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.

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SECTION A: (1 MARK)

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#### 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:
$$
n(A \Delta B) = n(A - B) + n(B - A)
$$
We already have:
- $ n(A - B) = n(A) - n(A \cap B) = 8 - 3 = 5 $
- $ n(B - A) = 4 $

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

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#### 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) $.

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

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} $

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

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

First, $ \frac{-11}{2} = -5.5 $, $ \frac{11}{2} = 5.5 $

Natural numbers $ \mathbb{N} = \{1, 2, 3, \dots\} $, so values satisfying $ x > -5.5 $ and $ x \leq 5.5 $ are:
$$
x = 1, 2, 3, 4, 5
$$
So,
$$
A = \boxed{\{1, 2, 3, 4, 5\}}
$$

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

Wait — this seems miswritten. Likely 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 = \left\{ x : x \text{ is a positive integer less than } 10 \text{ and } (2^x - 1) \text{ is odd} \right\} $

Check $ 2^x - 1 $ for $ x = 1 $ to $ 9 $:

Note: $ 2^x $ is even for $ x \geq 1 $, so $ 2^x - 1 $ is always odd for any positive integer $ x $.

So all $ x \in \{1, 2, 3, ..., 9\} $ satisfy the condition.

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

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

Note: $ A' = N - A $, $ (A \cup B)' = N - (A \cup B) $

We want:
$$
A' \cup (A \cup B)'
$$

But note:
- $ A' = N - A $
- $ (A \cup B)' = N - (A \cup B) \subseteq N - A = A' $

So $ (A \cup B)' \subseteq A' $

Hence,
$$
A' \cup (A \cup B)' = A'
$$

But we don’t know $ N $. However, since both $ A $ and $ B $ are subsets of $ \{1, 2, ..., 18\} $, assume $ N = \{1, 2, ..., 18\} $

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

So $ (A \cup B)' = N - (A \cup B) = \{1\} $

$ A' = N - A = \{1, 2, 4, 6, 8, 10, 12, 14, 16, 18\} $

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

So answer is $ \boxed{A'} $ or explicitly $ \boxed{\{1, 2, 4, 6, 8, 10, 12, 14, 16, 18\}} $

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#### 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 numerators: $ 0, 1, 2, 3, 4 $ → $ n = 0 $ to $ 4 $

Denominators: $ 6, 7, 8, 9, 10 $ → $ n + 6 $

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

Alternatively:
$$
A = \left\{ \frac{n}{n+6} \mid n = 0, 1, 2, 3, 4 \right\}
$$

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

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

Look at pattern:
- $ 0 = 1^2 - 1 $
- $ 3 = 2^2 - 1 = 4 - 1 $
- $ 8 = 3^2 - 1 = 9 - 1 $? No, $ 9 - 1 = 8 $ → yes!
- $ 15 = 4^2 - 1 = 16 - 1 $
- $ 24 = 5^2 - 1 = 25 - 1 $
- $ 35 = 6^2 - 1 = 36 - 1 $
- $ 48 = 7^2 - 1 = 49 - 1 $

So $ x = n^2 - 1 $, $ n = 1 $ to $ 7 $

But $ 0 $ corresponds to $ n=1 $: $ 1^2 - 1 = 0 $

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

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

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#### 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, 1.5) $ or $ \boxed{[1, \frac{3}{2})} $

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

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

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

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

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

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

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

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SECTION B: (2 MARKS)

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

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

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#### 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 $.

Given:
- $ \frac{2^m}{2^n} = 64 \Rightarrow 2^{m-n} = 64 = 2^6 \Rightarrow m - n = 6 $
- $ m + n = 32 $

Solve:
Add equations:
$$
(m - n) + (m + n) = 6 + 32 \Rightarrow 2m = 38 \Rightarrow m = 19
\Rightarrow n = 32 - 19 = 13
$$

So $ \boxed{m = 19, n = 13} $

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

Power set $ P(A) $ contains all subsets of $ A $.

Number of subsets: $ 2^3 = 8 $

List them:
- $ \emptyset $
- $ \{a\}, \{b\}, \{c\} $
- $ \{a,b\}, \{a,c\}, \{b,c\} $
- $ \{a,b,c\} $

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

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SECTION C: (4 MARKS)

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#### 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: Find $ A \cup B $

$ A \cup B = \{a, b, c, d, e, f, h, i, j\} $

So $ (A \cup B)' = U - (A \cup B) = \{g, k\} $

Step 2: Find $ A' $ and $ B' $

- $ 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' $ → ✔️ First law verified.

Second Law:

$ 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 as above → ✔️ Second law verified.

✔ De Morgan’s laws are verified.

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#### 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 of a set with $ m $ elements: $ 2^m $

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

Assume $ m > n $

Try small values:

Try $ n = 4 $: $ 2^4 = 16 $, $ 2^m = 112 + 16 = 128 = 2^7 $ → $ m = 7 $

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

Check: $ 2^7 = 128 $, $ 2^4 = 16 $, $ 128 - 16 = 112 $ → ✔

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

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✔ Final Answers Summary:

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

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#### Section B:
8. $ n(A') = 215 $
9. $ m = 19, n = 13 $
10. $ P(A) = \left\{ \emptyset, \{a\}, \{b\}, \{c\}, \{a,b\}, \{a,c\}, \{b,c\}, \{a,b,c\} \right\} $

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#### Section C:
11. Verified De Morgan’s laws using given sets.
12. $ m = 7, n = 4 $

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