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New answer posted•a year ago

Exemplar Solution Class 11th

Let F₁ be the set of parallelograms, F₂ the set of rectangles, F₃ the set of rhombuses, F₄the set of squares and F₅ the set of trapeziums in a plane. Then F₁may be equal to

(a) F₂ ∩ F₃

(b) F₃ ∩ F₄

(c) F₂ ∪ F₅

(d) F₂∪ F₃ ∪ F₄ ∪ F₁

9 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$\begin{array}{l} W e k n o w t h a t rectangles, rhombus a n d s q u a r e i n a p l a n e i s a parallelogram b u t t r a p e z i u m i s \\ n o t parallelogram . \\ ∴ F_{1} = F_{2} \cup F_{3} \cup F_{4} \cup F_{1} \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

If A = {x: x ∈ W, x < 2} B = {x: x ∈ N, 1 < x < 5} C = {3, 5} find

(i) A × (B ∩ C)

(ii) A × (B ∪ C)

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

$\begin{array}{l} G i v e n t h a t : A = {x : x \in W, x < 2} \Rightarrow A = {0, 1} \\ B = {x : x \in N, 1 < x < 5} \Rightarrow B = {2, 3, 4} \\ C = {3, 5} \\ (i) A \times (B \cap C) = {0, 1} \times {3} = {(0, 3), (1, 3)} \\ (i i) A \times (B \cup C) = {0, 1} \times {2, 3, 4, 5} \\ = {(0, 2), (0, 3), (0, 4), (0, 5), (1, 2), (1, 3), (1, 4), (1, 5)} \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

The set (A ∩ B′) ′ ∪ (B ∩ C) is equal to

(a) A′ ∪ B ∪ C (b) A′ ∪ B

(c) A′ ∪ C′ (d) A′ ∩ B

4 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$\begin{array}{l} W e k n o w t h a t : (A \cap B)' = A' \cup B' \\ ∴ (A \cap B')' \cup (B \cap C) = [A' \cup (B')'] \cup (B \cap C) \\ = (A' \cup B) \cup (B \cap C) [? (B')' = B] \\ = A' \cup B \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

Two finite sets have m and n elements. The number of subsets of the first set is112 more than that of the second set. The values of m and n are, respectively,

(a) 4, 7

(b) 7, 4

(c) 4, 4

(d) 7, 7

4 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$\begin{array}{l} N u m b e r o f s u b s e t s o f a g i v e n s e t h a v i n g m e l e m e n t = 2^{m} a n d \\ t h e n u m b e r o f s u b s e t s o f s e t c o n t a i n i n g n e l e m e n t s = 2^{n} \\ A s p e r t h e g i v e n c o n d i t i o n, w e h a v e \\ 2^{m} - 2^{n} = 112 \\ \Rightarrow 2^{n} (2^{m - n} - 1) = 112 \Rightarrow 2^{n} . (2^{m - n} - 1) = 2^{4} . 7 \\ ∴ 2^{n} = 2^{4} a n d 2^{m - n} - 1 = 7 \\ \Rightarrow n = 4 a n d 2^{m - n} = 1 + 7 = 8 = 2^{3} \\ \Rightarrow n = 4 a n d m - n = 3 \Rightarrow m - 4 = 3 \Rightarrow m = 3 + 4 = 7 \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

If $m s i n θ = n s i n (θ + 2 α)$ , then prove that $t a n (θ + α) c o t α = \frac{m + n}{m - n} .$ 

[Hint: Express $\frac{s i n (θ + 2 α)}{s i n θ}$ and apply componendo and dividendo.]

5 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:

$\begin{array}{l} G i v e n t h a t : m s i n θ = n s i n (θ + 2 α) \\ \Rightarrow \frac{s i n (θ + 2 α)}{s i n θ} = \frac{m}{n} \\ Using c o m p o n e n d o a n d d i v i d e n d o t h e o r m w e g e t \\ \Rightarrow \frac{s i n (θ + 2 α) + s i n θ}{s i n (θ + 2 α) - s i n θ} = \frac{m + n}{m - n} \\ \Rightarrow \frac{2 s i n (\frac{θ + 2 α + θ}{2}) . c o s (\frac{θ + 2 α - θ}{2})}{2 c o s (\frac{θ + 2 α + θ}{2}) . s i n (\frac{θ + 2 α - θ}{2})} = \frac{m + n}{m - n} \\ [\begin{array}{l} ? s i n A + s i n B = 2 s i n \frac{A + B}{2} . c o s \frac{A - B}{2} \\ s i n A - s i n B = 2 c o s \frac{A + B}{2} . s i n \frac{A - B}{2} \end{array}] \\ \Rightarrow \frac{s i n (θ + α) . c o s α}{c o s (θ + α) - s i n α} = \frac{m + n}{m - n} \\ \Rightarrow t a n (θ + α) . c o t α = \frac{m + n}{m - n} \\ H e n c e p r o v e d . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

If P = {x: x < 3, x ∈ N}, Q = {x: x ≤ 2, x ∈ W}. Find (P ∪ Q) × (P ∩ Q), where W is the set of whole numbers.

17 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

$\begin{array}{l} G i v e n t h a t : P = {x : x < 3, x \in N} \Rightarrow P = {1, 2} \\ Q = {x : x \leq 2, x \in W} \Rightarrow Q = {0, 1, 2} \\ N o w (P \cup Q) = {0, 1, 2} a n d (P \cap Q) = {1, 2} \\ ∴ (P \cup Q) \times (P \cap Q) = {(0, 1), (0, 2), (1, 1), (1, 2), (2, 1), (2, 2)} \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

Let A = {–1, 2, 3} and B = {1, 3}. Determine

(i) A × B

(ii) B × A

(iii) B × B

(iv) A × A

7 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

$\begin{array}{l} G i v e n t h a t : A = {- 1, 2, 3} a n d B = {1, 3} \\ (i) A \times B = {(- 1, 1), (- 1, 3), (2, 1), (2, 3), (3, 1), (3, 3)} \\ (i i) B \times A = {(1, - 1), (3, - 1), (1, 2), (3, 2), (1, 3), (3, 3)} \\ (i i i) B \times B = {1, 3} \times {1, 3} = {(1, 1), (1, 3), (3, 1), (3, 3)} \\ (i v) A \times A = {- 1, 2, 3} \times {- 1, 2, 3} \\ = {(- 1, - 1), (- 1, 2), (- 1, 3), (2, - 1), (2, 2), (2, 3), (3, - 1), (3, 2), (3, 3)} \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

Suppose A₁, A₂, ., A₃₀are thirty sets each having 5 elements and B₁, B₂, ., B_nare n sets each with 3 elements, let =and each element of belongs to exactly 10 of the A_i’s and exactly 9 of the B,’S. Then n is equal to

(a) 15

(b) 3

(c) 45

(d) 35

4 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$\begin{array}{l} N u m b e r o f e l e m e n t s i s A_{1} \cup A_{2} \cup A_{3} \dots \cup A_{30} \\ = 30 \times 5 = 150 (Repetition i s n o t d o n e) \\ B u t e a c h e l e m e n t i s u s e d 10 t i m e s \\ ∴ S = \frac{30 \times 5}{10} = 15 \\ I f t h e e l e m e n t s B_{1}, B_{2}, \dots, B_{n} a r e n o t r e p e a t e d . T h e n t h e t o t a l n u m b e r o f e l e m e n t s = 3 . \\ B u t e a c h e l e m e n t i s r e p e a t e d 9 t i m e s \\ ∴ S = \frac{3 n}{9} \Rightarrow 15 = \frac{n}{3} \Rightarrow n = 15 \times 3 = 45 \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

If $\frac{2 s i n α}{1 + c o s α + s i n α}, = 1$ then prove that $\frac{1 - c o s α + s i n α}{1 + s i n α} = i s a l s o e q u a l t o y .$ 

[Hint: Express $\frac{1 - c o s α + s i n α}{1 + s i n α} = \frac{1 - c o s α + s i n α}{1 + s i n α} = \frac{1 + c o s α + s i n α}{1 + c o s α + s i n α}$

2 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:

$\begin{array}{l} G i v e n t h a t : \\ y = \frac{2 s i n α}{1 + c o s α + s i n α} \\ = \frac{2 s i n α}{1 + c o s α + s i n α} \times \frac{1 + s i n α - c o s α}{1 + s i n α - c o s α} \\ = \frac{2 s i n α (1 - c o s α + s i n α)}{{(1 + s i n α)}^{2} - c o s^{2} α} = \frac{2 s i n α (1 - c o s α + s i n α)}{1 + s i n^{2} α + 2 s i n α - c o s^{2} α} \\ = \frac{2 s i n α (1 - c o s α + s i n α)}{(1 - c o s^{2} α) + s i n^{2} α + 2 s i n α} = \frac{2 s i n α (1 - c o s α + s i n α)}{s i n^{2} α + s i n^{2} α + 2 s i n α} \\ = \frac{2 s i n α (1 - c o s α + s i n α)}{2 s i n^{2} α + 2 s i n α} = \frac{2 s i n α (1 - c o s α + s i n α)}{2 s i n α (1 + s i n α)} \\ = \frac{1 - c o s α + s i n α}{1 + s i n α} = y \\ H e n c e p r o v e d . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

Prove that $\frac{t a n A + s e c A - 1}{t a n A - s e c A + 1} = \frac{1 + S i n A}{c o s A} .$

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:

$\begin{array}{l} L . H . S . \frac{t a n A + s e c A - 1}{t a n A - s e c A + 1} \\ = \frac{t a n A + (s e c A - 1)}{t a n A - (s e c A - 1)} \\ = \frac{[t a n A + (s e c A - 1)]}{[t a n A - (s e c A - 1)]} \times \frac{[t a n A + (s e c A - 1)]}{[t a n A - (s e c A - 1)]} [R a t i o n a l i z i n g t h e denominator] \\ = \frac{{[t a n A + (s e c A - 1)]}^{2}}{t a n^{2} A - {(s e c A - 1)}^{2}} \\ = \frac{t a n^{2} A + {(s e c A - 1)}^{2} + 2 t a n A (s e c A - 1)}{t a n^{2} A - (s e c^{2} A + 1 - 2 s e c A)} \\ = \frac{t a n^{2} A + s e c^{2} A + 1 - 2 s e c A + 2 t a n A s e c A - 2 t a n A}{t a n^{2} A - s e c^{2} A - 1 + 2 s e c A} \\ = \frac{s e c^{2} A + s e c^{2} A - 2 s e c A + 2 t a n A s e c A - 2 t a n A}{- 1 - 1 + 2 s e c A} \\ = \frac{2 s e c^{2} A - 2 s e c A + 2 t a n A s e c A - 2 t a n A}{2 s e c A - 2} \\ = \frac{s e c^{2} A - s e c A + t a n A s e c A - t a n A}{s e c A - 1} \\ = \frac{s e c A (s e c A - 1) + t a n A (s e c A - 1)}{s e c A - 1} \\ = \frac{(s e c A - 1) (s e c A + t a n A)}{(s e c A - 1)} \\ = s e c A + t a n A = \frac{1}{c o s A} + \frac{s i n A}{c o s A} = \frac{1 + s i n A}{c o s A} R . H . S . H e n c e p r o v e d . \end{array}$

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