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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter One Exemplar Solution

If X and Y are two sets and X′ denotes the complement of X, then X ∩ (X ∪ Y) ′is equal to

(a) X

(b) Y

(c) φ

(d) X ∩ Y

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Vishal Baghel

Contributor-Level 10

$\begin{array}{l} L e t x \in X \cap (X \cup Y)' \\ \Rightarrow x \in X \cap (X' \cap Y') \\ \Rightarrow x \in (X \cap X') \cap (X \cap Y') \\ \Rightarrow x \in ? \cap (X \cap Y') [? A \cap A' = ?] \\ \Rightarrow x \in ? \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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9 months ago

btech Takshashila University

Can Takshashila University B.Tech application form be submitted online?

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Shailja Rawat

Contributor-Level 10

Yes, Takshashila University BTech applicants can be submitted online. In order to apply online, students can visit the university's official website and follow the below-mentioned steps:

1. Click on 'Apply Now' on the home page.

2. Register by providing the required details.

3. Log in and fill out the application form.

4. Upload the documents and pay the application fees.

5. Preview and submit.

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter One Exemplar Solution

Let S = {x | x is a positive multiple of 3 less than 100}

P = {x | x is a prime number less than 20}. Then n(S) + n(P) is

(a) 34

(b) 31

(c) 33

(d) 30

0 Follower 2 Views

Vishal Baghel

Contributor-Level 10

$\begin{array}{l} G i v e n t h a t : S = {x | x i s a p o s i t i v e m u l t i p l e o f 3 < 100} \\ ∴ S = {3, 6, 9, 12, 15, 18, \dots, 99} \\ \Rightarrow n (S) = 33 \\ T = {x | x i s a p r i m e n u m b e r < 20} \\ ∴ T = {2, 3, 5, 7, 11, 13, 17, 19} \\ \Rightarrow n (T) = 8 \\ S o, n (S) + n (T) = 33 + 8 = 41 \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Two Exemplar Solution

Is the given relation a function? Give reasons for your answer.

(i) $h = {(4, 6), (3, 9), (- 11, 6), (3, 11)}$

(ii) $f = {(x, x) ? x is a real number}$

(iii) $g = {(n, \frac{1}{n}, n) ? n is a positive integer}$

(iv) $s = {(n, n^{2}) ? n is a positive integer}$

(v) $t = {(x, 3) ? x is a real number}$

0 Follower 6 Views

Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} (i) G i v e n t h a t : h = {(4, 6), (3, 9), (- 11, 6), (3, 11)} \\ Since i n t h e g i v e n r e l a t i o n 3 h a s t w o i m a g e s 9 a n d 11 . S o, h i s n o t a f u n c t i o n . \\ (i i) f = {(x, x) | x i s a r e a l n u m b e r} . H e r e, w e o b s e r v e t h a t f o r e v e r y e l e m e n t o f d o m a i n h a s a \\ u n i q u e i m a g e . S o, f i s a f u n c t i o n . \\ (i i i) G i v e n t h a t : g = {(n, \frac{1}{n}) | n i s a p o s i t i v e integer} . H e r e, w e o b s e r v e t h a t n i s a p o s i t i v e \\ integer so, f o r e v e r y e l e m e n t o f d o m a i n, t h e r e i s a u n i q u e \frac{1}{n} i m a g e . H e n c e, g i s a f u n c t i o n . \\ (i v) G i v e n t h a t : S = {(n, n^{2}) | n i s a p o s i t i v e integer} . H e r e, w e o b s e r v e t h a t t h e s q u a r e o f a n y \\ integer i s a u n i q u e n u m b e r . S o, e v e r y e l e m e n t i n t h e d o m a i n t h e r e i s a u n i q u e i m a g e . \\ H e n c e, S i s a f u n c t i o n . \\ (v) G i v e n t h a t : t = {(x, 3) | x i s a r e a l n u m b e r} . H e r e, w e o b s e r v e t h a t f o r e v e r y r e a l e l e m e n t \\ i n t h e d o m a i n, t h e r e i s a constant n u m b e r 3 . H e n c e, t i s a constant f u n c t i o n . \end{array}$

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Three Exemplar Solution

Prove that $s i n 4 A = 4 s i n A c o s^{3} A - 4 c o s A s i n^{3} A .$

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} L . H . S . s i n 4 A = s i n (A + 3 A) \\ = s i n A c o s 3 A + c o s A s i n 3 A \\ = s i n A (4 c o s^{3} A - 3 c o s A) + c o s A (3 s i n A - 4 s i n^{3} A) \\ = 4 s i n A c o s^{3} A - 3 s i n A c o s A + 3 s i n A c o s A - 4 c o s A s i n^{3} A \\ = 4 s i n A c o s^{3} A - 4 c o s A s i n^{3} A R . H . S . \\ L . H . S . = R . H . S . H e n c e p r o v e d . \end{array}$

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter One Exemplar Solution

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

0 Follower 10 Views

Vishal Baghel

Contributor-Level 10

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

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter One Exemplar Solution

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

(a) A

(b) B

(c) φ

(d) A ∩ B

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

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

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9 months ago

R.V. College of Engineering btech

What factors determine the B.Tech cutoff at RV College of Engineering?

0 Follower 5 Views

Tanisha Pandey

Contributor-Level 10

While preparing the cutoff for the RVCE BTech course, the authorities will consider examination and admission-related factors. The listed factors will be checked and considered by the authorities while determining the cutoff of RV College of Engineering BTech: previous years' cutoff trends, difficulty level of exam, total number of qualified candidates, total number of candidates applying for a specific course or institute, number of seats available for admission, etc. After checking all of these factors, the authorities will prepare the RVCE cutoff for BE admission.

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Two Exemplar Solution

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

0 Follower 4 Views

Payal Gupta

Contributor-Level 10

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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9 months ago

Teaching & Education Entrance Exam PTET Dates

What are the dates of PTET?

0 Follower 9 Views

Nishtha Taneja

Contributor-Level 7

The PTET (Pre-Teacher Education Test) 2025 was conducted on June 15, 2025, with admit cards were released earlier on June 9, 2025. The application window remained open from March 6 to May 1, 2025. Following the exam, the provisional answer key was released on June 19, 2025. Candidates must keep checking the official website regularly for updates and any changes in the schedule.

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter One Exemplar Solution

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

0 Follower 2 Views

Vishal Baghel

Contributor-Level 10

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

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9 months ago

btech Takshashila University Courses

What is the last date to apply for Takshashila University B.Tech courses?

0 Follower 3 Views

Shailja Rawat

Contributor-Level 10

Takshashila University's application portal for BTech is open for the academic year 2025. Candidates interested in taking admission must apply for the preferred course by submitting the application form available on the university's official website. However, the college has not announced any last date yet. To know more, students can get in touch with the university's admissions committee.

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Three Exemplar Solution

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

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

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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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter One Exemplar Solution

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

0 Follower 4 Views

Vishal Baghel

Contributor-Level 10

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

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9 months ago

Takshashila University

Are Takshashila University admissions open for 2025?

0 Follower 5 Views

Shailja Rawat

Contributor-Level 10

Yes, the admission process for various Takshashila University courses is open for the academic year 2025-2026. To get admission, students who meet the eligibility criteria can go to the official website of the university and submit an application form for the preferred course. Admissions are granted based on marks obtained in the qualifying examination.

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Two Exemplar Solution

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

0 Follower 4 Views

Payal Gupta

Contributor-Level 10

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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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter One Exemplar Solution

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

0 Follower 2 Views

Vishal Baghel

Contributor-Level 10

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

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter One Exemplar Solution

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

0 Follower 2 Views

Vishal Baghel

Contributor-Level 10

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

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Two Exemplar Solution

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 $R_{1}$ .

0 Follower 5 Views

Payal Gupta

Contributor-Level 10

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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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Three Exemplar Solution

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 $a^{2} + b^{2} = m^{2} + n^{2} .$

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

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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