-attribute relation, handled the same way as Q7.

    Reflexive: a polygon has the same number of sides as itself. Symmetric: if P1 and P2 have the same side count, so do P2 and P1. Transitive: if P1,P2 match and P2,P3 match, then P1,P3 match. So R is an equivalence relation.

    T is a triangle, which has 3 sides. The set of all polygons related to T is the set of all triangles — every polygon with exactly 3 sides.

    ✦ R is an equivalence relation. The set related to T is the set of all triangles in A.

14. 1.1.144 marksNCERT Class 12 Mathematics, Relations and Functions, Reprint 2026-27

    Let L be the set of all lines in XY plane and R be the relation in L defined as R={(L1,L2): L1 is parallel to L2}. Show that R is an equivalence relation. Find the set of all lines related to the line y=2x+4.

    Hint. A line is taken to be parallel to itself by convention; use that plus the standard geometric facts about parallel lines.

    Reflexive: a line is parallel to itself (by convention). Symmetric: if L1 is parallel to L2, then L2 is parallel to L1. Transitive: if L1 is parallel to L2 and L2 is parallel to L3, then L1 is parallel to L3. So R is an equivalence relation.

    The line y=2x+4 has slope 2. The set of all lines related to it is every line with slope 2, i.e. every line of the form y=2x+c for some constant c.

    ✦ R is an equivalence relation. The related set is {y=2x+c : c is any real number}.

15. 1.1.153 marksNCERT Class 12 Mathematics, Relations and Functions, Reprint 2026-27

    Let R be the 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)}. Choose the correct answer. (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.

    Hint. Check the four diagonal pairs first, then look for a single pair whose reverse is missing, then verify every chain closes.

    Reflexive: (1,1),(2,2),(3,3),(4,4) are all present, so reflexive holds.

    Symmetric: (1,2) is in R, but (2,1) is not, so symmetric fails.

    Transitive: check every chain — (1,3)&(3,2) needs (1,2): present. (1,3)&(3,3) needs (1,3): present. (3,2)&(2,2) needs (3,2): present. (1,2)&(2,2) needs (1,2): present. All chains close, so transitive holds.

    ✦ (B) R is reflexive and transitive but not symmetric.

16. 1.1.163 marksNCERT Class 12 Mathematics, Relations and Functions, Reprint 2026-27

    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) (8,7) in R.

    Hint. Test each option against both conditions of the rule: a=b-2 and b>6.

    (A) (2,4): b=4 is not greater than 6 — fails. (B) (3,8): check a=b-2, i.e. 3=8-2=6? No — fails. (C) (6,8): a=b-2 gives 6=8-2=6, true; b=8>6, true — both conditions hold. (D) (8,7): b=7>6 holds, but a=b-2 gives 8=7-2=5? No — fails.

    ✦ (C) (6,8) is in R.

Solutions written by the tuition.in editorial team and checked against the NCERT Class 12 Mathematics textbook, Reprint 2026-27 (lemh101.pdf) — Exercise 1.1 (16 questions), Exercise 1.2 (12 questions), plus the chapter's Miscellaneous Exercise (7 questions), 35 questions total. The old stub had no real exercise coverage at all (only an invented five-question 'Self-Test') and taught an entire 'binary operations' section that has no actual teaching content anywhere in the current book — the book's introduction and summary paragraphs still mention it, but the section itself, along with any example or exercise question on it, has been removed from this edition. Every proof-based answer in this file was independently re-derived from the definitions (not copied from a key), including the finite-set onto-functions count (n!) via the chapter's own one-one-iff-onto shortcut, and Miscellaneous Q5's function-equality check verified by direct computation of both f and g at all four points of the domain.. Questions are referenced from the NCERT textbook for identification.