NCERT Class 11 Maths Chapter 2 Relations and Functions Miscellaneous Exercise Solutions

Answer 18 Here f(0) = 10, f(1) = 11, f(2) = 12, …, f(10) = 20, etc., and f(–1) = 9, f(–2) = 8, …, f(–10) = 0 and so on.

Therefore, shape of the graph of the given function assumes the form as shown in figure

Answer 19. This problem asks you to prove three specific properties for a relation R defined on the set of rational numbers Q. The condition for two numbers to be related is that their difference must be an integer.

Step-by-step Solution:

(i) Prove that (a,a) ∈ R for all a ∈ Q

This is known as the Reflexive Property.

To check if (a,a) is in the relation, we look at the difference: a-a

• a−a = 0
• Since 0 is an integer ( 0∈Z ), the condition for the relation is satisfied.

Conclusion: (a,a)∈R for all a∈Q.

(ii) Prove that (a,b)∈R implies (b,a)∈R

This is known as the Symmetric Property.

• Assume (a,b)∈R. This means that a−b is an integer. Let a−b = k, where k∈Z.
• To check whether (b,a)∈R, consider the difference: b−a=−(a−b)=−k
• Since k is an integer, its negative (-k) is also an integer, the condition is satisfied.

Conclusion:
Because their difference is an integer, (b,a)∈R.

(iii) Prove that (a,b)∈R and (b,c)∈R implies (a,c)∈R

This is known as the Transitive Property.

• Assume (a,b)∈R and (b,c)∈R.
• This means:
  • a−b is an integer, let’s call it k₁ and b – c is an integer, let us call it k₂
• We need to check if (a, c) is in the relation by looking at a – c.
• We can write a-c as: (a-b) + (b-c)
• Substituting our integers: a-c = k₁ + k₂
• Since the sum of two integers k₁ + k₂ is always an integer, the condition is satisfied.

Conclusion : (a,c)∈R.

Final Conclusion

Since the relation R satisfies all three properties:

• Reflexive
• Symmetric
• Transitive

The relation R is an Equivalence Relation.

Answer 20 Since f is a linear function, f (x) = mx + c. Also, since (1, 1), (0, – 1) ∈ R, f(1) = m + c = 1 and f (0) = c = –1. This gives m = 2 and f(x) = 2x – 1.

Answer 21 Since x² – 5x + 4 = (x – 4) (x –1), the function f is defined for all real numbers except at x = 4 and x = 1.

Hence the domain of f is R – {1, 4}.

Answer 22

Here, f(x) = 1 – x, x < 0, this gives

f(– 4) = 1 – (– 4) = 5

f(– 3) =1 – (– 3) = 4,

f(– 2) = 1 – (– 2) = 3

f(–1) = 1 – (–1) = 2; etc,

and f(1) = 2, f (2) = 3, f (3) = 4

f(4) = 5 and so on for f(x) = x + 1, x > 0.

Thus, the graph of f is as shown in figure

Answer 1.

For a relation to be a function, every input x must have exactly one unique output y.

The domain of f is [0, 10].

For each x ∈ [0, 3), f(x) = x² is uniquely defined, i.e. each x ∈ [0, 3) has a unique image.

For each x ∈ (3, 10], f(x) = 3x is uniquely defined.

At x = 3, x² = 3² = 9. Also, 3x = 3 × 3 = 9, so that f is uniquely defined at 3.

Since every point of the domain has one and only one image under f, this relation is a function.

The domain of g is [0, 10].

For each x ∈ [0, 2), g(x) = x² is uniquely defined.

For each x ∈ (2, 10], g(x) = 3x is uniquely defined.

At x = 2, g(x) = x² = 2² = 4. Also, at x = 2, g(x) = 3x = 3 × 2 = 6.

Now, 2 is in the domain of g and the image of 2 under g is not unique since (2, 4), (2, 6) ∈ g.

Therefore, g is not a function.

Answer 2.

Answer 3.

Factor the numerator:
x² + 2x + 1 = (x + 1)²

Factor the denominator:
x² − 8x + 12 = (x − 2)(x − 6)

The function is not defined when the denominator is zero.

x − 2 = 0 ⇒ x = 2
x − 6 = 0 ⇒ x = 6

Domain of the function:

All real numbers except 2 and 6.

In interval notation:
(−∞, 2) ∪ (2, 6) ∪ (6, ∞)

Answer 4.

Given:
f(x) = √(x − 1)

Domain:

For the square root to be defined, the expression inside the root must be greater than or equal to zero.

x − 1 ≥ 0

x ≥ 1

So, the domain is:
[1, ∞)

Range:

The square root of any non-negative number is always non-negative.

√(x − 1) ≥ 0

So, the range is:
[0, ∞)

Final Answer:
Domain: [1, ∞)
Range: [0, ∞)

Answer 7.

(f + g)(x) = f(x) + g(x)
= (x + 1) + (2x − 3)
= 3x − 2

(f − g)(x) = f(x) − g(x)
= (x + 1) − (2x − 3)
= −x + 4

(f / g)(x) = f(x) / g(x), g(x) ≠ 0

= (x + 1) / (2x − 3), 2x − 3 ≠ 0

So,
(f / g)(x) = (x + 1) / (2x − 3), x ≠ 3/2

Answer 8.

Given:
f(x) = ax + b …(i)

Now, (1, 1) ∈ f
⇒ f(1) = 1

Replacing x by 1 in (i), we have
f(1) = a × 1 + b
⇒ 1 = a + b …(ii)

Also, (2, 3) ∈ f
⇒ f(2) = 3

Replacing x by 2 in (i), we have
f(2) = a × 2 + b
⇒ 3 = 2a + b …(iii)

Subtracting (ii) from (iii), we get
2 = a

Putting this value of a in (ii), we get
1 = 2 + b

or
b = −1

Therefore:
a = 2, b = −1

Answer 9.

(i) (a, a) ∈ R for all a ∈ N

For (a, a) to belong to R, we must have
a = a²

This gives
a² − a = 0
a(a − 1) = 0

So, a = 0 or a = 1.

Thus, the condition is not true for all natural numbers.

Conclusion:
Statement (i) is false.

(ii) (a, b) ∈ R ⇒ (b, a) ∈ R

Given (a, b) ∈ R,
a = b²

For (b, a) ∈ R, we must have
b = a²

Substituting a = b²,
b = (b²)² = b⁴

This is not true for all b ∈ N.

Example:
(4, 2) ∈ R since 4 = 2²
But (2, 4) ∉ R since 2 ≠ 4²

Conclusion:
Statement (ii) is false.

(iii) (a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R

Given:
(a, b) ∈ R ⇒ a = b²
(b, c) ∈ R ⇒ b = c²

Substituting b = c² in a = b²,
a = (c²)² = c⁴

For (a, c) ∈ R, we need
a = c²

But c⁴ ≠ c² in general.

Example:
(16, 4) ∈ R and (4, 2) ∈ R
But (16, 2) ∉ R

Conclusion:
Statement (iii) is false.