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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Three Exemplar Solution

If $t a n θ = \frac{s i n α - α}{c o s α + α},$ then show that $s i n α + c o s α = \sqrt[]{2} c o s θ .$

[Hint: Express the given equation as $t a n θ = t a n (α - \frac{π}{4})$ .]

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alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} G i v e n t h a t : t a n θ = \frac{s i n α - c o s α}{s i n α + c o s α} \\ \Rightarrow t a n θ = \frac{t a n α - 1}{t a n α + 1} = \frac{t a n α - t a n \frac{π}{4}}{1 + t a n \frac{π}{4} t a n α} \\ \Rightarrow t a n θ = t a n (α - \frac{π}{4}) \\ ∴ θ = α - \frac{π}{4} \Rightarrow c o s θ = c o s (α - \frac{π}{4}) \\ \Rightarrow c o s θ = c o s α c o s \frac{π}{4} + s i n α s i n \frac{π}{4} \\ \Rightarrow c o s θ = c o s α . \frac{1}{\sqrt[]{2}} + s i n α . \frac{1}{\sqrt[]{2}} \\ \Rightarrow \sqrt[]{2} c o s θ = c o s α + s i n α \\ \Rightarrow c o s α + s i n α = \sqrt[]{2} c o s θ . H e n c e p r o v e d . \end{array}$

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter One Exemplar Solution

Given that M = {1, 2, 3, 4, 5, 6, 7, 8, 9} and if B = {1, 2, 3, 4, 5, 6, 7, 8, 9}, then B ⊄ M

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Vishal Baghel

Contributor-Level 10

$\begin{array}{l} M = {1, 2, 3, 4, 5, 6, 7, 8, 9} \\ a n d B = {1, 2, 3, 4, 5, 6, 7, 8, 9} \\ Since M a n d B h a s s a m e e l e m e n t s \\ ∴ M = B, S o B ⊄ M ? ? i s' F a l s e' \end{array}$

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9 months ago

btech Takshashila University

Does Takshashila University offer scholarships to B.Tech students?

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Shailja Rawat

Contributor-Level 10

Yes, at Takshashila University, many scholarships are offered to students enrolled in the BTech courses. The scholarships offered by the university are allocated based on students' performance in the qualifying examination. Moreover, it is important to note that these scholarships are offered on a first-come, first-served basis. Thus, those who are interested in availing scholarships must apply timely.

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter One Exemplar Solution

If A is any set, then A ⊂ A

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Vishal Baghel

Contributor-Level 10

$Since e v e r y s e t i s a s u b s e t o f i t s e l f . S o, i t i s' T r u e' .$

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9 months ago

IIT Jodhpur - Indian Institute of Technology M.Sc

What is the eligibility criteria for M.Sc at IIT Jodhpur?

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Nishtha Panda

Contributor-Level 8

IIT Jodhpur offers a 2 year MSc course in 5 specialisations, including Chemistry, Mathematics, Physics, Digital Humanities, and Computational Social Science. Admission to IIT Jodhpur MSc course is through the all-India level Joint Admission Test for Master's (JAM) conducted by IITs. The reservation policy is adapted as suggested by the Ministry of Human Resource Development, Government of India. School of Liberal Arts, IITJ offers a 2-year flagship MSc course in Computational Social Science (CSS). Candidates must secure a minimum aggregate of 60% in graduation and qualify the counselling round to secure the final seat at the Institute.

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9 months ago

Gamics Institute Indore

How to get a seat at Gamics Institute?

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Parul Shukla

Contributor-Level 10

Gamics Institute offers various certificate courses. The selection criteria is merit-based. The application process at Gamics Institute conducted online. To secure a seat, students can check the following steps presented below:

Visit the official website and complete the online application process.
Ensure eligibility and selection criteria for all programme.
The selection criteria is based on merit.
Upon selected, pay the specified fee to confirm and secure your seat.

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9 months ago

btech Takshashila University

What are some of the important factors for B.Tech course at Takshashila University?

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Shailja Rawat

Contributor-Level 10

Takshashila University offers a BTech programme for students to pursue. Before taking admission, students can compare the course with BTech courses offered by other universities. Comparing, fees, ratings, and faculty can be some of the important factors to consider. Other than these, students may take into account secondary factors, such as infrastructure, hostel facilities, competition to get in, etc.

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Three Exemplar Solution

If $\frac{s i n (x + y)}{a + b} = \frac{s i n (x - y)}{a - b},$ then show that $t a n x \cdot a = t a n y \cdot b .$

[Hint: Use Componendo and Dividendo.]

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} G i v e n t h a t : \frac{s i n (x + y)}{s i n (x - y)} = \frac{a + b}{a - b} \\ \Rightarrow \frac{s i n (x + y) + s i n (x - y)}{s i n (x + y) - s i n (x - y)} = \frac{a + b + a - b}{a + b - a + b} [Using c o m p o n e n d o a n d d i v i d e n d o t h e o r e m] \\ \Rightarrow \frac{2 s i n (\frac{x + y + x - y}{2}) c o s (\frac{x + y - x + y}{2})}{2 c o s (\frac{x + y + x - y}{2}) s i n (\frac{x + y - x + y}{2})} = \frac{2 a}{2 b} \\ \Rightarrow \frac{s i n x . c o s y}{c o s x . s i n y} = \frac{a}{b} \Rightarrow t a n x . c o t y = \frac{a}{b} \\ \Rightarrow \frac{t a n x}{t a n y} = \frac{a}{b} . H e n c e p r o v e d . \end{array}$

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Two Exemplar Solution

Find the values of $x$ for which the functions $f (x) = 3 x^{2} - 1$ and $g (x) = 3 + x$ are equal.

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Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} G i v e n t h a t : f (x) = 3 x^{2} - 1 a n d g (x) = 3 + x \\ Since f (x) = g (x) (g i v e n) \\ \Rightarrow 3 x^{2} - 1 = 3 + x \Rightarrow 3 x^{2} - x - 4 = 0 \\ \Rightarrow 3 x^{2} - 4 x + 3 x - 4 = 0 \Rightarrow x (3 x - 4) + 1 (3 x - 4) = 0 \\ \Rightarrow (3 x - 4) (x + 1) = 0 \Rightarrow 3 x - 4 = 0 o r x + 1 = 0 \\ \Rightarrow 3 x = 4 o r x = - 1 \\ ∴ x = \frac{4}{3} \\ H e n c e, t h e v a l u e o f x a r e - 1 a n d \frac{4}{3} . \end{array}$

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter One Exemplar Solution

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)&nbs

...more

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Vishal Baghel

Contributor-Level 10

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

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9 months ago

Gamics Institute Indore Madhya Pradesh

What is the last date of application form of Gamics Institute?

0 Follower 3 Views

Nishtha

Contributor-Level 10

Gamics Institute admissions 2025 (enquiry form) is available at the official website. Till now, no schedule has been released by the institute. Since the admission is open, candidates can apply for certificate courses. To get admission to the Gamics Institute, candidates must meet the eligibility criteria set by the college. The application process at Gamics Institute is conducted online.

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9 months ago

Teaching & Education Entrance Exam PTET

How is the PTET cut-off calculated?

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Chanchal Sinha

Contributor-Level 7

The PTET cut-off is the minimum marks which is required to qualify in the counselling. The cut-off depends on factors such as:

Number of candidates appearing for the exam.
Difficulty level of the paper.
Total number of available seats.
Reservation criteria for different categories.

The PTET cut-off varies as per the different colleges, and candidates must check the previous years’ trends to estimate their chances of admission.

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Three Exemplar Solution

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

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

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

Sol:

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

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Two Exemplar Solution

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

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Payal Gupta

Contributor-Level 10

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

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

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9 months ago

Amrita Vishwa Vidyapeetham Amrita Vishwa Vidyapeetham Placement

What are the placement opportunities at Amrita Vishwa Vidyapeetham?

0 Follower 3 Views

Meitankeisangbam Vasundhara devi

Contributor-Level 10

Amrita Vishwa Vidyapeetham has not released the course-wise packages as of yet. Check out the table below to know the highlights of the placements for MBA during 2022-2023:

Particulars	Placement Statistics (2022)	Placement Statistics (2023)
the highest Package	INR 10 LPA	INR 12 LPA
Average Package	INR 6.39 LPA	INR 8.5 LPA
Placement Rate	98.15%	30% (so far placed in 2023 as per data released)

Note- The institute has not released the placement stats for 2024 as of yet.

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter One Exemplar Solution

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

0 Follower 2 Views

Vishal Baghel

Contributor-Level 10

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

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Two Exemplar Solution

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) $\frac{f}{g}$

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

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

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

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter One Exemplar Solution

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 ______________.

0 Follower 5 Views

Vishal Baghel

Contributor-Level 10

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

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9 months ago

Amrita Vishwa Vidyapeetham Amrita Vishwa Vidyapeetham Placement

What is the placement scenario at Amrita Vishwa Vidyapeetham?

0 Follower 2 Views

Meitankeisangbam Vasundhara devi

Contributor-Level 10

The Amrita Vishwa Vidyapeetham provides good placements to its graduating students. The students are given excellent placement support every year placing them in top companies. Check out the table below to know the key highlights of the Amrita Vishwa Vidyapeetham placements for MBA during 2022-2023:

Particulars	Placement Statistics (2022)	Placement Statistics (2023)
the highest Package	INR 10 LPA	INR 12 LPA
Average Package	INR 6.39 LPA	INR 8.5 LPA
Placement Rate	98.15%	30% (so far placed in 2023 as per data released)

Note- The institute has not released the placement stats for 2024 as of yet.

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Three Exemplar Solution

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 $2 A = (A + B) + (A - B)$ .]

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

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

Sol:

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

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