Here are the solutions to the questions: i) Resolve into partial fraction: $\frac{1}{x^2 - 1^2}$ Step 1: Factor the denominator using the difference of squares formula, $a^2 - b^2 = (a-b)(a+b)$. $$ x^2 - 1^2 = (x-1)(x+1) $$ Step 2: Set up the partial fraction decomposition. $$ \frac{1}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1} $$ Step 3: Multiply both sides by $(x-1)(x+1)$ to clear the denominators. $$ 1 = A(x+1) + B(x-1) $$ Step 4: Solve for $A$ and $B$. To find $A$, substitute $x=1$: $$ 1 = A(1+1) + B(1-1) $$ $$ 1 = 2A + 0 $$ $$ A = \frac{1}{2} $$ To find $B$, substitute $x=-1$: $$ 1 = A(-1+1) + B(-1-1) $$ $$ 1 = 0 + B(-2) $$ $$ B = -\frac{1}{2} $$ Step 5: Substitute the values of $A$ and $B$ back into the decomposition. $$ \frac{1}{x^2 - 1} = \frac{\frac{1}{2}}{x-1} + \frac{-\frac{1}{2}}{x+1} $$ $$ \frac{1}{x^2 - 1} = \frac{1}{2(x-1)} - \frac{1}{2(x+1)} $$ The partial fraction decomposition is $\boxed{\frac{1}{2(x-1)} - \frac{1}{2(x+1)}}$. ii) What is the difference between proper and improper fraction? A proper fraction is a rational expression where the degree of the numerator polynomial is less than the degree of the denominator polynomial. An improper fraction is a rational expression where the degree of the numerator polynomial is greater than or equal to the degree of the denominator polynomial. iii) Define union of two sets. Give an example. The union of two sets $A$ and $B$, denoted by $A \cup B$, is the set containing all elements that are members of $A$, or members of $B$, or members of both. Example: If $A = \{1, 2, 3\}$ and $B = \{3, 4, 5\}$, then $A \cup B = \{1, 2, 3, 4, 5\}$. iv) If $L=\{a,b\}$ and $M=\{c,d\}$ then find $M \times L$. The Cartesian product $M \times L$ is the set of all ordered pairs $(m, l)$ where $m \in M$ and $l \in L$. Given $M = \{c, d\}$ and $L = \{a, b\}$. $$ M \times L = \{(c, a), (c, b), (d, a), (d, b)\} $$ The Cartesian product is $\boxed{\{(c, a), (c, b), (d, a), (d, b)\}}$. v) Write any two subsets of the set $\{a,b\}$. A subset is a set containing some or all of the elements of the original set. Two subsets of $\{a,b\}$ are $\boxed{\emptyset}$ and $\boxed{\{a\}}$. (Other valid subsets include $\{b\}$ and $\{a,b\}$). vi) If $U=\{1,2,3,...,10\}$ and $A=\{2,3,5,7\}$, then find $A'$. $A'$ (the complement of $A$) consists of all elements in the universal set $U$ that are not in $A$. Given $U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$ and $A = \{2, 3, 5, 7\}$. To find $A'$, we remove the elements of $A$ from $U$. $$ A' = \{1, 4, 6, 8, 9, 10\} $$ The complement of $A$ is $\boxed{\{1, 4, 6, 8, 9, 10\}}$. vii) What is the difference between median and mode? The median is the middle value in a dataset when the values are arranged in numerical order. If there is an even number of data points, the median is the average of the two middle values. The mode is the value that appears most frequently in a dataset. A dataset can have one mode, multiple modes, or no mode. viii) Find arithmetic mean of the data: $x = [60, 70, 30, 90, 80, 42]$. The arithmetic mean ($\bar{x}$) is the sum of all values divided by the number of values ($n$). Step 1: Sum the data values. $$ \sum x = 60 + 70 + 30 + 90 + 80 + 42 = 372 $$ Step 2: Count the number of values. $$ n = 6 $$ Step 3: Calculate the mean. $$ \bar{x} = \frac{\sum x}{n} = \frac{372}{6} = 62 $$ The arithmetic mean is $\boxed{62}$. ix) Find the range of the data: $12, 14, 17, 20, 24, 29, 35, 45$. The range is the difference between the maximum and minimum values in a dataset. Step 1: Identify the maximum value. Maximum value = $45$ Step 2: Identify the minimum value. Minimum value = $12$ Step 3: Calculate the range. $$ \text{Range} = \text{Maximum value} - \text{Minimum value} = 45 - 12 = 33 $$ The range of the data is $\boxed{33}$. Send me the next one 📸