Miscellaneous Exercise — Relations and Functions

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

   Show that the function f: R -> {x in R: -1<x<1} defined by f(x)=x/(1+|x|), x in R is one-one and onto function.

   Hint. Split into cases x>=0 and x<0, since |x| behaves differently in each; for onto, solve y=f(x) for x separately in each case.

   One-one: for x>=0, f(x)=x/(1+x), which is strictly increasing, so distinct non-negative x give distinct outputs. For x<0, f(x)=x/(1-x), also strictly increasing (can be checked by the same monotonicity argument), so distinct negative x give distinct outputs. A non-negative x and a negative x can never share an image, since f(x)>=0 for x>=0 and f(x)<0 for x<0 (equal only at x=0, which is unique to that case). So f is one-one across all of R.

   Onto: given y in (-1,1), if y>=0, solve x/(1+x)=y for x>=0, giving x=y/(1-y), which is well-defined and non-negative since 0<=y<1. If y<0, solve x/(1-x)=y for x<0, giving x=y/(1+y), well-defined and negative since -1<y<0. Either way, a preimage in R exists.

   ✦ f is one-one and onto.

2. 1.M.23 marksNCERT Class 12 Mathematics, Relations and Functions, Reprint 2026-27

   Show that the function f: R -> R given by f(x)=x^3 is injective.

   Hint. Factor x1^3-x2^3 and examine when the quadratic factor can vanish.

   Suppose f(x1)=f(x2), i.e. x1^3=x2^3, so x1^3-x2^3=0. Factoring: (x1-x2)(x1^2+x1x2+x2^2)=0. The second factor, x1^2+x1x2+x2^2, can be rewritten as (x1+x2/2)^2+3x2^2/4, a sum of squares that is zero only when x1=x2=0. In every other case the second factor is strictly positive, forcing x1-x2=0, i.e. x1=x2. Either way, x1=x2.

   ✦ f is injective.

3. 1.M.34 marksNCERT Class 12 Mathematics, Relations and Functions, Reprint 2026-27

   Given a non empty set X, consider P(X) which is the set of all subsets of X. Define the relation R in P(X) as follows: For subsets A, B in P(X), ARB if and only if A is a subset of B. Is R an equivalence relation on P(X)? Justify your answer.

   Hint. Check all three properties honestly — subset containment is not a two-way relationship in general.

   Reflexive: A is a subset of A always, so this holds. Transitive: if A is a subset of B and B is a subset of C, then A is a subset of C, so this holds. Symmetric: A being a subset of B does not generally mean B is a subset of A — for example, {1} is a subset of {1,2}, but {1,2} is not a subset of {1}.

   ✦ No, R is not an equivalence relation, since it fails symmetry (it is reflexive and transitive but not symmetric).

4. 1.M.44 marksNCERT Class 12 Mathematics, Relations and Functions, Reprint 2026-27

   Find the number of all onto functions from the set {1,2,3,...,n} to itself.

   Hint. Use the chapter's own finite-set fact: for a function from a finite set to itself, onto is the same condition as one-one.

   Since the domain and co-domain are the same finite set {1,...,n}, a function from this set to itself is onto if and only if it is one-one (the finite-set shortcut proved earlier in the chapter). So counting onto functions is the same as counting bijections, which is the same as counting permutations of n objects.

   ✦ n! (n factorial).

5. 1.M.55 marksNCERT Class 12 Mathematics, Relations and Functions, Reprint 2026-27

   Let A={-1,0,1,2}, B={-4,-2,0,2} and f,g: A -> B be functions defined by f(x)=x^2-x, x in A and g(x)=2|x-1/2|-1, x in A. Are f and g equal? Justify your answer.

   Hint. Two functions are equal exactly when they agree at every point of their shared domain — compute both at all four points of A.

   f(-1)=(-1)^2-(-1)=1+1=2. f(0)=0-0=0. f(1)=1-1=0. f(2)=4-2=2.

   g(-1)=2|-1-0.5|-1=2(1.5)-1=2. g(0)=2|0-0.5|-1=2(0.5)-1=0. g(1)=2|1-0.5|-1=2(0.5)-1=0. g(2)=2|2-0.5|-1=2(1.5)-1=2.

   Comparing: f(-1)=g(-1)=2, f(0)=g(0)=0, f(1)=g(1)=0, f(2)=g(2)=2. The two functions agree at every point of A.

   ✦ Yes, f and g are equal, since f(a)=g(a) for every a in A.

6. 1.M.63 marksNCERT Class 12 Mathematics, Relations and Functions, Reprint 2026-27

   Let A={1,2,3}. Then number of relations containing (1,2) and (1,3) which are reflexive and symmetric but not transitive is (A) 1 (B) 2 (C) 3 (D) 4.

   Hint. Work out the smallest relation forced by the requirements, then check whether it is already non-transitive, and whether any larger valid relation exists.

   Reflexive forces (1,1),(2,2),(3,3) into R. Since (1,2) is required, symmetry forces (2,1) in too; since (1,3) is required, symmetry forces (3,1) in too. This gives the smallest possible relation R0={(1,1),(2,2),(3,3),(1,2),(2,1),(1,3),(3,1)}, 7 pairs. Checking transitivity of R0: (2,1) and (1,3) are in R0, but (2,3) is not — so R0 is already not transitive, satisfying the requirement.

   The only pairs not yet in R0 are (2,3) and (3,2); by symmetry these can only be added together, never alone. Adding both makes R the full universal relation {1,2,3}x{1,2,3}, which is transitive — disqualified. So R0 is the only relation meeting every condition.

   ✦ (A) 1.

7. 1.M.73 marksNCERT Class 12 Mathematics, Relations and Functions, Reprint 2026-27

   Let A={1,2,3}. Then number of equivalence relations containing (1,2) is (A) 1 (B) 2 (C) 3 (D) 4.

   Hint. Since an equivalence relation containing (1,2) must also contain (2,1) by symmetry, this reduces to counting equivalence relations containing both (1,2) and (2,1) together.

   For an equivalence relation, containing (1,2) forces (2,1) in as well (symmetry). The smallest such equivalence relation is R1={(1,1),(2,2),(3,3),(1,2),(2,1)}, keeping 3 as its own separate class. The only other possibility is merging 3 into the same class as 1 and 2, which forces the universal relation {1,2,3}x{1,2,3} (also an equivalence relation). No relation strictly between these two can stay an equivalence relation, since partially connecting 3 to only one of 1,2 would break symmetry or transitivity.

   ✦ (B) 2.

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.