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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Two Exemplar Solution

If $P = {1, 2}$ , then $P \times P \times P = {(1, 1, 1), (2, 2, 2), (1, 2, 2), (2, 1, 1)}$ .

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Payal Gupta

Contributor-Level 10

This is a True or False Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : P = {1, 2} \\ ∴ P \times P = {1, 2} \times {1, 2} = {(1, 1), (1, 2), (2, 1), (2, 2)} \\ P \times P \times P = {(1, 1), (1, 2), (2, 1), (2, 2)} \times {1, 2} \\ = {(1, 1, 1), (1, 1, 2), (1, 2, 1), (1, 2, 2), (2, 1, 1), (2, 1, 2), (2, 2, 1), (2, 2, 2)} \\ H e n c e, t h e s t a t e m e n t i s' F a l s e' . \end{array}$

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Two Exemplar Solution

The ordered pair $(5, 2)$ belongs to the relation $R = {(x, y) : y = x - 5, x, y \in ?}$ .

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Payal Gupta

Contributor-Level 10

This is a True or False Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : R = {(x, y) : y = x - 5, x, y \in Z} \\ F o r (5, 2), y = x - 5 \\ P u t x = 5, y = 5 - 5 = 0 \neq 2 \\ S o, (5, 2) i s n o t t h e o r d e r e d p a i r o f R . \\ H e n c e, t h e s t a t e m e n t i s' F a l s e' . \end{array}$

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Two Exemplar Solution

If ${[x]}^{2} - 5 [x] + 6 = 0$ , where [.] denotes the greatest integer function, then

(a) $x \in [3, 4]$

(b) $x \in (2, 3]$

(c) $x \in [2, 3]$

(d) $x \in [2, 4]$

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Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} W e h a v e {[x]}^{2} - 5 [x] + 6 = 0 \\ \Rightarrow {[x]}^{2} - 3 [x] - 2 [x] + 6 = 0 \\ \Rightarrow [x] ([x] - 3) - 2 ([x] - 3) = 0 \\ \Rightarrow ([x] - 3) ([x] - 2) = 0 \Rightarrow [x] = 2, 3 \\ S o, x \in [2, 3] \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Two Exemplar Solution

Let $n (A) = m$ , and $n (B) = n$ . Then the total number of non-empty relations that can be defined from $A$ to $B$ is:

(a) $m^{n}$

(b) $n^{m} - 1$

(c) $m^{n} - 1$

(d) $2^{m n} - 1$

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Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : n (A) = m a n d n (B) = n \\ ∴ n (A \times B) = n (A) . n (B) = m n \\ S o, t h e t o t a l n u m b e r o f r e l a t i o n s f r o m A t o B = 2^{m n} - 1 . \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

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9 months ago

Medicine Entrance Exam NEET Counselling

Do I need to get my documents verified for Gujarat NEET 2025 counselling?

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Nishtha Choudhury

Beginner-Level 5

Yes. Candidates need to verify their documents at help centres. It is mandatory for candidates to verify all original documents at designated help centers and obtain registration slip, without which candidate shall not be eligible for further admission process. Following document verification, a merit list is prepared who are allowed to declare choices for Gujarat NEET 2025 counselling.

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9 months ago

DTU - Delhi Technological University MBA/PGDM

What are the packages offered to MBA students at DTU?

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Shoaib Mehdi

Contributor-Level 10

The salary package offered during DTU MBA placements over the past three years is mentioned below:

Particulars	MBA DSM Placement Statistics 2023	MBA DSM Placement Statistics 2024	MBA DSM Placement Statistics 2025
Highest package	INR 14.5 LPA	INR 37 LPA	INR 16.42 LPA
Average package	INR 8.96 LPA	INR 9.48 LPA	INR 7.74 LPA
Particulars	MBA USME Placement Statistics 2023	MBA USME Placement Statistics 2024	MBA USME Placement Statistics 2025
Highest package	INR 23.78 LPA	INR 14.13 LPA	INR 22 LPA
Average package	INR 9.02 LPA	INR 9 LPA	INR 7.55 LPA

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Two Exemplar Solution

Let $f = {(2, 4), (5, 6), (8, - 1), (10, - 3)}$ $g = {(2, 5), (7, 1), (8, 4), (10, 13), (11, 5)$ be two real functions. Then Match the following:

(a) $f - g$ (i) ${(2, \frac{4}{5})} (8, \frac{- 1}{4}) (10, \frac{- 3}{13})$

(b) $f + g$ (ii) ${(2, 20), (8, - 4), (10, - 39)}$

(c) $f \cdot g$ (iii) ${(2, - 1), (8, - 5), (10, - 16)}$

(d) $\frac{f}{g}$ (iv) ${(2, 9), (8, 3), (10, 10)}$

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Payal Gupta

Contributor-Level 10

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : f = {(2, 4), (5, 6), (8, - 1), (10, - 3)} \\ a n d g = {(2, 5), (7, 1), (8, 4), (10, 13), (11, 5)} \\ f - g, f + g, f . g, \frac{f}{g} a r e d e f i n e d i n t h e d o m a i n \\ (d o m a i n o f f \cap d o m a i n o f g) \\ \Rightarrow {2, 5, 8, 10} \cap {2, 7, 8, 10, 11} \Rightarrow {2, 8, 10} \\ (i) (f - g) 2 = f (2) - g (2) = 4 - 5 = - 1 \\ (f - g) 8 = f (8) - g (8) = - 1 - 4 = - 5 \\ (f - g) 10 = f (10) - g (10) = - 3 - 13 = - 16 \\ ∴ (f - g) = {(2, - 1), (8, - 5), (10, - 16)} \\ (i i) (f + g) 2 = f (2) + g (2) = 4 + 5 = 9 \\ (f + g) 8 = f (8) + g (8) = - 1 + 4 = 3 \\ (f + g) 10 = f (10) + g (10) = - 3 + 13 = 10 \\ ∴ (f + g) = {(2, 9), (8, 3), (10, 10)} \\ (i i i) (f . g) 2 = f (2) . g (2) = 4.5 = 20 \\ (f . g) 8 = f (8) . g (8) = (- 1) . 4 = - 4 \\ (f . g) 10 = f (10) . g (10) = - 3.13 = - 39 \\ ∴ (f . g) = {(2, 20), (8, - 4), (10, - 39)} \\ (i v) (\frac{f}{g}) (2) = \frac{f (2)}{g (2)} = \frac{4}{5} \\ (\frac{f}{g}) (8) = \frac{f (8)}{g (8)} = \frac{- 1}{4} \\ (\frac{f}{g}) (10) = \frac{f (10)}{g (10)} = \frac{- 3}{13} \\ (\frac{f}{g}) = {(2, \frac{4}{5}), (8, \frac{- 1}{4}), (10, \frac{- 3}{13})} \\ H e n c e, t h e c o r r e c t o p t i o n i s \\ (a) \leftrightarrow (i i i), (b) \leftrightarrow (i v), (c) \leftrightarrow (i i), (d) \leftrightarrow (i) \end{array}$

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Two Exemplar Solution

Let f and g be two real functions given by $f = {(0, 1), (2, 0), (3, - 4), (4, 2), (5, 1)}$ $g = {(1, 0), (2, 2), (3, - 1), (4, 4), (5, 3)}$ then the domain of $f ? g$ is given by _________.

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Payal Gupta

Contributor-Level 10

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : f (x) = {(0, 1), (2, 0), (3, - 4), (4, 2), (5, 1)} \\ a n d g (x) = {(1, 0), (2, 2), (3, - 1), (4, 4), (5, 3)} \\ ∴ d o m a i n o f f = {0, 2, 3, 4, 5} \\ a n d d o m a i n o f g = {1, 2, 3, 4, 5} \\ S o, d o m a i n o f f . g = D o m a i n o f f \cap D o m a i n o f g \\ = {2, 3, 4, 5} \\ H e n c e, t h e f i l l e r i s {2, 3, 4, 5} . \end{array}$

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9 months ago

Silver Oak University Silver Oak University Courses

What is the total fee for the Silver Oak University BBA course in 2025?

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Chikat Marchang

Contributor-Level 8

The fee structure for the BBA course at Silver Oak University consists of INR 1.1 to 1.6 Lakhs. The fee structure consists not only of tuition fees but also, any refundable security deposits. The total fees may vary for students enrolled in the Hostel, students should consult the official university website for exact information on the Silver Oak University fee structure.

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Two Exemplar Solution

If $f (x) = y = \frac{a x - b}{c x - a}$ , then prove that $f (y) = x$ .

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Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} G i v e n t h a t : f (x) = y = \frac{a x - b}{c x - a} \\ f (y) = \frac{a y - b}{c y - a} = \frac{a [\frac{a x - b}{c x - a}] - b}{c [\frac{a x - b}{c x - a}] - a} \\ \Rightarrow f (y) = \frac{a^{2} x - a b - b c x + a b}{c a x - b c - c a x + a^{2}} = \frac{a^{2} x - b c x}{a^{2} - b c} \\ \Rightarrow f (y) = \frac{(a^{2} - b c)}{a^{2} - b c} = x \\ H e n c e, f (y) = x . \end{array}$

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9 months ago

Silver Oak University Silver Oak University Eligibility

What is the eligibility criteria for the BBA course at Silver Oak University?

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Himanshu Singh

Contributor-Level 10

To be eligible for the BBA course at Silver Oak University, students must have passed Class 12 from a recognised board. This basic qualification is mandatory for admission to the Silver Oak University BBA programme across all specialisations and formats.

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9 months ago

Silver Oak University BBA

What specialisations are available under the Silver Oak University for BBA programme?

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Chikat Marchang

Contributor-Level 8

Silver Oak University offers various specialisations under the BBA programme. A few of the popular specialisations with industry relevance are:

BBA in Aviation.
BBA in Information Technology

Candidates will find that the various specialisations under the BBA programme allow students to make flexible and career-oriented choice that will benefit their future careers.

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9 months ago

Gujarat NEET Medicine Entrance Exam

What is the official website to submit application form for Gujarat NEET counselling?

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Abhishek Divya

Beginner-Level 5

The official website to submit application form for Gujarat NEET counselling is medadmgujarat.org. Candidates will have to generate 14-digit PIN to complete registration process. It must be note all rounds of counselling will be held online by Admission Committee for Professional Under Graduate Medical Educational Courses (ACPUGMEC).

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Two Exemplar Solution

Find the domain and range of the function $f (x) = \frac{1}{\sqrt[]{x - 5}}$ .

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Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} G i v e n t h a t : f (x) = \frac{1}{\sqrt[]{x - 5}} \\ H e r e, i t i s c l e a r t h a t f (x) i s r e a l w h e n x - 5 > 0 \Rightarrow x > 5 \\ H e n c e, t h e d o m a i n = (5, \infty) \\ N o w t o f i n d t h e r a n g e p u t \\ f (x) = y = \frac{1}{\sqrt[]{x - 5}} \\ \Rightarrow \sqrt[]{x - 5} = \frac{1}{y} \Rightarrow x - 5 = \frac{1}{y^{2}} \\ \Rightarrow x = \frac{1}{y^{2}} + 5 \\ F o r x \in (5, \infty), y \in R^{+} . \\ H e n c e, t h e r a n g e o f f = R^{+} . \end{array}$

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Three Exemplar Solution

If $c o s (θ + ϕ) = m c o s (θ - ϕ)$ , then prove that tan $θ = \frac{1 - m}{1 + m} . c o t ϕ$

[Hint: Express $\frac{c o s (θ + α)}{c o s (θ - α)} = \frac{m}{1}$ .]and apply componendo and dividendo.]

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alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} G i v e n t h a t : c o s (θ + ?) = m c o s (θ - ?) \\ \Rightarrow \frac{c o s (θ + ?)}{c o s (θ - ?)} = \frac{m}{1} \\ Using c o m p o n e n d o a n d d i v i d e n d o t h e o r e m, w e g e t \\ \frac{c o s (θ + ?) + c o s (θ - ?)}{c o s (θ + ?) - c o s (θ - ?)} = \frac{m + 1}{m - 1} \\ \Rightarrow \frac{2 c o s (\frac{θ + ? + θ - ?}{2}) . c o s (\frac{θ + ? - θ + ?}{2})}{- 2 s i n (\frac{θ + ? + θ - ?}{2}) . s i n (\frac{θ + ? - θ + ?}{2})} = \frac{m + 1}{m - 1} \\ \Rightarrow \frac{c o s θ . c o s ?}{- s i n θ . s i n ?} = \frac{m + 1}{m - 1} \Rightarrow - c o t θ . c o t ? = \frac{m + 1}{m - 1} \\ \Rightarrow \frac{- c o t ?}{t a n θ} = \frac{m + 1}{m - 1} = - \frac{1 + m}{1 - m} \\ \Rightarrow t a n θ = \frac{1 + m}{1 - m} c o t ? . \\ H e n c e p r o v e d . \end{array}$

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9 months ago

Silver Oak University Silver Oak University Courses

What is the duration of the BBA course at Silver Oak University?

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Himanshu Singh

Contributor-Level 10

BBA course at Silver Oak University is a three-year undergraduate programme divided into six semesters. In addition to the regular BBA course, Silver Oak University also offers an integrated BBA+MBA option for students seeking a seamless transition into postgraduate business studies.

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9 months ago

Gamics Institute Indore Exam

Are there any entrance exam or interviews required for admissions at Gamics Institute?

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Pragati Taneja

Contributor-Level 10

No, certificare course admission is based on merit at Gamics Institute. The selection criteria is merit-based. The college offers various programmes such as 2D Animation, Motion Graphics, UI/UX, etc. The selection criteria is merit-based. The eligibility criteria is to complete Class 12.

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9 months ago

Paramedical Entrance Exam UP CPET Counselling

When did UP CPET counselling begin?

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Shruti Khanna

Beginner-Level 5

The UP CPET 2025 counselling started on July 2, 2025. Applicants who are declared qualified as per the UP CPET results are eligible to participate in the counselling process. The last date to fill for the counselling process is July 9, 2025.

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9 months ago

MICA MICA Admission Process

What is the last date to apply at MICA?

0 Follower 2 Views

Nishtha Shukla

Guide-Level 15

MICA admission 2025 is open. The institute opens applications online on the official website of MICA. The institute has not released any application deadline for the now. The applications are closed for Post Graduate Programme. Aspirants have to pay an application fee to apply for a course at MICA.

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Three Exemplar Solution

If $s i n (θ + α) = a$ and $s i n (θ + β) = b$ , then prove that $c o s 2 (α - β) - 4 a b c o s (α - β) = 1 - 2 a^{2} - 2 b^{2} .$

Hint: Express cos $(α - β) =$ cos (( $θ +$ $α$ ) $-$ $(θ + β)$ )]

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alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} G i v e n t h a t : \\ s i n (θ + α) = a a n d s i n (θ + β) = b \\ c o s (α - β) = c o s [θ + α - θ - β] = c o s [(θ + α) - (θ + β)] \\ = c o s (θ + α) c o s (θ + β) + s i n (θ + α) s i n (θ + β) \\ = \sqrt[]{1 - s i n^{2} (θ + α)} \sqrt[]{1 - s i n^{2} (θ + β)} + a . b \\ = \sqrt[]{(1 - a^{2}) (1 - b^{2})} + a b = a b + \sqrt[]{1 - a^{2} - b^{2} + a^{2} b^{2}} \\ N o w c o s 2 (α - β) - 4 a b c o s (α - β) \\ = 2 c o s^{2} (α - β) - 1 - 4 a b c o s (α - β) [? c o s 2 θ = 2 c o s^{2} θ - 1] \\ = 2 {[a b + \sqrt[]{1 - a^{2} - b^{2} + a^{2} b^{2}}]}^{2} - 1 - 4 a b [a b + \sqrt[]{1 - a^{2} - b^{2} + a^{2} b^{2}}] \\ = 2 [a^{2} b^{2} + 1 - a^{2} - b^{2} + a^{2} b^{2} + 2 a b \sqrt[]{1 - a^{2} - b^{2} + a^{2} b^{2}}] - 1 - 4 a^{2} b^{2} - 4 a b \sqrt[]{1 - a^{2} - b^{2} + a^{2} b^{2}} \\ = 2 a^{2} b^{2} + 2 - 2 a^{2} - 2 b^{2} + 2 a^{2} b^{2} + 4 a b \sqrt[]{1 - a^{2} - b^{2} + a^{2} b^{2}} - 1 - 4 a^{2} b^{2} - 4 a b \sqrt[]{1 - a^{2} - b^{2} + a^{2} b^{2}} \\ = 1 - 2 a^{2} - 2 b^{2} \\ H e n c e, c o s 2 (α - β) - 4 a b c o s (α - β) = 1 - 2 a^{2} - 2 b^{2} \\ H e n c e p r o v e d . \end{array}$

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