Relations & Functions is an important chapter in Class 12 Maths Chapter 1 and lays the foundation for higher-level topics like calculus, matrices, and mapping-based questions.
This chapter focuses on key concepts such as relations, types of functions (one-one, onto, invertible), composition of functions, and properties of relations, explained using clear logic and proofs.
These NCERT Solutions for Class 12 Maths Chapter 1 – Relations & Functions include important board-level questions and are explained step by step in simple language to help students build strong conceptual clarity and score well in examinations.
NCERT Solutions for Class 12 Maths Chapter 1 – Relations & Functions
Q.
LetrmArm=rmR−{3}rmandrmBrm=rmR−{1}.rmConsiderrmthermfunctionrmf:rmA→Brmdefinedrmbyf(x)=x−3x−2.rmIsrmfrmone−onermandrmonto?rmJustifyrmyourrmanswer.
Q.
Determine whether or not each of the definition of given below gives a binary operation. In the event that * is not a binary operation, give justification for this.
(i) On Z+, define * by a * b = a − b
(ii) On Z+, define * by a * b = ab
(iii) On R, define * by a * b = ab2
(iv) On Z+, define * by a * b = |a − b|
(v) On Z+, define * by a * b = a
Q.
Given two functions f:N→N and g:N→N suchthat f(x)=x+1and g(x)={x−1, ifx>1x+1, ifx<1. Show that
gof is onto but
f is not onto.
Q.
Let f:R→R be defined as f(x)=10x+7.Find the functiong:R→R such that go f=fog=IR.
Q.
LetA=N×Nand∗bethebinaryoperationonAdefinedby(a,b)∗(c,d)=(a+c,b+d)Showthat∗iscommutativeandassociative.Findtheidentityelementfor∗onA,ifany.
Q.
Let * be the binary operation on N defined by
a * b = H.C.F. of a and b. Is * commutative? Is * associative? Does there exist identity for this binary operation on N?
Q.
Let∗bethebinaryoperationonNgivenbya∗b=L.C.M.ofaandb.Find(i)5∗7,20∗16(ii)Is∗commutative?(iii)Is∗associative?(iv)Findtheidentityof∗inN(v)WhichelementsofNareinvertiblefortheoperation∗?
Q.
Consider the binary operation and on theset {1, 2, 3, 4, 5}definedby aandb=min{a,b}.Write the operation tableof the operation and.
Q.
Foreachbinaryoperation∗definedbelow,determinewhether∗iscommutativeorassociative.(i)OnZ,definea∗b=a−b(ii)OnQ,definea∗b=ab+1(iii)OnQ,definea∗b=2ab(iv)OnZ+,definea∗b=2ab(v)OnZ+,definea∗b=ab(vi)OnR−{−1},definea∗b=b+1a
Q.
Letf:R−{−34}→Rbeafunctiondefinedasf(x)=3x+44x.Theinverseoffismapg:Rangef→R−{−34}givenby(A)g(y)=3−4y3y(B)g(y)=4−3y4y(C)g(y)=3−4y4y(D)g(y)=4−3y3y
Q.
Letf:R→Rbedefinedasf(x)=x4.Choosethecorrectanswer.(A)fisone−oneonto(B)fismany−oneonto.(C)fisone−onebutnotonto(D)fisneitherone−onenoronto.
Q.
Iff:R→Rbegivenby,f(x)=(3−x3)31thenfof(x)is(A)x31(B)x3(C)x(D)(3−x3).
Q.
Letf:X→Ybeaninvertiblefunction.Showthattheinverseoff−1isf,i.e.,(f−1)−1=f.
Q.
Let f :X→Y be an invertible function.Show that f has unique inverse.
Q.
Considerrmf:rmR+→rm[−5,rm∞)rmgivenrmbyrmf(x)rm=rm9x2rm+rm6x−5.rmShowrmthatrmfrmisrminvertiblermwithrmf−1(y)=(3(y+6)−1).
Q.
Considerf:R→Rgivenbyf(x)=4x+3.Showthatfisinvertible.Findtheinverseoff.
Q.
Let f, g and h be functions from R to R. Show that
(f + g)oh = foh + goh
(f. g)oh = (foh) . (goh)
Q.
Letf:{1,3,4}→{1,2,5}andg:{1,2,5}→{1,3}begivenbyf={(1,2),(3,5),(4,1)}andg={(1,3),(2,3),(5,1)}.Writedowngof.
Q.
Letf:R→Rbedefinedasf(x)=3x.Choosethecorrectanswer.(A)fisone−oneonto(B)fismany−oneonto(C)fisone−onebutnotonto(D)fisneitherone−onenoronto.
Q.
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
A⊂B. Is R an equivalence relatio on P(X)? Justify your answer.
NCERT Solutions for Class 12 Maths Chapter 1 – Relations & Functions
Q1.
Given two functions f:N→N and g:N→N such that f(x)=x+1 and
g(x)=x−1, if x>1x+1, if x<1.
Show that gof is onto but f is not onto.
Ans:
Given f(x)=x+1.
Step 1: Show that f is not onto
Range of f is {2,3,4,...} because for any x∈N, f(x)=x+1 ≥ 2.
But 1 ∈ N and there is no x ∈ N such that f(x)=1 (since x+1=1 gives x=0, not in N).
Hence, f is not onto.
Step 2: Find (g∘f)(x)
(g∘f)(x)=g(f(x))=g(x+1).
For x∈N, x+1 > 1 always, so we use g(t)=t−1 when t>1.
Therefore, g(x+1)=(x+1)−1=x.
So, (g∘f)(x)=x for all x∈N.
Step 3: Show that g∘f is onto
Let y∈N. Choose x=y. Then (g∘f)(x)=x=y.
So every y∈N has a preimage.
Hence, g∘f is onto.
Q2.
Let A = R − {3} and B = R − {1}. Consider the function f:A→B defined by
f(x) = (x − 1)/(x − 3). Is f one-one and onto? Justify your answer.
Ans:
Given f(x)= (x−1)/(x−3), domain A=R−{3}, codomain B=R−{1}.
1) Check one-one (injective)
Assume f(x1)=f(x2). Then
(x1−1)/(x1−3) = (x2−1)/(x2−3).
Cross-multiplying:
(x1−1)(x2−3) = (x2−1)(x1−3).
Expand both sides:
x1x2 − 3x1 − x2 + 3 = x1x2 − 3x2 − x1 + 3
Cancel x1x2 and 3 from both sides:
−3x1 − x2 = −3x2 − x1
Bring like terms together:
−3x1 + x1 = −3x2 + x2
−2x1 = −2x2
x1 = x2
Hence, f is one-one.
2) Check onto (surjective)
Let y ∈ B (i.e., y ≠ 1). We must find x ∈ A such that f(x)=y.
y = (x−1)/(x−3)
⇒ y(x−3)=x−1
⇒ yx − 3y = x − 1
⇒ yx − x = 3y − 1
⇒ x(y−1)=3y−1
⇒ x = (3y−1)/(y−1).
Since y ≠ 1, denominator (y−1) ≠ 0, so x is defined.
Now check x ≠ 3 (so that x ∈ A):
If x=3, then (3y−1)/(y−1)=3
⇒ 3y−1 = 3y−3, which gives −1 = −3 (impossible).
So x ≠ 3 always.
Therefore, for every y ∈ B, there exists x ∈ A such that f(x)=y.
Hence, f is onto.
Conclusion: f is one-one and onto.
Q3.
Let f, g and h be functions from R to R. Show that
(f + g)∘h = f∘h + g∘h
(f · g)∘h = (f∘h) · (g∘h)
Ans:
Let x ∈ R.
(i) Prove (f+g)∘h = f∘h + g∘h
LHS:
((f+g)∘h)(x) = (f+g)(h(x))
= f(h(x)) + g(h(x))
RHS:
(f∘h + g∘h)(x) = (f∘h)(x) + (g∘h)(x)
= f(h(x)) + g(h(x))
Hence, LHS = RHS. Therefore, (f + g)∘h = f∘h + g∘h.
(ii) Prove (f·g)∘h = (f∘h)·(g∘h)
LHS:
((f·g)∘h)(x) = (f·g)(h(x))
= f(h(x)) · g(h(x))
RHS:
((f∘h)·(g∘h))(x) = (f∘h)(x) · (g∘h)(x)
= f(h(x)) · g(h(x))
Hence, LHS = RHS. Therefore, (f·g)∘h = (f∘h)·(g∘h).
Q4.
Consider f:R+ → [−5, ∞) given by f(x)=9x²+6x−5.
Show that f is invertible with f⁻¹(y) = ( √(y+6) − 1 ) / 3.
Ans:
Given f(x)=9x²+6x−5, where x ∈ R+.
Step 1: Show f is one-one on R+
f'(x)=18x+6.
For x>0, 18x+6 > 0, so f is strictly increasing on R+.
Hence, f is one-one on R+ and therefore invertible on its range.
Step 2: Find inverse
Put y = 9x²+6x−5.
y+5 = 9x²+6x
Complete the square:
9x²+6x = 9(x² + (2/3)x)
= 9[(x + 1/3)² − (1/9)]
= 9(x + 1/3)² − 1
So, y+5 = 9(x + 1/3)² − 1
⇒ y+6 = 9(x + 1/3)²
⇒ (y+6)/9 = (x + 1/3)²
Since x ∈ R+, take positive square root:
√(y+6)/3 = x + 1/3
⇒ x = ( √(y+6) − 1 )/3
Therefore, f⁻¹(y) = ( √(y+6) − 1 ) / 3.
Q5.
Consider the binary operation * on the set {1,2,3,4,5} defined by
a*b = min{a,b}. Write the operation table of *.
Ans:
Since a*b = min{a,b}, each entry is the smaller of the row number and column number.
| * |
1 |
2 |
3 |
4 |
5 |
| 1 |
1 |
1 |
1 |
1 |
1 |
| 2 |
1 |
2 |
2 |
2 |
2 |
| 3 |
1 |
2 |
3 |
3 |
3 |
| 4 |
1 |
2 |
3 |
4 |
4 |
| 5 |
1 |
2 |
3 |
4 |
5 |
Q6.
Let f:R→R be defined as f(x)=10x+7. Find g:R→R such that g∘f = f∘g = IR.
Ans:
Since g∘f = IR, g is the inverse of f.
Let y = f(x) = 10x+7.
Solve for x:
10x = y−7
x = (y−7)/10
Therefore, f⁻¹(y) = (y−7)/10.
Hence, g(x) = (x−7)/10.
Now check:
g(f(x)) = g(10x+7) = (10x+7−7)/10 = x.
f(g(x)) = f((x−7)/10) = 10((x−7)/10)+7 = x.
So, g∘f = f∘g = IR.
Q7.
Given a non-empty set X, consider P(X) (set of all subsets of X). Define relation R in P(X) as:
For A, B ∈ P(X), ARB if and only if A ⊂ B.
Is R an equivalence relation on P(X)? Justify your answer.
Ans:
For R to be an equivalence relation, it must be reflexive, symmetric, and transitive.
Reflexive: For any A, we must have ARA i.e., A ⊂ A.
But A ⊂ A is false (a set is not a proper subset of itself).
So, R is not reflexive.
Since reflexive property fails, R cannot be an equivalence relation.
Conclusion: R is not an equivalence relation on P(X).
Q8.
Let f:R→R be defined as f(x)=x⁴. Choose the correct answer:
(A) f is one-one onto
(B) f is many-one onto
(C) f is one-one but not onto
(D) f is neither one-one nor onto
Ans:
f(x)=x⁴.
Not one-one: Because f(1)=1 and f(−1)=1, but 1 ≠ −1. So f is many-one.
Not onto: Range of x⁴ is [0, ∞). But codomain is R, so negative numbers are not achieved.
Hence, f is not onto.
Therefore, the correct option is (D) f is neither one-one nor onto.
Q9.
Let f:R→R be defined as f(x)=3x. Choose the correct answer:
(A) f is one-one onto
(B) f is many-one onto
(C) f is one-one but not onto
(D) f is neither one-one nor onto
Ans:
f(x)=3x.
One-one: If 3x1=3x2 then x1=x2.
Onto: For any y ∈ R, choose x = y/3 ∈ R. Then f(x)=3(y/3)=y.
Hence, f is one-one and onto.
Correct option: (A)
Q10.
Let f:{1,3,4}→{1,2,5} and g:{1,2,5}→{1,3} be given by
f={(1,2),(3,5),(4,1)} and g={(1,3),(2,3),(5,1)}. Write down g∘f.
Ans:
We need g(f(x)) for each x in domain {1,3,4}.
f(1)=2 ⇒ g(2)=3, so (g∘f)(1)=3
f(3)=5 ⇒ g(5)=1, so (g∘f)(3)=1
f(4)=1 ⇒ g(1)=3, so (g∘f)(4)=3
Therefore,
g∘f = {(1,3),(3,1),(4,3)}.
FAQs – Relations & Functions
1) One-one aur onto function mein difference kya hai?
One-one: different inputs → different outputs. Onto: codomain ka har element image hota hai.
2) Invertible function kaise pehchane?
Function invertible tab hota hai jab wo one-one ho (aur given codomain/range ke hisaab se inverse define ho sake).
3) Composition (g∘f) ka matlab kya hota hai?
Pehle f apply hota hai, phir uske output par g: (g∘f)(x)=g(f(x)).
4) Equivalence relation ke 3 conditions kya hain?
Reflexive, Symmetric, Transitive — teeno satisfy kare tab equivalence relation hota hai.
5) x⁴ function R→R onto kyun nahi hota?
Kyuki x⁴ ka output kabhi negative nahi aata, isliye R ke negative numbers cover nahi hote.