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New answer posted•a year ago

Exemplar Solution Class 11th

In a town of 840 persons, 450 persons read Hindi, 300 read English and 200 read both. Then the number of persons who read neither is

(a) 210

(b) 290

(c) 180

(d) 260

3 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$\begin{array}{l} T o t a l n u m b e r o f p e r s o n s i n a t o w n = 840 \Rightarrow n (U) = 840 \\ N u m b e r o f p e r s o n s w h o r e a d H i n d i = 450 \Rightarrow n (H) = 450 \\ N u m b e r o f p e r s o n s w h o r e a d E n g l i s h = 300 \Rightarrow n (E) = 300 \\ N u m b e r o f p e r s o n s w h o r e a d b o t h = 200 \Rightarrow n (H \cap E) = 200 \\ ∴ n (H \cup E) = n (H) + n (E) - n (H \cap E) \\ = 450 + 300 - 200 = 550 \\ ∴ n (H' \cap E') = n (U) - n (H \cup E) \\ = 840 - 550 = 290 \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

In a class of 60 students, 25 students play cricket and 20 students play tennis, and 10 students play both the games. Then, the number of students who play neither is

(a) 0

(b) 25

(c) 35

(d) 45

4 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$\begin{array}{l} T o t a l n u m b e r o f s t u d e n t s = 60 \Rightarrow n (U) = 60 \\ N u m b e r o f s t u d e n t s w h o p l a y C r i c k e t = 25 \Rightarrow n (C) = 25 \\ N u m b e r o f s t u d e n t s w h o p l a y T e n n i s = 20 \Rightarrow n (T) = 20 \\ N u m b e r o f s t u d e n t s w h o p l a y C r i c k e t a n d T e n n i s b o t h = 10 \\ \Rightarrow n (C \cap T) = 10 \\ ∴ n (C \cup T) = n (C) + n (T) - n (C \cap T) \\ = 25 + 20 - 10 = 45 - 10 = 35 \\ ∴ n (C' \cap T') = n (U) - n (C \cup T) \\ = 60 - 35 = 25 \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

Given R = {(x, y): x, y ∈ W, $x^{2} + y^{2} = 25$ }. Find the domain and Range of R.

2 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

$\begin{array}{l} G i v e n t h a t : R = {(x, y) : x, y \in W, x^{2} + y^{2} = 25} \\ S o, t h e o r d e r e d p a i r s s a t i s f y i n g t h e g i v e n c o n d i t i o n x^{2} + y^{2} = 25 a r e \\ (0, 5), (3, 4), (5, 0), (4, 3) [? x, y \in W] \\ H e n c e, t h e d o m a i n = {0, 3, 4, 5} a n d r a n g e = {0, 3, 4, 5} . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

Prove that $c o s θ + c o s \frac{θ}{2} - c o s 3 θ c o s \frac{9 θ}{2} = s i n 7 θ s i n 8 θ .$

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:

$\begin{array}{l} L . H . S . c o s θ c o s \frac{θ}{2} - c o s 3 θ c o s \frac{9 θ}{2} \\ = \frac{1}{2} [2 c o s θ c o s \frac{θ}{2}] - \frac{1}{2} [2 c o s 3 θ c o s \frac{9 θ}{2}] \\ = \frac{1}{2} [c o s (θ + \frac{θ}{2}) + c o s (θ - \frac{θ}{2})] - \frac{1}{2} [c o s (3 θ + \frac{9 θ}{2}) + c o s (3 θ - \frac{9 θ}{2})] \\ = \frac{1}{2} [c o s \frac{3 θ}{2} + c o s \frac{θ}{2} - c o s \frac{15 θ}{2} - c o s (\frac{- 3 θ}{2})] \\ = \frac{1}{2} [c o s \frac{3 θ}{2} + c o s \frac{θ}{2} - c o s \frac{15 θ}{2} - c o s \frac{3 θ}{2}] [? c o s (- θ) = c o s θ] \\ = \frac{1}{2} [c o s \frac{θ}{2} - c o s \frac{15 θ}{2}] = \frac{1}{2} [- 2 s i n (\frac{θ}{2} + \frac{15 θ}{2}) . s i n (\frac{θ}{2} - \frac{15 θ}{2})] \\ = - s i n 8 θ s i n (- 7 θ) = s i n 7 θ s i n 8 θ [? s i n (- θ) = - s i n θ] \\ L . H . S . = R . H . S . H e n c e p r o v e d . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

Let R be set of points inside a rectangle of sides a and b (a, b > 1) with two sides along the positive direction of x-axis and y-axis. Then

(a) R = {(x, y): 0 ≤ x ≤ a, 0 ≤ y ≤ b}

(b) R = {(x, y): 0 ≤ x < a, 0 ≤ y ≤ b}

(c) R = {(x, y): 0 ≤ x ≤ a, 0 < y < b}

(d) R = {(x, y): 0 < x < a, 0 < y < b}

3 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

Kindly go through the solution

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New answer posted•a year ago

Exemplar Solution Class 11th

Given A = {1, 2, 3, 4, 5}, S = {(x, y): x ∈ A, y ∈ A}. Find the ordered pairs which satisfy the conditions given below:

(i) x + y = 5

(ii) x + y < 5

(iii) x + y > 8

9 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

$\begin{array}{l} G i v e n t h a t : A = {1, 2, 3, 4, 5} \\ a n d S = {(x, y) : x \in A, y \in A} \\ (i) x + y = 5, s o, t h e o r d e r e d p a i r s s a t i s f y i n g t h e g i v e n c o n d i t i o n s a r e (1, 4), (4, 1), (2, 3), (3, 2) . \\ (i i) x + y < 5, s o, t h e o r d e r e d p a i r s s a t i s f y i n g t h e g i v e n c o n d i t i o n s a r e \\ (1, 1), (1, 2), (2, 1), (1, 3), (2, 2), (3, 1) . \\ (i i i) x + y > 8, s o, t h e o r d e r e d p a i r s s a t i s f y i n g t h e g i v e n c o n d i t i o n s a r e (4, 5), (5, 4), (5, 5) . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

If $t a n x = \frac{b}{a}$ , then find the value of $\sqrt[]{\frac{a + b}{a - b}} + \sqrt[]{\frac{a - b}{a + b}} .$

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:

$\begin{array}{l} G i v e n t h a t : t a n x = \frac{b}{a} \\ \sqrt[]{\frac{a + b}{a - b}} + \sqrt[]{\frac{a - b}{a + b}} = \frac{\sqrt[]{a + b}}{\sqrt[]{a - b}} + \frac{\sqrt[]{a - b}}{\sqrt[]{a + b}} \\ = \frac{a + b + a - b}{\sqrt[]{(a - b) (a + b)}} = \frac{2 a}{\sqrt[]{a^{2} - b^{2}}} = \frac{2 a}{a \sqrt[]{1 - \frac{b^{2}}{a^{2}}}} \\ = \frac{2}{\sqrt[]{1 - t a n^{2} x}} [? t a n x = \frac{b}{a}] \\ = \frac{2}{\sqrt[]{1 - \frac{s i n^{2} x}{c o s^{2} x}}} = \frac{2}{\frac{\sqrt[]{c o s^{2} x - s i n^{2} x}}{c o s x}} \\ = \frac{2 c o s x}{\sqrt[]{c o s 2 x}} [? c o s 2 x = c o s^{2} x - s i n^{2} x] \\ H e n c e, \sqrt[]{\frac{a + b}{a - b}} + \sqrt[]{\frac{a - b}{a + b}} = \frac{2 c o s x}{\sqrt[]{c o s 2 x}} \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

In each of the following cases, find a and b.

(i) (2a + b, a – b) = (8, 3)

(ii) $(\frac{a}{4}, a - b) = (0,6 + b)$  <!-- [if gte msEquation 12]>a4,a-b=0,6+b

2 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

$\begin{array}{l} (i) G i v e n t h a t : (2 a + b, a - b) = (8, 3) \\ C o m p a r i n g t h e d o m a i n s a n d r a n g e s, w e g e t \\ 2 a + b = 8 \dots (i) \\ a - b = 3 \dots (i i) \\ S o l v i n g (i) a n d (i i) w e g e t a = \frac{11}{3} a n d b = \frac{2}{3} \\ (i i) G i v e n t h a t : (\frac{a}{4}, a - 2 b) = (0, 6 + b) \\ C o m p a r i n g t h e d o m a i n s a n d r a n g e s, w e g e t \\ \frac{a}{4} = 0 \Rightarrow a = 0 \\ \Rightarrow a - 2 b = 6 + b \Rightarrow a - 3 b = 6 \Rightarrow 0 - 3 b = 6 \\ ∴ b = - 2 \\ S o, a = 0, b = - 2 . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

Let S = set of points inside the square, T = the set of points inside the triangle and C = the set of points inside the circle. If the triangle and circle intersect each other and are contained in a square. Then

(a) S ∩ T ∩ C = φ

(b) S ∪ T ∪ C = C

(c) S ∪ T ∪ C = S

(d) S ∪ T = S ∩ C

3 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$\begin{array}{l} T h e g i v e n c o n d i t i o n s o f t h e q u e s t i o n m a y b e r e p r e s e n t e d b y t h e f o l l o w i n g V e n n d i a g r a m . \\ F r o m t h e g i v e n V e n n d i a g r a m, \\ W e c l e a r l y c o n c l u d e t h a t \\ S \cup T \cup C = s \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

If $c o s (α + β) = \frac{4}{5}$ and $s i n (α - β) = \frac{5}{13}$ , where $α$ lies between $0$ and $\frac{π}{4}$ , find the value of $t a n^{2} α$ .
[Hint: Express $t a n 2 α$ as $t a n (α + β + α - β)$ .]

4 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:

$\begin{array}{l} G i v e n t h a t : \\ c o s (α + β) = \frac{4}{5} ∴ t a n (α + β) = \frac{3}{4} \\ a n d s i n (α - β) = \frac{5}{13} ∴ t a n (α - β) = \frac{5}{12} \\ N o w t a n 2 α = t a n [α + β + α - β] \\ = t a n [(α + β) + (α - β)] \\ = \frac{t a n (α + β) + t a n (α - β)}{1 - t a n (α + β) . t a n (α - β)} \\ = \frac{\frac{3}{4} + \frac{5}{12}}{1 - \frac{3}{4} \times \frac{5}{12}} = \frac{\frac{9 + 5}{12}}{\frac{48 - 15}{48}} = \frac{14}{12} \times \frac{48}{33} = \frac{56}{33} \\ H e n c e, t a n 2 α = \frac{56}{33} . \end{array}$

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