GSEB Solutions Class 12 Statistics Part 2 Chapter 4 Limit Ex 4.1

Gujarat Board Statistics Class 12 GSEB Solutions Part 2 Chapter 4 Limit Ex 4.1 Textbook Exercise Questions and Answers.

Gujarat Board Textbook Solutions Class 12 Statistics Part 2 Chapter 4 Limit Ex 4.1

Question 1.
Express the following in modulus and interval form:
(1) 0.4 neighbourhood of 4:
Answer:
Here, a = 4; δ = 0.4
In Modulus form: |x – a| < δ
∴ N(4, 0.4) = |x – 4| < 0.4
In Interval form:(a – δ, a + δ)
∴ N(4, 0.4) = (4 – 0.4, 4 + 0.4)
= (3.6, 4.4)

(2) 0.02 neighbourhood of 2:
Answer:
Here. a = 2; δ = 0.02
In Modulus form: |x – a| < δ
∴ N(2. 0.02) = | x – 2| < 0.02
In Interval form: (a – 8, a + δ)
∴ N (2, 0.02) = (2 – 0.02, 2 + 0.02)
= (1.98, 2.02)

(3) 0.05 neighbourhood of 0:
Answer:
Here, a = 0; δ = 0.05
In Modulus form : | x – a | < δ
∴ N (0, 0.05) = |x – 0| < 0.05
= | x | < 0.05
In Interval form : (a – δ, a + δ)
∴ N(0, 0.05) = (0 – 0.05, 0 + 0.05)
= (- 0.05, 0.05)

(4) 0.001 neighbourhood of – 1:
Answer:
Here, a = – 1; δ = 0.001
In Modulus form: | x – a | < δ
∴ N (- 1, 0.001) = |x – (- 1)| < 0.001
= |x + 1|< 0.001
In Interval form : (a – δ, a + δ)
∴ N (- 1, 0.001) = (- 1 – 0.001, – 1 + 0.001)
= (- 1.001, – 0.999)

Question 2.
Express the following in interval and neighbourhood form:
(1) |x – 2|< 0.01
Answer:
Here, a = 2; δ = 0.01
In Interval form: (a – δ, a + δ)
|x – 2|< 0.01 = (2 – 0.01, 2 + 0.01)
= (1.99, 2.01)
In Neighbourhood form : N (a, δ)
|x – 2| < 0.01 = N (2, 0.01)

(2) |x + 5|< 0.1
Answer:
Here, a = – 5; δ = 0.1
In Interval form : (a – δ, a + δ)
|x + 5|< 0.1 = (- 5 – 0.1, – 5 + 0.1)
= (- 5.1, – 4.9)
In Neighbourhood form: N (a, δ)
|x + 51 < 0.1 = N (- 5, 0.1)

(3) |x| < \(\frac{1}{3}\)
Answer:
Here, a = 0; δ = \(\frac{1}{3}\)
In Interval form : (a – δ, a + δ)
|x| < \(\frac{1}{3}\) = (0 – \(\frac{1}{3}\), 0 + \(\frac{1}{3}\)) = (- \(\frac{1}{3}\), \(\frac{1}{3}\))
In Neighbourhood form : N (a, δ)
|x| < \(\frac{1}{3}\) = N(0, \(\frac{1}{3}\))

(4) |x + 3| < 0.15
Answer:
Here, a = – 3: δ = 0.15
In Interval form : (a – δ, a + δ)
|x + 3| < 0.15 = (- 3 – 0.15, – 3 + 0.15)
= (- 3.15, – 2.85)
In Neighbourhood form : N (a, δ)
|x + 3| < 0.15 = N(-3, 0.15)

Question 3.
Express the following in modulus and neighbourhood form:
(1) 3.8 < x < 4.8
Answer:
Here, a – δ = 3.8; a + δ = 4.8
By adding, 2a = 8.6
∴ a = \(\frac{8.6}{2}\) = 4.3
Putting a = 4.3 in a + δ = 4.8,
4.3 + 8 = 4.8
∴ δ = 4.8 – 4.3 = 0.5

In Modulus form : | x – a| < δ
3.8 < x < 4.8 = |x – 4.31 < 0.5

In Neighbourhood form: N (a, δ)
3.8 < x < 4.8 = N (4.3, 0.5)

(2) 1.95 < x < 2.05
Answer:
Here, a – δ = 1.95; a + δ = 2.05
By adding, 2a = 1.95 + 2.05 = 4
∴ a = 2
Putting a = 2 in a + δ = 2.05,
δ = 0.05

In Modulus form: | x – a | < 8
Putting a = 2; δ = 0.05,
1.95 < x < 2.05 = |x – 2| < 0.05

In Neighbourhood form : N (a, δ)
Putting a = 2; δ = 0.05,
1.95 < x < 2.05 = N (2, 0.05)

(3) – 0.4 < x < 1.4
Answer:
Here, a – δ = – 0.4; a + δ = 1.4
By adding, 2a = 1
∴ a = \(\frac{1}{2}\) = 0.5
Putting a = 0.5 in a +δ = 1.4,
δ = 0.9

In Modulus form: |x – a| < δ
Putting a = 0.5; δ = 0.9,
-0.4 < x < 1.4 = |x – 0.51 < 0.9

In Neighbourhood form : N (a, <S)
Putting a = 0.5; 8 = 0.9,
– 0.4 < x < 1.4 = N (0.5, 0.9)

(4) 1.998 < x < 2.002
Answer:
Here, a – δ = 1.998; a + δ = 2.002
By adding, 2a = 4
∴ a = 2
Putting a = 2 in a + δ = 2.002,
δ = 0.002
In Modulus form: |x – a| < δ
Putting a = 2; δ = 0.002,
1.998 < x < 2.002 = |x – 2| < 0.002

In Neighbourhood form : N (a, δ)
Putting a = 2; δ = 0.002,
1.998 < x < 2.002 = N (2, 0.002)

Question 4.
Express N (16, 0.5) in the interval and modulus form.
Answer:
N (16, 0.5) ∴ a = 16, δ = 0.5

In Interval form: (a – δ, a + δ)
Putting a = 16; δ = 0.5,
N (16, 0.5) = (16 – 0.5, 16 + 0.5)
= (15.5, 16.5)

In Modulus form: |x – a| < δ
Putting a = 16; δ = 0.5,
N (16, 0.5) = |x – 16| < 0.5

Question 5.
If N (3, b) = (2.95, k), then find the values of b and k.
Answer:
N (3, b) = (2.95, k)
∴ (a = 3, δ = b) = (a – δ= 2.95, a + δ = k)
Putting a = 3 in a – δ = 2.95,
3 – δ = 2.95
∴ 3 – 2.95 = δ
δ = 0.05
But δ = b ∴ b = 0.05
Now, a + δ = k
Putting a = 3; δ = 0.05,
3 + 0.05 = k
∴ k = 3.05
Hence, b = 0.05 and k = 3.05

Question 6.
If |x – 10| < k₁ = (k₂, 10.01), then find the values of k₁ and k₂.
Answer:
|x – 10| < k₁ = (k₂, 10.01)
∴ (10 – k₁, 10 + k₁) = (k₂, 10.01)
∴ 10 + k₁ = 10.01
∴ k₁ = 10.01 – 10 = 0.01
Now, putting k₁ = 0.01 in 10 – k₁ = k₂
10 – 0.01 = k₂
∴ k₂ = 9.99
Hence, k₁ = 0.01 and k₂ = 9.99