Exemplar Solution

Get insights from 1.3k questions on Exemplar Solution, answered by students, alumni, and experts. You may also ask and answer any question you like about Exemplar Solution

1.3k
Questions
0
Discussions
0
Active Users
0
Followers

Have a question about Exemplar Solution? Ask our experts

Ask Your Question

Keep it short & simple. Type complete word. Avoid abusive language.

New answer posted•a year ago

Exemplar Solution Class 11th

Match the following sets for all sets A, B and C

i. ((A′ ∪ B′) – A) ′                     (a) A – B

ii. [B′ ∪ (B′ – A)] ′                    (b) A

iii. (A – B) – (B – C)                 (c) B

iv. (A – B) ∩ (C – B)                 (d) (A × B) ∩ (A × C)

v. A × (B ∩ C)                          (e) (A × B) ∪ (A × C)

vi. A × (B ∪ C)                         (f) (A ∩ C) – B

18 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$\begin{array}{l} (i) ((A' \cup B') - A)' = [(A' \cup B') \cap A']' [? A - B = A \cap B'] \\ = [(A \cap B)' \cap A']' [? A' \cup B' = (A \cap B)'] \\ = [(A \cap B)']' \cup (A')' [? (A')' = A] \\ = (A \cap B) \cup A = A . S o, (i) i s m a t c h e d w i t h (b) . \\ (i i) [B' \cup (B' - A)]' = [B' \cup (B' \cap A')]' [? A - B = A \cap B'] \\ = (B')' \cap (B' \cap A')' [? A' \cap B' = (A \cup B)'] \\ = B \cap (B \cup A) = B . S o, (i i) i s m a t c h e d w i t h (c) . \\ (i i i) (A - B) - (B - C) = (A \cap B') - (B \cap C') [? A - B = A \cap B'] \\ = (A \cap B') \cap (B \cap C')' \\ = (A \cap B') \cap (B' \cup (C')') [? A' \cup B' = (A \cap B)'] \\ = (A \cap B') \cap (B' \cup C) \\ = [A \cap (B' \cup C)] \cap [B' \cap (B' \cup C)] \\ = [A \cap (B' \cup C)] \cap B' \\ = (A \cap B') \cap (B' \cup C) \cap B' \\ = (A \cap B') \cap B' = (A \cap B') = A - B \\ S o, (i i i) i s m a t c h e d w i t h (a) . \\ (i v) (A - B) \cap (C - B) = (A \cap B') \cap (C \cap B') [? A - B = A \cap B'] \\ = (A \cap C) \cap B' = (A \cap C) - B [? A \cap B' = A - B] \\ S o, (i v) i s m a t c h e d w i t h (f) . \\ (v) A \times (B \cap C) = (A \times B) \cap (A \times C) \\ S o, (v) i s m a t c h e d w i t h (d) . \\ (v i) A \times (B \cup C) = (A \times B) \cup (A \times C) \\ S o, (v i) i s m a t c h e d w i t h (e) . \end{array}$

Upvote

New answer posted•a year ago

Exemplar Solution Class 11th

If $c o s α + c o s β = 0$ and $s i n α + s i n β = 0$ , then prove that $c o s 2 α + c o s 2 β = - 2 c o s (α + β) .$ 

[Hint: Use ${(c o s α + c o s β)}^{2} - {(s i n α + s i n β)}^{2} = 0$ .]

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:

$\begin{array}{l} G i v e n t h a t : c o s α + c o s β = 0 a n d s i n α + s i n β = 0 \\ ∴ {(c o s α + c o s β)}^{2} - {(s i n α + s i n β)}^{2} = 0 \\ \Rightarrow (c o s^{2} α + c o s^{2} β + 2 c o s α c o s β) - (s i n^{2} α + s i n^{2} β + 2 s i n α s i n β) = 0 \\ \Rightarrow c o s^{2} α + c o s^{2} β + 2 c o s α c o s β - s i n^{2} α - s i n^{2} β - 2 s i n α s i n β = 0 \\ \Rightarrow (c o s^{2} α - s i n^{2} α) + (c o s^{2} β - s i n^{2} β) + 2 (c o s α c o s β - s i n α s i n β) = 0 \\ \Rightarrow c o s 2 α + c o s 2 β + 2 c o s (α + β) = 0 \\ H e n c e, c o s 2 α + c o s 2 β = - 2 c o s (α + β) . \\ H e n c e p r o v e d . \end{array}$

Upvote

New answer posted•a year ago

Exemplar Solution Class 11th

Express the following functions as a set of ordered pairs and determine their range. $f : X \to R, f (x) = x^{3} + 1, where X = {- 1, 0, 3, 9, 7}$

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

$\begin{array}{l} G i v e n t h a t : f : X \to R, f (x) = x^{3} + 1, w h e r e X = {- 1, 0, 3, 9, 7} \\ H e r e, X = {- 1, 0, 3, 9, 7} \\ F o r x = - 1, f (- 1) = {(- 1)}^{3} + 1 = 0 \\ F o r x = 0, f (0) = {(0)}^{3} + 1 = 1 \\ F o r x = 3, f (3) = {(3)}^{3} + 1 = 28 \\ F o r x = 9, f (9) = {(9)}^{3} + 1 = 730 \\ F o r x = 7, f (7) = {(7)}^{3} + 1 = 344 \\ ∴ T h e o r d e r e d p a i r s a r e (- 1, 0), (0, 1), (3, 28), (7, 344), (9, 730) \\ a n d t h e r a n g e = {0, 1, 28, 344, 730} . \end{array}$

Upvote

New answer posted•a year ago

Exemplar Solution Class 11th

For all sets A and B, A – (A ∩ B) is equal to ______________.

2 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

^{$\begin{array}{l} Since A - B = A \cap B' \\ \Rightarrow A - (A \cap B) = A \cap B' \\ H e n c e, t h e f i l l e r i s A \cap B' . \end{array}$}

Upvote

New answer posted•a year ago

Exemplar Solution Class 11th

If  $f$ and  $g$ are two real valued functions defined as $f (x) = 2 x + 1$ , $g (x) = x^{2} + 1$ , then find:

(i) $f + g$  

(ii) $f - g$  

(iii) $f g$  

(iv) $\frac{f}{g}$

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

$\begin{array}{l} G i v e n t h a t : f (x) = 2 x + 1 a n d g (x) = x^{2} + 1 \\ (i) f + g = f (x) + g (x) \Rightarrow 2 x + 1 + x^{2} + 1 \Rightarrow x^{2} + 2 x + 2 \\ (i i) f - g = f (x) - g (x) \Rightarrow (2 x + 1) - (x^{2} + 1) = 2 x + 1 - x^{2} - 1 \Rightarrow 2 x - x^{2} \\ (i i i) f . g = f (x) . g (x) \Rightarrow (2 x + 1) . (x^{2} + 1) \Rightarrow 2 x^{3} + x^{2} + 2 x + 1 \\ (i v) \frac{f}{g} = \frac{f (x)}{g (x)} = \frac{2 x + 1}{x^{2} + 1} \end{array}$

Upvote

New answer posted•a year ago

Exemplar Solution Class 11th

If U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, A = {1, 2, 3, 5}, B = {2, 4, 6, 7} and C = {2, 3, 4, 8}. Then

a) (B ∪ C) ′ is .

b) (C – A) ′ is .

6 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$\begin{array}{l} G i v e n t h a t : U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} \\ A = {1, 2, 3, 5}, B = {2, 4, 6, 7} a n d C = {2, 3, 4, 8} \\ (i) (B \cup C)' = {2, 3, 4, 6, 7, 8}' = {1, 5, 9, 10} \\ H e n c e, t h e f i l l e r i s {1, 5, 9, 10} . \\ (i) (C - A)' = {4, 8}' = {1, 2, 3, 5, 6, 7, 9, 10} \\ H e n c e, t h e f i l l e r i s {1, 2, 3, 5, 6, 7, 9, 10} . \end{array}$

Upvote

New answer posted•a year ago

Exemplar Solution Class 11th

If $t a n (A + B) = p$ , $t a n (A - B) = q$ , then show that $t a n 2 A = \frac{1 + p q}{p - q} .$ 

[Hint: Use $2 A = (A + B) + (A - B)$ .]

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:

$\begin{array}{l} G i v e n t h a t : t a n (A + B) = p a n d t a n (A - B) = q \\ L . H . S . t a n 2 A = t a n (A + B + A - B) = t a n [(A + B) + (A - B)] \\ = \frac{t a n (A + B) + t a n (A - B)}{1 - t a n (A + B) . t a n (A - B)} \\ = \frac{p + q}{1 - p q} R . H . S . \\ L . H . S . = R . H . S . H e n c e p r o v e d . \end{array}$

Upvote

New answer posted•a year ago

Exemplar Solution Class 11th

Given the sets A = {1, 3, 5}. B = {2, 4, 6} and C = {0, 2, 4, 6, 8}. Then the universal set of all the three sets A, B and C can be ______________.

3 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$\begin{array}{l} G i v e n t h a t : A = {1, 3, 5}, B = {2, 4, 6} a n d C = {0, 2, 4, 6, 8} \\ ∴ U n i v e r s a l s e t o f a l l t h e g i v e n s e t s i s \\ U = {0, 1, 2, 3, 4, 5, 6, 8} \\ H e n c e, t h e f i l l e r i s {0, 1, 2, 3, 4, 5, 6, 8} . \end{array}$

Upvote

New answer posted•a year ago

Exemplar Solution Class 11th

Power set of the set A = {1, 2} is ______________.

4 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$\begin{array}{l} P o w e r s e t o f A = P (A) = {{1}, {2}, {1, 2}, ?} \\ H e n c e, t h e f i l l e r i s {{1}, {2}, {1, 2}, ?} . \end{array}$

Upvote

New answer posted•a year ago

Exemplar Solution Class 11th

Let  $f$ and  $g$ be real functions defined by $f (x) = 2 x + 1$ and $g (x) = 4 x - 7$ .

(a) For what real numbers $x$ , $f (x) = g (x) ?$ 

(b) For what real numbers $x$ , $f (x) < g (x) ?$

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

$\begin{array}{l} G i v e n t h a t : f (x) = 2 x + 1 a n d g (x) = 4 x - 7 \\ (i) F o r f (x) = g (x), w e g e t \\ 2 x + 1 = 4 x - 7 \Rightarrow 2 x - 4 x = - 7 - 1 \\ \Rightarrow - 2 x = - 8 \\ \Rightarrow x = 4 . H e n c e, t h e r e q u i r e d r e a l n u m b e r i s 4 . \\ (i i) F o r f (x) < g (x), w e g e t \\ 2 x + 1 < 4 x - 7 \Rightarrow 2 x - 4 x < - 7 - 1 \\ \Rightarrow - 2 x < - 8 \\ \Rightarrow x > 4 . H e n c e, t h e r e q u i r e d r e a l n u m b e r i s x > 4 . \end{array}$

Upvote

Exemplar Solution

Questions

Discussions

Active Users

Followers