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New answer posted•a year ago

Exemplar Solution Class 11th

If A = {1, 3, 5, 7, 9, 11, 13, 15, 17} B = {2, 4, ., 18} and N the set of natural numbers is the universal set, then A′ ∪(A ∪ B) ∩ B′) is

(a) φ

(b) N

(c) A

(d) B

12 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$\begin{array}{l} G i v e n t h a t : A = {1, 3, 5, 7, 9, 11, 13, 15, 17} \\ B = {2, 4, \dots, 18} \\ U = N = {1, 2, 3, 4, 5, \dots} \\ A' \cup (A \cup B) \cap B' = A' \cup [(A \cap B') \cup (B \cap B')] \\ = A' \cup (A \cap B') \cup ? [? B \cap B' = ?] \\ = A' \cup (A \cap B') \\ = (A' \cup A) \cap (A' \cup B') \\ = N \cap (A' \cup B') [? A' \cup A = N] \\ = A' \cup B' = (A \cap B)' = (?)' = N [? A \cap B = ?] \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

If A and B are two sets, then A ∩ (A ∪ B) equals

(a) A

(b) B

(c) φ

(d) A ∩ B

3 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$\begin{array}{l} G i v e n t h a t : A \cap (A \cup B) \\ L e t x \in A \cap (A \cup B) \\ \Rightarrow x \in A a n d x \in (A \cup B) \\ \Rightarrow x \in A a n d (x \in A o r x \in B) \\ \Rightarrow (x \in A a n d x \in A) o r (x \in A a n d x \in B) \\ \Rightarrow x \in A o r x \in (A \cap B) \Rightarrow x \in A \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

If  $R_{3} = {(x, x^{2}) ? x is a real number}$ is a relation. Then find the domain and range of $R_{3}$ .

4 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : R_{3} = {(x, | x) i s a r e a l n u m b e r} \\ H e n c e, D o m a i n o f R_{3} = R \\ a n d R a n g e o f R_{3} i s (0, \infty) [? | x | = R_{+}] \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

If sets A and B are defined as

A = {(x, y) | y = 1/x, 0 ≠ x ∈ R} B = {(x, y) | y = – x, x ∈ R}, then

(a) A ∩ B = A

(b) A ∩ B = B

(c) A ∩ B = φ

(d) A ∪ B = A

2 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$\begin{array}{l} G i v e n t h a t : A = {(x, y) | y = \frac{1}{x}, 0 \neq x \in R} \\ B = {(x, y) | y = - x, x \in R} \\ H e r e, i t i s c l e a r t h a t y = \frac{1}{x} a n d y = - x \\ ? \frac{1}{x} \neq - x \\ ∴ A \cap B = ? \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

Find the value of $t a n 2 2^{\circ} 3 0^{'}$ .
[Hint: Use $θ = 4 5^{\circ}$ .]

3 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t 2 2^{0} 3 0^{'} = \frac{θ}{2} ∴ θ = 4 5^{0} \\ t a n 2 2^{0} 3 0^{'} = t a n \frac{θ}{2} = \frac{s i n \frac{θ}{2}}{c o s \frac{θ}{2}} = \frac{2 s i n \frac{θ}{2} c o s \frac{θ}{2}}{2 c o s^{2} \frac{θ}{2}} = \frac{s i n θ}{1 + c o s θ} \\ P u t θ = 4 5^{0} \\ ∴ \frac{s i n θ}{1 + c o s θ} = \frac{s i n 4 5^{0}}{1 + c o s 4 5^{0}} = \frac{\frac{1}{\sqrt[]{2}}}{1 + \frac{1}{\sqrt[]{2}}} = \frac{1}{\sqrt[]{2} + 1} \\ = \frac{1 \times (\sqrt[]{2} - 1)}{(\sqrt[]{2} + 1) (\sqrt[]{2} - 1)} = \sqrt[]{2} - 1 \\ H e n c e, t a n 2 2^{0} 3 0^{'} = \sqrt[]{2} - 1 . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

A survey shows that 63% of the people watch a News Channel whereas 76%watch another channel. If x% of the people watch both channels, then

(a) x = 35

(b) x = 63

(c) 39 ≤ x ≤ 63

(d) x = 39

4 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$\begin{array}{l} L e t p % o f t h e p e o p l e w a t c h a c h a n n e l a n d q % o f t h e p e o p l e w a t c h a n o t h e r c h a n n e l \\ n (p \cap q) = x % a n d n (p \cup q) \leq 100 \\ n (p \cup q) \geq n (p) + n (q) - n (p \cap q) \\ 100 \geq 63 + 76 - x \Rightarrow 100 \geq 139 - x \\ x \geq 139 - 100 \Rightarrow x \geq 39 \\ N o w n (p) = 63 \\ ∴ n (p \cap q) \leq n (p) \Rightarrow x \leq 63 \\ S o, 39 \leq x \leq 63 \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

If $R_{2} = {(x, y) ? x and y are integers and x^{2} + y^{2} = 64}$  is a relation. Then find $R_{2}$ .

5 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : x^{2} + y^{2} = 64, x, y \in Z \\ Since t h e s u m o f t h e s q u a r e s o f t w o integers i s 64 \\ ∴ F o r x = 0, y = \pm 8 \\ F o r x = \pm 8, y = 0 \\ H e n c e, R_{2} = {(0, 8), (0, - 8), (8, 0), (- 8, 0)} \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

If X = {8n – 7n – 1 | n ∈ N} and Y = {49n – 49 | n ∈ N}. Then

(a) X ⊂ Y

(b) Y ⊂ X

(c) X = Y

(d) X ∩ Y = φ

2 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$\begin{array}{l} G i v e n t h a t : X = {8^{n} - 7 n - 1 | n \in N} = {0, 49, 490, \dots} \\ Y = {49 n - 49 | n \in N} = {0, 49, 98, \dots} \\ H e r e, i t i s c l e a r t h a t e v e r y e l e m e n t b e l o n g i n g s t o X i s a l s o p r e s e n t i n Y . \\ ∴ X \subset Y . \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

If $a c o s θ + b s i n θ = m$ and $a s i n θ - b c o s θ = n$ , then show that $a^{2} + b^{2} = m^{2} + n^{2} .$

3 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : a c o s θ + b s i n θ = m a n d a s i n θ - b c o s θ = n \\ R . H . S . m^{2} + n^{2} = {(a c o s θ + b s i n θ)}^{2} + {(a s i n θ - b c o s θ)}^{2} \\ = a^{2} c o s^{2} θ + b^{2} s i n^{2} θ + 2 a b s i n θ c o s θ + a^{2} s i n^{2} θ + b^{2} c o s^{2} θ - 2 a b s i n θ c o s θ \\ = a^{2} c o s^{2} θ + b^{2} s i n^{2} θ + a^{2} s i n^{2} θ + b^{2} c o s^{2} θ \\ = a^{2} (c o s^{2} θ + s i n^{2} θ) + b^{2} (s i n^{2} θ + c o s^{2} θ) \\ = a^{2} . 1 + b^{2} . 1 = a^{2} + b^{2} L . H . S . \\ L . H . S . = R . H . S . H e n c e p r o v e d . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

If $R_{1} = {(x, y) ? y = 2 x + 7, where x \in R and - 5 \leq x \leq 5}$  is a relation. Then find the domain and Range of $R_{1}$ .

6 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : R_{1} = {(x, y) | y = 2 x + 7 w h e r e x \in R a n d - 5 \leq x \leq 5} \\ H e n c e, d o m a i n i s - 5 \leq x \leq 5 \Rightarrow {- 5, - 4, - 3, - 2, - 1, 0, 1, 2, 3, 4, 5} \\ a n d y = 2 x + 7 . \\ S o, t h e v a l u e s o f y f o r t h e c o r r e s p o n d i n g g i v e n v a l u e s o f x a r e \\ {- 3, - 1, 1, 3, 5, 7, 9, 11, 13, 15, 17} \\ H e n c e, t h e d o m a i n o f R_{1} = [- 5, 5] a n d r a n g e o f R_{1} = [- 3, 17] \end{array}$

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