GSEB Class 12 Statistics Solutions Chapter 4 Limit Exercise 4.1

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: For this question, the center 'a' is 4 and the radius 'δ' is 0.4. In modulus form, a neighborhood is written as \(|x - a| < \delta\). So, for N(4, 0.4), it is \(|x - 4| < 0.4\). In interval form, it is written as \((a - \delta, a + \delta)\). Therefore, for N(4, 0.4), the interval is \((4 - 0.4, 4 + 0.4)\), which simplifies to \((3.6, 4.4)\).
In simple words: A neighborhood of 4 with radius 0.4 means numbers very close to 4. In modulus form, this is \(|x - 4| < 0.4\), and as an interval, it is from 3.6 to 4.4.

🎯 Exam Tip: Always clearly identify 'a' (center) and 'δ' (radius) first. This makes converting between modulus and interval forms straightforward.

 

Question 1.
(2) 0.02 neighbourhood of 2:
Answer: Here, the center 'a' is 2 and the radius 'δ' is 0.02. The modulus form is \(|x - a| < \delta\), so N(2, 0.02) becomes \(|x - 2| < 0.02\). The interval form is \((a - \delta, a + \delta)\). Thus, for N(2, 0.02), the interval is \((2 - 0.02, 2 + 0.02)\), which results in \((1.98, 2.02)\).
In simple words: For a neighborhood around 2 with a radius of 0.02, the modulus form is \(|x - 2| < 0.02\). This means numbers between 1.98 and 2.02.

🎯 Exam Tip: Remember that modulus form \((|x-a| < \delta)\) and interval form \(((a-\delta, a+\delta))\) are two ways to represent the same concept of a neighborhood.

 

Question 1.
(3) 0.05 neighbourhood of 0:
Answer: In this case, the center 'a' is 0 and the radius 'δ' is 0.05. The modulus form \(|x - a| < \delta\) translates to \(|x - 0| < 0.05\), which simplifies to \(|x| < 0.05\). The interval form \((a - \delta, a + \delta)\) for N(0, 0.05) is \((0 - 0.05, 0 + 0.05)\), leading to \((-0.05, 0.05)\).
In simple words: A neighborhood of 0 with radius 0.05 is written as \(|x| < 0.05\) in modulus form. In interval form, it means all numbers between -0.05 and 0.05.

🎯 Exam Tip: When the center 'a' is 0, the modulus form simplifies to \(|x| < \delta\), making calculations easier.

 

Question 1.
(4) 0.001 neighbourhood of - 1:
Answer: Here, the center 'a' is -1 and the radius 'δ' is 0.001. Using the modulus form \(|x - a| < \delta\), N(-1, 0.001) becomes \(|x - (-1)| < 0.001\), which is \(|x + 1| < 0.001\). In interval form \((a - \delta, a + \delta)\), N(-1, 0.001) is \((-1 - 0.001, -1 + 0.001)\), resulting in \((-1.001, -0.999)\).
In simple words: A neighborhood around -1 with radius 0.001 is written as \(|x + 1| < 0.001\). This means numbers very close to -1, specifically from -1.001 to -0.999.

🎯 Exam Tip: Be careful with negative values for 'a'; \(x - (-a)\) becomes \(x + a\).

 

Question 2. Express the following in interval and neighbourhood form:
(1) \(|x - 2| < 0.01\)
Answer: For this expression, the center 'a' is 2 and the radius 'δ' is 0.01. To convert it into interval form \((a - \delta, a + \delta)\), we get \((2 - 0.01, 2 + 0.01)\), which is \((1.99, 2.01)\). In neighbourhood form, which is N(a, δ), the expression \(|x - 2| < 0.01\) becomes N(2, 0.01).
In simple words: The expression \(|x - 2| < 0.01\) means 'x' is less than 0.01 units away from 2. This creates an interval from 1.99 to 2.01, and is called the neighborhood N(2, 0.01).

🎯 Exam Tip: When converting from modulus to interval, remember to subtract delta from 'a' for the lower bound and add delta to 'a' for the upper bound.

 

Question 2.
(2) \(|x + 5| < 0.1\)
Answer: Here, the value for 'a' is -5 and 'δ' is 0.1. To write this in interval form, which is \((a - \delta, a + \delta)\), we calculate \((-5 - 0.1, -5 + 0.1)\). This results in the interval \((-5.1, -4.9)\). In neighbourhood form, expressed as N(a, δ), the given expression \(|x + 5| < 0.1\) translates to N(-5, 0.1).
In simple words: The expression \(|x + 5| < 0.1\) shows that 'x' is close to -5. The interval is from -5.1 to -4.9, and the neighborhood is N(-5, 0.1).

🎯 Exam Tip: Note that \(|x + 5|\) means \(|x - (-5)|\), so 'a' is -5, not 5.

 

Question 2.
(3) \(|x| < \frac{1}{3}\)
Answer: For this problem, 'a' is 0 and 'δ' is \( \frac{1}{3} \). The interval form, \((a - \delta, a + \delta)\), means \((0 - \frac{1}{3}, 0 + \frac{1}{3})\), which simplifies to \((-\frac{1}{3}, \frac{1}{3})\). In neighbourhood form, N(a, δ), the expression \(|x| < \frac{1}{3}\) is written as N(0, \( \frac{1}{3} \)).
In simple words: \(|x| < \frac{1}{3}\) means 'x' is less than \( \frac{1}{3} \) unit away from 0. This gives an interval from \(-\frac{1}{3}\) to \( \frac{1}{3} \) and is the neighborhood N(0, \( \frac{1}{3} \)).

🎯 Exam Tip: Fractions should be handled carefully in calculations. Ensure the final interval is simplified.

 

Question 2.
(4) \(|x + 3| < 0.15\)
Answer: In this example, the center 'a' is -3 and the radius 'δ' is 0.15. For the interval form \((a - \delta, a + \delta)\), we have \((-3 - 0.15, -3 + 0.15)\), which results in \((-3.15, -2.85)\). In neighbourhood form, N(a, δ), the inequality \(|x + 3| < 0.15\) is represented as N(-3, 0.15).
In simple words: \(|x + 3| < 0.15\) means 'x' is near -3, with a distance less than 0.15. This forms the interval \((-3.15, -2.85)\) and the neighborhood N(-3, 0.15).

🎯 Exam Tip: Always convert \(|x+c|\) to \(|x-(-c)|\) to find 'a' correctly, especially when 'c' is positive.

 

Question 3. Express the following in modulus and neighbourhood form:
(1) \(3.8 < x < 4.8\)
Answer: Given the interval \(3.8 < x < 4.8\), we know that \(a - \delta = 3.8\) and \(a + \delta = 4.8\). Adding these two equations gives \(2a = 8.6\), so \(a = \frac{8.6}{2} = 4.3\). Substituting \(a = 4.3\) into \(a + \delta = 4.8\), we get \(4.3 + \delta = 4.8\), which means \(\delta = 4.8 - 4.3 = 0.5\). In modulus form, \(|x - a| < \delta\), so the interval \(3.8 < x < 4.8\) becomes \(|x - 4.3| < 0.5\). In neighbourhood form, N(a, δ), the interval \(3.8 < x < 4.8\) is expressed as N(4.3, 0.5).
In simple words: If 'x' is between 3.8 and 4.8, the center 'a' is 4.3 and the radius 'δ' is 0.5. So, in modulus form, it's \(|x - 4.3| < 0.5\), and in neighborhood form, it's N(4.3, 0.5).

🎯 Exam Tip: To find 'a' from an interval \((L, U)\), use \(a = \frac{L+U}{2}\). To find 'δ', use \(\delta = \frac{U-L}{2}\).

 

Question 3.
(2) \(1.95 < x < 2.05\)
Answer: For the interval \(1.95 < x < 2.05\), we have \(a - \delta = 1.95\) and \(a + \delta = 2.05\). Adding these equations gives \(2a = 4\), so \(a = 2\). Substituting \(a = 2\) into \(a + \delta = 2.05\) results in \(\delta = 0.05\). To write this in modulus form, \(|x - a| < \delta\), we get \(|x - 2| < 0.05\). In neighbourhood form, N(a, δ), this is expressed as N(2, 0.05).
In simple words: For numbers between 1.95 and 2.05, the middle point 'a' is 2, and the spread 'δ' is 0.05. This means \(|x - 2| < 0.05\) in modulus form, or N(2, 0.05) as a neighborhood.

🎯 Exam Tip: Always double-check your 'a' and 'δ' calculations before converting to other forms to avoid errors.

 

Question 3.
(3) \(-0.4 < x < 1.4\)
Answer: Given the interval \(-0.4 < x < 1.4\), we deduce that \(a - \delta = -0.4\) and \(a + \delta = 1.4\). Adding these gives \(2a = 1\), so \(a = \frac{1}{2} = 0.5\). Substituting \(a = 0.5\) into \(a + \delta = 1.4\) gives \(\delta = 0.9\). In modulus form, \(|x - a| < \delta\), this becomes \(|x - 0.5| < 0.9\). In neighbourhood form, N(a, δ), this is written as N(0.5, 0.9).
In simple words: For the interval \(-0.4 < x < 1.4\), the center is 0.5 and the radius is 0.9. In modulus form, it's \(|x - 0.5| < 0.9\), and in neighborhood form, it's N(0.5, 0.9).

🎯 Exam Tip: When the interval spans across zero, 'a' can be positive or negative depending on the bounds. Use the average formula for 'a' and difference for 'δ'.

 

Question 3.
(4) \(1.998 < x < 2.002\)
Answer: For the interval \(1.998 < x < 2.002\), we have \(a - \delta = 1.998\) and \(a + \delta = 2.002\). Adding these equations gives \(2a = 4\), so \(a = 2\). Substituting \(a = 2\) into \(a + \delta = 2.002\) results in \(\delta = 0.002\). In modulus form, \(|x - a| < \delta\), this interval is expressed as \(|x - 2| < 0.002\). In neighbourhood form, N(a, δ), it is written as N(2, 0.002).
In simple words: For 'x' between 1.998 and 2.002, the center 'a' is 2 and the radius 'δ' is 0.002. This is represented as \(|x - 2| < 0.002\) in modulus form and N(2, 0.002) as a neighborhood.

🎯 Exam Tip: Precision in calculations is vital when dealing with small decimal values for delta.

 

Question 4. Express N (16, 0.5) in the interval and modulus form.
Answer: Given the neighbourhood N(16, 0.5), we identify 'a' as 16 and 'δ' as 0.5. To convert this to interval form, \((a - \delta, a + \delta)\), we substitute the values: \((16 - 0.5, 16 + 0.5)\), which gives \((15.5, 16.5)\). For the modulus form, \(|x - a| < \delta\), we substitute 'a' and 'δ' to get \(|x - 16| < 0.5\).
In simple words: The neighborhood N(16, 0.5) has a center of 16 and a radius of 0.5. This means 'x' is between 15.5 and 16.5, and can be written as \(|x - 16| < 0.5\).

🎯 Exam Tip: Always remember that N(a, δ) directly provides 'a' and 'δ' for conversion to other forms.

 

Question 5. If N (3, b) = (2.95, k), then find the values of b and k.
Answer: We are given N(3, b) = (2.95, k). This means that \(a = 3\) and \( \delta = b \). Also, the interval form is \((a - \delta, a + \delta)\), so \(a - \delta = 2.95\) and \(a + \delta = k\). Substituting \(a = 3\) into \(a - \delta = 2.95\), we get \(3 - \delta = 2.95\). Solving for \( \delta \), we find \( \delta = 0.05 \). Since \( \delta = b \), then \( b = 0.05 \). Next, using \(a + \delta = k\), and substituting \(a = 3\) and \(\delta = 0.05\), we have \(3 + 0.05 = k\), which gives \(k = 3.05\). Therefore, the values are \(b = 0.05\) and \(k = 3.05\).
In simple words: Comparing N(3, b) with \((2.95, k)\), we know the center is 3 and the radius is 'b'. From the interval, the center is \((2.95+k)/2\) and the radius is \((k-2.95)/2\). By matching these, we find 'b' is 0.05 and 'k' is 3.05.

🎯 Exam Tip: Equate the components of the neighborhood form N(a, δ) and the interval form (a-δ, a+δ) to solve for unknown variables like 'b' and 'k'.

 

Question 6. If \(|x - 10| < k_1 = (k_2, 10.01)\), then find the values of \(k_1\) and \(k_2\).
Answer: We are given \(|x - 10| < k_1 = (k_2, 10.01)\). From the modulus form \(|x - a| < \delta\), we see that \(a = 10\) and \( \delta = k_1 \). The interval form is \((a - \delta, a + \delta)\), so \((10 - k_1, 10 + k_1)\) is equivalent to \((k_2, 10.01)\). By comparing the upper bounds, \(10 + k_1 = 10.01\), which means \(k_1 = 10.01 - 10 = 0.01\). Now, substitute \(k_1 = 0.01\) into the lower bound comparison: \(10 - k_1 = k_2\), which becomes \(10 - 0.01 = k_2\). This gives \(k_2 = 9.99\). So, the final values are \(k_1 = 0.01\) and \(k_2 = 9.99\).
In simple words: From \(|x - 10| < k_1\), we know the center is 10 and the radius is \(k_1\). The interval is \((10 - k_1, 10 + k_1)\). Comparing this to \((k_2, 10.01)\), we find \(k_1 = 0.01\) and \(k_2 = 9.99\).

🎯 Exam Tip: Always compare the corresponding parts of the modulus or neighborhood form with the interval form (lower bound with lower bound, upper bound with upper bound) to set up equations for unknowns.