Question
Method to Find the Sets When Cartesian Product is Given For finding these two sets, we write first element of each ordered pair in first set say A and corresponding second element in second set B (say). Number of Elements in Cartesian Product of Two Sets If there are p elements in set A and g elements in set B, then there will be pq elements in A . B i.e. if n(A) = p and n(B) = q, then n(A . B) = pq.
Based on the above two topic, answer the following questions.
  1. If A . B = {(a, 1), (b, 3), (a, 3), (b, 1), (a, 2), (b, 2)}. Then, A and B are:
  1. {1, 3, 2}, {a, b}
  2. {a, b}, {1, 3}
  3. {a, b}, {1, 3, 2}
  4. None of these
  1. If the set A has 3 elements and set B has 4 elements, then the number of elements in A . B is:
  1. 3
  2. 4
  3. 7
  4. 12
  1. A and B are two sets given in such a way that A . B contains 6 elements. If three elements of A . B are (1, 3), (2, 5) and (3, 3), then A, B are:
  1. {1, 2, 3}, {3, 5}
  2. {3, 5,}, {1, 2, 3}
  3. {1, 2}, {3, 5}
  4. {1, 2, 3}, {5}
  1. The remaining elements of A . B in (iii) is:
  1. (5, 1), (3, 2), (3, 5)
  2. (1, 5), (2, 3), (3, 5)
  3. (1, 5), (3, 2), (5, 3)
  4. None of the above
  1. The cartesian product P . P has 16 elements among which are found (a, 1) and (b, 2). Then, the set P is:
  1. {a, b}
  2. {1, 2}
  3. {a, b,1, 2}
  4. {0, b, 1, 2, 4}

Answer

  1. (c) {a, b}, {1, 3, 2}

Solution:

Here, first element of each ordered pair of A . B gives the elements of set A and corresponding second element gives the elements of set B.

$\therefore$ A = {a, b} and B = {1, 3, 2}

Note We write each element only one time in set, if it occurs more than one time.

  1. (d) 12

Solution: 

Given, n (A) = 3 and n (B) = 4.

$\therefore$ The number of elements in A . B is:

n(A . B) = n(A) . n(B) = 3 . 4 = 12

  1. (a) {1, 2, 3}, {3, 5}

Solution: 

It is given that (1, 3), (2, 5) and (3, 3) are in A . B. It follows that 1, 2, 3 are elements of A and 3, 5 are elements of B.

$\therefore$ A = {1, 2, 3} and B = {3, 5}

  1. (b) (1, 5), (2, 3), (3, 5)

Solution:

$\because$ A = {1, 2, 3} and B = {3, 5}

$\therefore$ A = {1, 2, 3} and B = {3, 5}

= {(1, 3), (1, 5), (2 3), (2, 5), (3, 3), (3, 5)}

Hence, the remaining elements of (A . B) are (1, 5), (2, 3), (3, 5).

  1. {a, b,1, 2}

Solution:

Given, n(P . P) = 16

⇒ n(P) . n(P) = 16

⇒ n(P) = 4 ....(i)

Now, as $(\text{a},1)\in\text{P}\cdot\text{P}$

$\therefore\text{a}\in\text{P}$ and $1\in\text{P}$

Again, $(\text{b},2)\in\text{P}\cdot\text{P}$

$\therefore\text{b}\in\text{P}$ and $2\in\text{P}$

$\Rightarrow\text{a},\text{b},1,2\in\text{P}$

From Eq. (i), it is clear that P has exactly four elements.

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Republic day is a national holiday of India. It honours the date on which the constitution of India came into effect on 26 January 1950 replacing the Government of India Act (1935) as the governing document of India and thus, turning the nation into a newly formed republic.

Answer the following question, which are based on the word "REPUBLIC".


(i) Find the number of arrangements of the letters of the word 'REPUBLIC'.
(a) 40300     (b) 30420    (c) 40320     (d) 40400

(ii) How many arrangements start with a vowel?
(a) 12015     (b) 15120     (c) 12018     (d) 15100

(iii) Which concept is used for finding the arrangements start with a vowel?
(a) Permutation     (b) FPM     (c) Combination     (d) FPA

(iv) If the number of arrangements of the letters of the word 'REPUBLIC' is abcde, the (a + b + $\mathbf{c}+\mathbf{d}+\mathbf{e})$ is
(a) 10     (b) 9     (c) 8     (d) 15

(v) If the number of arrangements start with a vowel is abcde, then $(\mathbf{a}+\mathbf{b})-(\mathbf{d}+\mathbf{e})$ is
(a) 2     (b) 3     (c) 4     (d) 5
The school organised a cultural event for 100 students. In the event, 15 students participated in dance, drama and singing. 25 students participated in dance and drama; 20 students participated in drama and singing; 30 students participated in dance and singing. 8 students participated in dance only; 5 students in drama only and 12 students in singing only.

Based on the above information, answer the following questions.
  1. The number of students who participated in dance, is:
  1. 18
  2. 30
  3. 40
  4. 48
  1. The number of students who participated in drama, is:
  1. 35
  2. 30
  3. 25
  4. 20
  1. The number of students who participated in singing, is:
  1. 42
  2. 45
  3. 47
  4. 37
  1. The number of students who participated in dance and drama but not in singing, is:
  1. 20
  2. 5
  3. 10
  4. 15
  1. The number of students who did not participate in any of the events, is:
  1. 20
  2. 30
  3. 25
  4. 35
The logarithmic function expressed as $\log _e R^{+} \rightarrow R$ and given by $\log _e x=y$ iff $e^y=x$. The graph of the function is given below :
Image
(i) Domain of $f(x)=(0, \infty)$ or $R^{+}$
(ii) Range of $f(x)=(-\infty, \infty)$ or $R$

To find the limit of functions involving logarithmic function, we use the following theorem Theorem $\lim _{x \rightarrow 0} \frac{\log _e(1+x)}{x}=1$

Based on above information, answer the following questions.

(i) $\lim _{x \rightarrow 0} \frac{\log _e(1+5 x)}{x}$ is equal to
    (a) 5     (b) 4     (c) 3     (d) 1

(ii) $\lim _{x \rightarrow 0} \frac{\log _e(1+6 x)-5 x^2}{x}$ is equal to
    (a) 1     (b) 2     (c) 3     (d) 6

(iii) $\quad \lim _{x \rightarrow 0} \frac{\sqrt{1+x}-1}{\log (1+x)}$ is equal to
    (a) 1     (b) $\frac{1}{2}$     (c) $\frac{1}{3}$     (d) $\frac{3}{2}$

(iv) $\quad \lim _{x \rightarrow 5} \frac{\log x-\log 5}{x-5}$ is equal to
    (a) $\frac{1}{5}$     (b) $\frac{3}{5}$     (c) $\frac{1}{4}$     (d) $\frac{2}{3}$

(v) $\quad \lim _{x \rightarrow 0} \frac{\log (5+x)-\log (5-x)}{x}$ is equal to
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In a class test of class XI, a teacher asked to students to consider $\mathbf{A}+\mathbf{B}=\frac{\pi}{4}$, where $\mathbf{A}$ and $\mathbf{B}$ are acute angles.
Based on the above information, answer the following questions.
(i) Find the value of $(1+\tan A)(1+\tan B)$ ?
(ii) Find the value of $(\cot \mathbf{A}-1)(\cot \mathbf{B}-1)$ ?
(iii) Find the value of
$
\sin (A+B)-\cos (A+B)+\tan (A+B) .
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Complex Number in Polar Form 
Image

i. Find the principal argument of Z. (1)
ii. Find the value of $z \overline { Z }$ ? (1)
iii. Find the value of $| Z |$. (2) 
OR
Find the real part of Z. (2)
Ordered Pairs The ordered pair of two elements a and 3 is denoted by (a, b) : a is first element (or first component) and d is second element (or second component). Two ordered pairs are equal if their corresponding elements are equal. ie. (a, b) = (c, d)

⇒ a = c and b = d

Cartesian Product of Two Sets For two non-empty sets A and B, the cartesian product A . B is the set of all ordered pairs of elements from sets Aand B. In symbolic form, it can be written as 

$\text{A}\cdot\text{B}=\{(\text{a},\text{b}):\text{a}\in\text{A},\text{b}\in\text{B}\}$

Based on the above topics, answer the following questions.

If (a - 3, 6 + 7) = (3, 7), then the value of aand d are:

6, 0

3, 7

7, 0

3, -7

If (x + 6, y - 2) = (0, 6), then the value of x and y are:

6, 8

-6, -8

-6, 8

6, -8

If (x + 2, 4) = (5, 2x + y), then the value of x and y are:

-3, 2

3, 2

-3, -2

Let A and B be two sets such that A . B consists of 6 elements. If three elements of A . B are (1, 4), (2, 6) and (3, 6), then

(A . B) = (B . A)

$(\text{A}\cdot\text{B})\neq(\text{B}\cdot\text{A})$

A . B = {(1, 4), (1, 6), (2, 4)}

None of the above

If m(A . B) = 45, then n(A) cannot be

15

17

5

9

Five students Ajay, Shyam, Yojana, Rahul and Akansha are sitting in a playground in a line.
Image
Based on the above information, answer the following questions.

(i) Total number of ways of sitting arrangement of five students is
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(iii) Total number of arrangement 'Yojana and Rahul sitting at extreme position' is
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(iv) Total number of arrangement, if shyam is sitting in the middle, is
    (a) 24     (b) 12     (c) 6     (d) 36

(v) Total number of arrangement sitting Yojana and Rahul not sit together, is
    (a) 72     (b) 120     (c) 60     (d) 144
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