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Factor the denominator using the difference of squares formula, a² − b² = (a−b)(a+b).

Mathematics

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Part i· 5 steps

Resolve into partial fraction: 1x2−12\frac{1}{x^2 - 1^2}

  1. Factor the denominator using the difference of squares formula, a2−b2=(a−b)(a+b)a^2 - b^2 = (a-b)(a+b).

    x2−12=(x−1)(x+1)x^2 - 1^2 = (x-1)(x+1)
  2. Set up the partial fraction decomposition.

    1(x−1)(x+1)=Ax−1+Bx+1\frac{1}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1}
  3. Multiply both sides by (x−1)(x+1)(x-1)(x+1) to clear the denominators.

    1=A(x+1)+B(x−1)1 = A(x+1) + B(x-1)
  4. Solve for AA and BB.

    To find AA, substitute x=1x=1:

    1=A(1+1)+B(1−1)1 = A(1+1) + B(1-1) 1=2A+01 = 2A + 0 A=12A = \frac{1}{2}

    To find BB, substitute x=−1x=-1:

    1=A(−1+1)+B(−1−1)1 = A(-1+1) + B(-1-1) 1=0+B(−2)1 = 0 + B(-2) B=−12B = -\frac{1}{2}
  5. Substitute the values of AA and BB back into the decomposition.

    1x2−1=12x−1+−12x+1\frac{1}{x^2 - 1} = \frac{\frac{1}{2}}{x-1} + \frac{-\frac{1}{2}}{x+1} 1x2−1=12(x−1)−12(x+1)\frac{1}{x^2 - 1} = \frac{1}{2(x-1)} - \frac{1}{2(x+1)}

    The partial fraction decomposition is 12(x−1)−12(x+1)\boxed{\frac{1}{2(x-1)} - \frac{1}{2(x+1)}}.

Part 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.

Part iii

Define union of two sets. Give an example.

The union of two sets AA and BB, denoted by A∪BA \cup B, is the set containing all elements that are members of AA, or members of BB, or members of both.
Example: If A={1,2,3}A = \{1, 2, 3\} and B={3,4,5}B = \{3, 4, 5\}, then A∪B={1,2,3,4,5}A \cup B = \{1, 2, 3, 4, 5\}.

Part iv

If L={a,b}L=\{a,b\} and M={c,d}M=\{c,d\} then find M×LM \times L.

The Cartesian product M×LM \times L is the set of all ordered pairs (m,l)(m, l) where m∈Mm \in M and l∈Ll \in L.
Given M={c,d}M = \{c, d\} and L={a,b}L = \{a, b\}.

M×L={(c,a),(c,b),(d,a),(d,b)}M \times L = \{(c, a), (c, b), (d, a), (d, b)\}

The Cartesian product is {(c,a),(c,b),(d,a),(d,b)}\boxed{\{(c, a), (c, b), (d, a), (d, b)\}}.

Part v

Write any two subsets of the set {a,b}\{a,b\}.

A subset is a set containing some or all of the elements of the original set.
Two subsets of {a,b}\{a,b\} are ∅\boxed{\emptyset} and {a}\boxed{\{a\}}. (Other valid subsets include {b}\{b\} and {a,b}\{a,b\}).

Part vi

If U={1,2,3,...,10}U=\{1,2,3,...,10\} and A={2,3,5,7}A=\{2,3,5,7\}, then find A′A'.

A′A' (the complement of AA) consists of all elements in the universal set UU that are not in AA.
Given U={1,2,3,4,5,6,7,8,9,10}U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\} and A={2,3,5,7}A = \{2, 3, 5, 7\}.
To find A′A', we remove the elements of AA from UU.

A′={1,4,6,8,9,10}A' = \{1, 4, 6, 8, 9, 10\}

The complement of AA is {1,4,6,8,9,10}\boxed{\{1, 4, 6, 8, 9, 10\}}.

Part 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.

Part viii· 3 steps

Find arithmetic mean of the data: x=[60,70,30,90,80,42]x = [60, 70, 30, 90, 80, 42].

The arithmetic mean (xˉ\bar{x}) is the sum of all values divided by the number of values (nn).

  1. Sum the data values.

    ∑x=60+70+30+90+80+42=372\sum x = 60 + 70 + 30 + 90 + 80 + 42 = 372
  2. Count the number of values.

    n=6n = 6
  3. Calculate the mean.

    xˉ=∑xn=3726=62\bar{x} = \frac{\sum x}{n} = \frac{372}{6} = 62

    The arithmetic mean is 62\boxed{62}.

Part ix· 3 steps

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.

  1. Identify the maximum value.

    Maximum value = 45

  2. Identify the minimum value.

    Minimum value = 12

  3. Calculate the range.

    Range=Maximum value−Minimum value=45−12=33\text{Range} = \text{Maximum value} - \text{Minimum value} = 45 - 12 = 33

    The range of the data is 33\boxed{33}.

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