|
| 1 | +--- |
| 2 | +title: Mathematics Graded Assignment |
| 3 | +categories: |
| 4 | +- graded assignment |
| 5 | +image: /images/tree.jpg |
| 6 | +excludeSearch: false |
| 7 | +width: wide |
| 8 | +--- |
| 9 | + |
| 10 | +## 1. Multiple Choice Questions (MSQ) |
| 11 | + |
| 12 | +**Question 1:** |
| 13 | +Which of the following are irrational numbers? |
| 14 | + |
| 15 | +(a) $3^{1/3}$ |
| 16 | + |
| 17 | +(b) $(\sqrt{8}+\sqrt{2})(\sqrt{12}-\sqrt{3})$ |
| 18 | + |
| 19 | +(c) $\frac{\sqrt{18}-3}{\sqrt{2}-1}$ |
| 20 | + |
| 21 | +(d) $\frac{\sqrt{8}+\sqrt{2}}{\sqrt{8}-\sqrt{2}}$ |
| 22 | + |
| 23 | +**Solution:** |
| 24 | + |
| 25 | +(a) $3^{1/3}$ is irrational as it cannot be written as a fraction $p/q$ with $p, q \in \mathbb{Z}, q \neq 0$. |
| 26 | + |
| 27 | +(b) $(\sqrt{8}+\sqrt{2})(\sqrt{12}-\sqrt{3}) = (2\sqrt{2}+\sqrt{2})(2\sqrt{3}-\sqrt{3}) = 3\sqrt{2} \cdot \sqrt{3} = 3\sqrt{6}$, which is irrational. |
| 28 | + |
| 29 | +(c) $\frac{\sqrt{18}-3}{\sqrt{2}-1} = \frac{3\sqrt{2}-3}{\sqrt{2}-1} = \frac{3(\sqrt{2}-1)}{\sqrt{2}-1} = 3$ (rational). |
| 30 | + |
| 31 | +(d) $\frac{\sqrt{8}+\sqrt{2}}{\sqrt{8}-\sqrt{2}} = \frac{3\sqrt{2}}{\sqrt{2}} = 3$ (rational). |
| 32 | + |
| 33 | +**Answer:** (a), (b) |
| 34 | + |
| 35 | +--- |
| 36 | + |
| 37 | +## 2. Examples and Additional Solutions |
| 38 | + |
| 39 | +**Example (QL):** |
| 40 | +Consider a table of materials and their dielectric constants: |
| 41 | + |
| 42 | + |
| 43 | +| Material | Dielectric constant | |
| 44 | +| :-- | :-- | |
| 45 | +| Air | 1 | |
| 46 | +| Vacuum | 2 | |
| 47 | +| Paper | 3 | |
| 48 | +| Glass | 8 | |
| 49 | +| Nerve | 7 | |
| 50 | +| Membrane | 7 | |
| 51 | +| Silicon | 13 | |
| 52 | + |
| 53 | +**Solution:** |
| 54 | +Each material maps to a unique dielectric constant, so the function is bijective. |
| 55 | + |
| 56 | +--- |
| 57 | + |
| 58 | +**Example (Q3):** |
| 59 | +Given: |
| 60 | +$A = \{x \in \mathbb{N} \mid x \bmod 2 = 0 \text{ and } 1 \leq x \leq 10\} = \{2,4,6,8,10\}$ |
| 61 | +$B = \{x \in \mathbb{N} \mid x \bmod 5 = 0 \text{ and } 6 \leq x \leq 25\} = \{10,15,20,25\}$ |
| 62 | +$C = \{x \in \mathbb{N} \mid x \bmod 7 = 0 \text{ and } 7 \leq x \leq 29\} = \{7,14,21,28\}$ |
| 63 | + |
| 64 | +Find $A \setminus (B \cup C)$, $B \setminus (A \cup C)$, $C \setminus (B \cup A)$. |
| 65 | + |
| 66 | +**Solution:** |
| 67 | +$A \setminus (B \cup C) = \{2,4,6,8\}$ |
| 68 | +$B \setminus (A \cup C) = \{15,20,25\}$ |
| 69 | +$C \setminus (B \cup A) = \{7,14,21,28\}$ (note: likely a typo in the original for "2δ", should be "28") |
| 70 | + |
| 71 | +Thus, |
| 72 | +$A \setminus (B \cup C) \cup B \setminus (A \cup C) \cup C \setminus (B \cup A) = \{2,4,6,8,15,20,25,7,14,21,28\}$ |
| 73 | +**Cardinality:** 11 |
| 74 | + |
| 75 | +--- |
| 76 | + |
| 77 | +**Example (Q4):** |
| 78 | +Total number of people = 180 |
| 79 | +Number who watched Dabang $N(D) = 95$ |
| 80 | +Number who watched Avatar and RRR $N(A \cup R) = 40$ |
| 81 | +Number who watched Dabang and RRR $N(D \cup R) = 55$ |
| 82 | +Let $x$ be the number who watched all three movies. |
| 83 | + |
| 84 | +**Solution:** |
| 85 | + |
| 86 | +$$ |
| 87 | +(x-10) + (10+x) + (5+x) + (55-x) + (50-x) + x + (40-x) = 180 |
| 88 | +$$ |
| 89 | + |
| 90 | +$$ |
| 91 | +x + 150 = 180 \implies x = 30 |
| 92 | +$$ |
| 93 | + |
| 94 | +Number who watched only RRR and Avatar = $40 - x = 10$ |
| 95 | + |
| 96 | +--- |
| 97 | + |
| 98 | +## 3. Numerical Answer Type (NAT) |
| 99 | + |
| 100 | +**Question 5:** |
| 101 | +Suppose $f: D \longrightarrow \mathbb{R}$ is a function defined by $f(x) = \frac{\sqrt{x^2-9}}{x+3}$, where $D \subset \mathbb{Z}$. Let $A$ be the set of integers not in the domain of $f$. Find the cardinality of $A$. |
| 102 | + |
| 103 | +**Solution:** |
| 104 | +Domain requires $x^2 - 9 \geq 0$ and $x+3 \neq 0$, so $x \notin \{-3,-2,-1,0,1,2\}$. |
| 105 | +Thus, $A = \{-3,-2,-1,0,1,2\}$; **cardinality = 6**. |
| 106 | + |
| 107 | +**Answer:** 6 |
| 108 | + |
| 109 | +--- |
| 110 | + |
| 111 | +**Question 6:** |
| 112 | +Consider $S = \{a \mid a \in \mathbb{N}, a \leq 18\}$. |
| 113 | +Let $R_1 = \{(x,y) \mid x, y \in S, y = 3x\}$ |
| 114 | +Let $R_2 = \{(x,y) \mid x, y \in S, y = x^2\}$ |
| 115 | + |
| 116 | +Find the cardinality of $R_1 \setminus (R_1 \cap R_2)$. |
| 117 | + |
| 118 | +**Solution:** |
| 119 | +$S = \{0,1,2,\ldots,18\}$ |
| 120 | +$R_1 = \{(0,0),(1,3),(2,6),(3,9),(4,12),(5,15),(6,18)\}$ |
| 121 | +$R_2 = \{(0,0),(1,1),(2,4),(3,9),(4,16)\}$ |
| 122 | +$R_1 \cap R_2 = \{(0,0),(3,9)\}$ |
| 123 | +$R_1 \setminus (R_1 \cap R_2) = \{(1,3),(2,6),(4,12),(5,15),(6,18)\}$ |
| 124 | + |
| 125 | +**Answer:** 5 |
| 126 | + |
| 127 | +--- |
| 128 | + |
| 129 | +**Question 7:** |
| 130 | +In a Zoo, there are 6 Bengal white tigers and 7 Bengal royal tigers. Out of these, 5 are males and 10 are either Bengal royal tigers or males. Find the number of female Bengal white tigers. |
| 131 | + |
| 132 | +**Solution:** |
| 133 | +Let BW = Bengal White, BR = Bengal Royal, M = Male. |
| 134 | +$n(BR) = 7$, $n(M) = 5$, $n(BR \cup M) = 10$ |
| 135 | +Using inclusion-exclusion: |
| 136 | + |
| 137 | +$$ |
| 138 | +n(BR \cup M) = n(BR) + n(M) - n(BR \cap M) |
| 139 | +$$ |
| 140 | + |
| 141 | +$$ |
| 142 | +10 = 7 + 5 - n(BR \cap M) \implies n(BR \cap M) = 2 |
| 143 | +$$ |
| 144 | + |
| 145 | +So, male Bengal Royal tigers = 2, male Bengal White tigers = 3 |
| 146 | +Female Bengal White tigers = 6 - 3 = **3** |
| 147 | + |
| 148 | +**Answer:** 3 |
| 149 | + |
| 150 | +--- |
| 151 | + |
| 152 | +## 4. Additional Examples and Incorrect Questions (Not for Marks) |
| 153 | + |
| 154 | +**Example (Q8):** |
| 155 | +Define $R = \{(A,B) \mid A \text{ and } B \text{ are cousins}\}$, |
| 156 | +and $S = \{(A,B) \mid A \text{ is son of } B\}$. |
| 157 | + |
| 158 | +**Solution:** |
| 159 | +(Not fully solved, but shows how relations are defined for family relationships.) |
| 160 | + |
| 161 | +--- |
| 162 | + |
| 163 | +**Example (Q9):** |
| 164 | +Given $R(m)$ has cardinality $m = 8$, $S(n)$ has cardinality $n = 8$, so $m + n = 16$. |
| 165 | + |
| 166 | +**Example (Q10):** |
| 167 | +Define $f = \{(A,B) \mid A \text{ is son of } B\}$ as a function from $P$ to $Q$. |
| 168 | + |
| 169 | +**Analysis:** |
| 170 | + |
| 171 | +- **Option 1:** Not a function if some $L \in P$ has no image. |
| 172 | +- **Option 2:** Function, not injective (not one-one). |
| 173 | +- **Option 3:** Function, onto (every element in codomain has preimage). |
| 174 | +- **Option 4:** Bijective (both injective and surjective). |
| 175 | + |
| 176 | +--- |
| 177 | + |
| 178 | +## Summary Table |
| 179 | + |
| 180 | +| Question | Type | Solution/Answer | |
| 181 | +| :-- | :-- | :-- | |
| 182 | +| 1 | MSQ | (a), (b) | |
| 183 | +| QL | Ex. | Bijective function example | |
| 184 | +| Q3 | Ex. | Cardinality = 11 | |
| 185 | +| Q4 | Ex. | $x = 30$, only Avatar \& RRR = 10 | |
| 186 | +| 5 | NAT | 6 | |
| 187 | +| 6 | NAT | 5 | |
| 188 | +| 7 | NAT | 3 | |
| 189 | +| Q8-Q10 | Ex. | Family relations and function examples | |
| 190 | + |
| 191 | + |
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