NCERT Solutions for Class 12 Maths Chapter 1 – Relations & Functions

Show that gof is onto but f is not onto.

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.

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.

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)

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).

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.

Since a*b = min{a,b}, each entry is the smaller of the row number and column number.

Since g∘f = I_R, 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 = I_R.

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).

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.

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)

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)}.