CBSE Class 11 Mathematics Sets Worksheet Set C - Free Printable

Let's solve each question step by step from Section A, Section B, and Section C as per the image you've provided.

---

SECTION A: (1 MARK)

---

#### 1. If $ n(A) = 8 $, $ n(B) = 7 $, and $ n(A \cup B) = 12 $, find:

(i) $ n(B - A) $

We use the formula:
$$
n(A \cup B) = n(A) + n(B) - n(A \cap B)
$$
$$
12 = 8 + 7 - n(A \cap B) \Rightarrow n(A \cap B) = 3
$$

Now,
$$
n(B - A) = n(B) - n(A \cap B) = 7 - 3 = \boxed{4}
$$

(ii) $ n(P(A)) $

Power set of $ A $ has $ 2^{n(A)} = 2^8 = \boxed{256} $

(iii) $ n[(A - B) \cup (B - A)] $

This is the symmetric difference:
$$
n(A \Delta B) = n(A - B) + n(B - A)
$$
- $ n(A - B) = n(A) - n(A \cap B) = 8 - 3 = 5 $
- $ n(B - A) = 4 $ (from part i)

So,
$$
n[(A - B) \cup (B - A)] = 5 + 4 = \boxed{9}
$$

---

#### 2. If $ n(A - B) = 14 + x $, $ n(B - A) = 3x $, $ n(A \cap B) = x $, and $ n(A \cup B) = 74 $. Find $ n(A) $ and $ n(B) $.

Use:
$$
n(A \cup B) = n(A - B) + n(B - A) + n(A \cap B)
$$
$$
74 = (14 + x) + 3x + x = 14 + 5x
\Rightarrow 5x = 60 \Rightarrow x = 12
$$

Now:
- $ n(A - B) = 14 + 12 = 26 $
- $ n(B - A) = 3(12) = 36 $
- $ n(A \cap B) = 12 $

So:
- $ n(A) = n(A - B) + n(A \cap B) = 26 + 12 = \boxed{38} $
- $ n(B) = n(B - A) + n(A \cap B) = 36 + 12 = \boxed{48} $

---

#### 3. Describe the following sets in roster form:

(i) $ A = \left\{ x : x \in \mathbb{N}, -\frac{11}{2} \leq x \leq \frac{11}{2} \right\} $

Since $ x \in \mathbb{N} $ (natural numbers), and $ -\frac{11}{2} = -5.5 $, $ \frac{11}{2} = 5.5 $

So natural numbers $ x $ such that $ 1 \leq x \leq 5.5 $ → $ x = 1, 2, 3, 4, 5 $

$$
A = \boxed{\{1, 2, 3, 4, 5\}}
$$

(ii) $ B = \left\{ x : x = \frac{n}{n^2 + 1} \leq n \leq 5, n \in \mathbb{N} \right\} $

Note: This seems poorly written. Probably meant:
$$
B = \left\{ x : x = \frac{n}{n^2 + 1}, 1 \leq n \leq 5, n \in \mathbb{N} \right\}
$$

Compute for $ n = 1 $ to $ 5 $:

- $ n=1 $: $ \frac{1}{1+1} = \frac{1}{2} $
- $ n=2 $: $ \frac{2}{4+1} = \frac{2}{5} $
- $ n=3 $: $ \frac{3}{9+1} = \frac{3}{10} $
- $ n=4 $: $ \frac{4}{16+1} = \frac{4}{17} $
- $ n=5 $: $ \frac{5}{25+1} = \frac{5}{26} $

So,
$$
B = \boxed{\left\{ \frac{1}{2}, \frac{2}{5}, \frac{3}{10}, \frac{4}{17}, \frac{5}{26} \right\}}
$$

(iii) $ C = \{ x : x \text{ is a positive integer less than 10 and } (2^x - 1) \text{ is an odd number} \} $

Check for $ x = 1 $ to $ 9 $:

- $ 2^x - 1 $ is always odd because $ 2^x $ is even for $ x \geq 1 $, so $ 2^x - 1 $ is odd.

So all $ x = 1, 2, ..., 9 $ satisfy the condition.

$$
C = \boxed{\{1, 2, 3, 4, 5, 6, 7, 8, 9\}}
$$

---

#### 4. If $ A = \{3, 5, 7, 9, 11, 13, 15, 17\} $, $ B = \{2, 4, 6, ..., 18\} $, and $ N $ is universal set, then find $ A' \cup (A \cup B)' $

First, note:
- $ A $: odd numbers from 3 to 17
- $ B $: even numbers from 2 to 18

So $ A \cup B = \{2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18\} $

Assuming $ N = \{1, 2, 3, ..., 18\} $ (since max element is 18)

Then:
- $ A' = N - A = \{1, 2, 4, 6, 8, 10, 12, 14, 16, 18\} $
- $ (A \cup B)' = N - (A \cup B) = \{1\} $

So:
$$
A' \cup (A \cup B)' = \{1, 2, 4, 6, 8, 10, 12, 14, 16, 18\} \cup \{1\} = \boxed{\{1, 2, 4, 6, 8, 10, 12, 14, 16, 18\}}
$$

---

#### 5. Describe the following sets in set-builder form:

(i) $ A = \left\{ 0, \frac{1}{6}, \frac{2}{7}, \frac{3}{8}, \frac{4}{9} \right\} $

Look at pattern: numerator: $ 0,1,2,3,4 $; denominator: $ 6,7,8,9,10 $ → $ n+6 $

So $ x = \frac{n}{n+6} $, $ n = 0 $ to $ 4 $

$$
A = \left\{ x : x = \frac{n}{n+6}, n \in \mathbb{W}, 0 \leq n \leq 4 \right\}
$$

Or better: $ n \in \{0,1,2,3,4\} $

$$
A = \boxed{\left\{ x : x = \frac{n}{n+6}, n \in \mathbb{Z}, 0 \leq n \leq 4 \right\}}
$$

(ii) $ B = \{0, 3, 8, 15, 24, 35, 48\} $

Check differences:
- $ 3-0=3 $
- $ 8-3=5 $
- $ 15-8=7 $
- $ 24-15=9 $
- $ 35-24=11 $
- $ 48-35=13 $

Odd numbers increasing — suggests $ x = n^2 - 1 $

Check:
- $ n=1 $: $ 1-1=0 $
- $ n=2 $: $ 4-1=3 $
- $ n=3 $: $ 9-1=8 $
- $ n=4 $: $ 16-1=15 $
- $ n=5 $: $ 25-1=24 $
- $ n=6 $: $ 36-1=35 $
- $ n=7 $: $ 49-1=48 $

Perfect!

$$
B = \boxed{\left\{ x : x = n^2 - 1, n \in \mathbb{N}, 1 \leq n \leq 7 \right\}}
$$

---

#### 6. Write each of the following sets as an interval:

(i) $ \{ x : x \in \mathbb{R}, -1 < x \leq 0 \} $

This is $ (-1, 0] $

(ii) $ \{ x : x \in \mathbb{R}, 3 \leq 2x + 1 < 4 \} $

Solve inequality:
$$
3 \leq 2x + 1 < 4
\Rightarrow 2 \leq 2x < 3
\Rightarrow 1 \leq x < \frac{3}{2}
$$

So interval: $ [1, \frac{3}{2}) $

---

#### 7. Write each of the following intervals as a set:

(i) $ \left[ \frac{3}{2}, \infty \right) $

$ \left\{ x : x \in \mathbb{R}, x \geq \frac{3}{2} \right\} $

(ii) $ \left[ -2, \frac{7}{2} \right] $

$ \left\{ x : x \in \mathbb{R}, -2 \leq x \leq \frac{7}{2} \right\} $

(iii) $ \left( 0, \frac{1}{4} \right) $

$ \left\{ x : x \in \mathbb{R}, 0 < x < \frac{1}{4} \right\} $

---

SECTION B: (2 MARKS)

---

#### 8. Given $ n(A) = 285 $, $ n(B) = 195 $, $ n(U) = 500 $, $ n(A \cup B) = 400 $. Find $ n(A') $

We know:
$$
n(A') = n(U) - n(A) = 500 - 285 = \boxed{215}
$$

---

#### 9. Let $ A $ and $ B $ be two finite sets such that $ n(A) = m $, $ n(B) = n $. If ratio of cardinal numbers of power sets of $ A $ and $ B $ is 64 and $ n(A) + n(B) = 32 $. Find $ m $ and $ n $.

Power set sizes: $ 2^m $ and $ 2^n $

Given:
$$
\frac{2^m}{2^n} = 64 = 2^6 \Rightarrow 2^{m-n} = 2^6 \Rightarrow m - n = 6
$$

Also:
$$
m + n = 32
$$

Add equations:
- $ m - n = 6 $
- $ m + n = 32 $

Add: $ 2m = 38 \Rightarrow m = 19 $, $ n = 13 $

Answer: $ \boxed{m = 19, n = 13} $

---

#### 10. Find power set of the set $ A = \{a, b, c\} $

Power set contains all subsets:

- $ \emptyset $
- $ \{a\}, \{b\}, \{c\} $
- $ \{a,b\}, \{a,c\}, \{b,c\} $
- $ \{a,b,c\} $

Total $ 2^3 = 8 $ elements.

$$
P(A) = \boxed{
\left\{
\emptyset, \{a\}, \{b\}, \{c\}, \{a,b\}, \{a,c\}, \{b,c\}, \{a,b,c\}
\right\}
}
$$

---

SECTION C: (4 MARKS)

---

#### 11. Verify De Morgan’s laws:

Given:
- $ U = \{a, b, c, d, e, f, g, h, i, j, k\} $
- $ A = \{c, e, f, h, i, j\} $
- $ B = \{a, b, d, f, i\} $

De Morgan’s Laws:
1. $ (A \cup B)' = A' \cap B' $
2. $ (A \cap B)' = A' \cup B' $

Step 1: Compute $ A \cup B $
$$
A \cup B = \{a, b, c, d, e, f, h, i, j\}
\Rightarrow (A \cup B)' = U - (A \cup B) = \{g, k\}
$$

Step 2: $ A' = U - A = \{a, b, d, g, k\} $
$ B' = U - B = \{c, e, g, h, j, k\} $

Now $ A' \cap B' = \{g, k\} $

So $ (A \cup B)' = A' \cap B' = \{g, k\} $ → Verified

Step 3: $ A \cap B = \{f, i\} $
So $ (A \cap B)' = U - \{f,i\} = \{a, b, c, d, e, g, h, j, k\} $

Now $ A' \cup B' = \{a, b, d, g, k\} \cup \{c, e, g, h, j, k\} = \{a, b, c, d, e, g, h, j, k\} $

Same → Verified

✔ Both laws are verified.

---

#### 12. Two finite sets have $ m $ and $ n $ elements. The number of subsets of the first set is 112 more than that of the second set. Find $ m $ and $ n $.

Number of subsets: $ 2^m $ and $ 2^n $

Given:
$$
2^m - 2^n = 112
$$

Assume $ m > n $. Try small values:

Try $ n = 5 $: $ 2^5 = 32 $, $ 2^m = 112 + 32 = 144 $ → not power of 2

$ n = 6 $: $ 64 $, $ 2^m = 176 $ → no

$ n = 4 $: $ 16 $, $ 2^m = 128 $ → $ 2^7 = 128 $ → yes!

So $ m = 7 $, $ n = 4 $

Check: $ 128 - 16 = 112 $ ✔

So $ \boxed{m = 7, n = 4} $

---

✔ Final Answers Summary:

---

Section A:

1. (i) 4 (ii) 256 (iii) 9
2. $ n(A) = 38 $, $ n(B) = 48 $
3. (i) {1,2,3,4,5} (ii) {1/2, 2/5, 3/10, 4/17, 5/26} (iii) {1,2,...,9}
4. {1,2,4,6,8,10,12,14,16,18}
5. (i) $ \left\{ x : x = \frac{n}{n+6}, n = 0 \text{ to } 4 \right\} $ (ii) $ \left\{ x : x = n^2 - 1, 1 \leq n \leq 7 \right\} $
6. (i) $ (-1, 0] $ (ii) $ [1, 1.5) $
7. (i) $ \{x \in \mathbb{R} : x \geq 1.5\} $ (ii) $ \{x \in \mathbb{R} : -2 \leq x \leq 3.5\} $ (iii) $ \{x \in \mathbb{R} : 0 < x < 0.25\} $

Section B:

8. $ n(A') = 215 $
9. $ m = 19, n = 13 $
10. Power set of $ \{a,b,c\} $: All 8 subsets listed above.

Section C:

11. Verified both De Morgan’s laws.
12. $ m = 7, n = 4 $

Let me know if you'd like this formatted as a PDF or printed version!