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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Four Exemplar Solution

Prove that, for all $n \in ℕ$ :

$c o s α + c o s (α + β) + c o s (α + 2 β) + \dots + c o s (α + (n - 1) β) = \frac{c o s (α + \frac{(n - 1) β}{2}) s i n \frac{n β}{2}}{s i n \frac{β}{2}} .$

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Vishal Baghel

Contributor-Level 10

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

$\begin{array}{l} L e t P (n) : c o s α + c o s (α + β) + c o s (α + 2 β) + \dots + c o s [α + (n - 1) β] \\ = \frac{c o s [α + (\frac{n - 1}{2}) β] [s i n \frac{n β}{2}]}{s i n \frac{β}{2}} \\ S t e p 1 : P (1) : c o s α = \frac{(c o s α) (s i n \frac{β}{2})}{s i n \frac{β}{2}} = c o s α w h i c h i s t r u e f o r P (1) . \\ S t e p 2 : P (k) : c o s α + c o s (α + β) + c o s (α + 2 β) + \dots + c o s [α + (k - 1) β] \\ = \frac{c o s [α + (\frac{k - 1}{2}) β] [s i n \frac{k β}{2}]}{s i n \frac{β}{2}} L e t i t b e t r u e . \\ S t e p 3 : P (k + 1) : c o s α + c o s (α + β) + c o s (α + 2 β) + \dots + c o s [α + (k - 1) β] + c o s [α + (k + 1 - 1) β] \\ = \frac{c o s [α + (\frac{k - 1}{2}) β] [s i n \frac{k β}{2}]}{s i n \frac{β}{2}} + c o s (α + k β) [F r o m S t e p 2] \\ = \frac{2 c o s [α + (\frac{k - 1}{2}) β] (s i n \frac{k β}{2}) + 2 c o s (α + k β) . s i n \frac{β}{2}}{2 s i n \frac{β}{2}} \\ = \frac{s i n [α + k β - \frac{β}{2}] - s i n [α - \frac{β}{2}] + s i n [α + k β + \frac{β}{2}] - s i n [α + k β - \frac{β}{2}]}{2 s i n \frac{β}{2}} \\ [? s i n A - s i n B = 2 c o s \frac{A + B}{2} . s i n \frac{A - B}{2}] \\ = \frac{c o s (α + \frac{k β}{2}) . s i n (k + 1) \frac{β}{2}}{s i n \frac{β}{2}} \\ = \frac{c o s [α + (\frac{k + 1 - 1}{2}) β] . s i n (\frac{k + 1}{2}) β}{s i n \frac{β}{2}} w h i c h i s t r u e f o r P (k + 1) . \\ H e n c e, P (k + 1) i s t r u e w h e n e v e r P (k) i s t r u e . \end{array}$

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Two Exemplar Solution

The domain for which the functions defined by $f (x) = 3 x^{2} - 1$ and $g (x) = 3 + x$ are equal is

(a) $\{- 1, \frac{4}{3}\}$

(b) $\{- 1, \frac{4}{3}\}$

(c) $\{- 1, \frac{4}{3}\}$

(d) $\{- 1, \frac{4}{3}\}$

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Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

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

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Four Exemplar Solution

A sequence $d_{1}, d_{2}, d_{3}, \dots$ is defined by letting $d_{1} = 2$ and $d_{k} = \frac{d_{k - 1}}{k}$ , for all $k \geq 2$ . Show that $d_{n} = \frac{2}{n!}$ , for all $n \in ℕ$ .

0 Follower 4 Views

Vishal Baghel

Contributor-Level 10

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

$\begin{array}{l} G i v e n t h a t : d_{1} = 2 a n d d_{k} = \frac{d_{k - 1}}{k} \\ L e t P (n) : d_{n} = \frac{2}{n!} \\ S t e p 1 : P (1) : d_{1} = \frac{2}{1!} = 2 w h i c h i s t r u e f o r P (1) . \\ S t e p 2 : P (k) : d_{k} = \frac{2}{k!} . L e t i t b e t r u e f o r P (k) . \\ S t e p 3 : G i v e n t h a t : d_{k} = \frac{d_{k - 1}}{k} \\ ∴ d_{k + 1} = \frac{d_{k + 1 - 1}}{k + 1} = \frac{d_{k}}{k + 1} \\ \Rightarrow d_{k + 1} = \frac{1}{k + 1} . d_{k} = \frac{1}{k + 1} . \frac{2}{k!} \\ \Rightarrow d_{k + 1} = \frac{2}{(k + 1)!} w h i c h i s t r u e f o r P (k + 1) . \\ H e n c e, P (k + 1) i s t r u e w h e n e v e r P (k) i s t r u e . \end{array}$

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Four Exemplar Solution

A sequence $b_{0}, b_{1}, b_{2}, \dots$ is defined by letting $b_{0} = 5$ and $b_{k} = 4 + b_{k - 1}$ , for all $k \geq 1$ . Show that $b_{n} = 5 + 4 n$ , for all $n \in ℕ$ .

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Vishal Baghel

Contributor-Level 10

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

$\begin{array}{l} W e h a v e b_{0} = 5 a n d b_{k} = 4 + b_{k - 1} \\ \Rightarrow b_{0} = 5, b_{1} = 4 + b_{0} = 4 + 5 = 9 a n d b_{2} = 4 + b_{1} = 4 + 9 = 13 \\ L e t P (n) : b_{n} = 5 + 4 n \\ S t e p 1 : P (1) : b_{1} = 5 + 4 = 9 \Rightarrow 9 = 9 w h i c h i s t r u e f o r P (1) . \\ S t e p 2 : P (k) : b_{k} = 5 + 4 k . L e t i t b e t r u e \forall k \in N . \\ S t e p 3 : G i v e n t h a t : \\ P (k) = 4 + b_{k - 1} \\ \Rightarrow P (k + 1) = 4 + b_{k + 1 - 1} \\ \Rightarrow P (k + 1) = 4 + b_{k} = 4 + 5 + 4 k \\ \Rightarrow P (k + 1) = 4 + b_{k} = 5 + 4 (k + 1) w h i c h i s t r u e f o r P (k + 1) . \\ H e n c e, P (k + 1) i s t r u e w h e n e v e r P (k) i s t r u e . \end{array}$

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Two Exemplar Solution

The domain and range of the function $f$ given by $f (x) = 2 - \sqrt[]{x - 5}$ is

(a) Domain = $?^{+}$ , Range = $(- \infty, 1]$

(b) Domain = $?,$ Range = $(- \infty, 2]$

(c) Domain = $?,$ Range = $(- \infty, 2)$

(d) Domain = $?^{+}$ , Range = $(- \infty, 2]$

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : f (x) = 2 - | x - 5 | \\ f (x) i s d e f i n e d f o r x \in R \\ ∴ D o m a i n o f f (x) = R \\ N o w, | x - 5 | \geq 0 \Rightarrow - | x - 5 | \leq 0 \\ \Rightarrow 2 - | x - 5 | \leq 2 \Rightarrow f (x) \leq 2 \\ ∴ R a n g e o f f (x) = (- \infty, 2] \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Four Exemplar Solution

A sequence $a_{1}, a_{2}, a_{3}, \dots$ is defined by letting $a_{1} = 3$ and $a_{k} = 7 a_{k - 1}$ , for all $k \geq 2$ . Show that $a_{n} = 3 \cdot 7^{n - 1}$ , for all $n \in ℕ$ .

0 Follower 4 Views

Vishal Baghel

Contributor-Level 10

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

$\begin{array}{l} G i v e n t h a t : a_{1} = 3 \\ a_{2} = 7 a_{2 - 1} = 7 . a_{1} = 7.3 = 21 \\ a_{3} = 7 . a_{3 - 1} = 7 . a_{2} = 7.21 = 147 \\ \dots \\ L e t P (n) : a_{n} = 3 . 7^{n - 1}, \forall n \in N . \\ S t e p 1 : P (2) : a_{2} = 3 . 7^{2 - 1} = 21 \Rightarrow 21 = 21 w h i c h i s t r u e f o r P (2) . \\ S t e p 2 : P (k) : a_{k} = 3 . 7^{k - 1} . L e t i t b e t r u e . \\ S t e p 3 : a_{k} = 7 a_{k - 1} (G i v e n) \\ P u t k = k + 1 \\ a_{k + 1} = 7 a_{k} = 7 (3 . 7^{k - 1}) = 3 . 7^{k + 1 - 1} = 3 . 7^{(k + 1) - 1} \\ w h i c h i s t r u e f o r P (k + 1) . \\ H e n c e, P (k + 1) i s t r u e w h e n e v e r P (k) i s t r u e . \end{array}$

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9 months ago

Kerala University B.Sc.

Can we get direct admission in B.Sc at Kerala Agricultural University?

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Indrani Kumar

Contributor-Level 10

Kerala Agricultural University BSc admission is completely entrance-based. The university offers several specialisations in the BSc course and accepts different entrance examination scores. The accepted entrance exams are KEAM Medical, ICAR CUET (the Common University Entrance Test (CUET) conducted for admissions to undergraduate programmes in agriculture and allied sciences at various Agricultural Universities (AUs) in India, etc. Therefore, candidates are not allowed to get direct admission to BSc courses at KAU.

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Two Exemplar Solution

The domain $f$ of the function given by $f (x) = \frac{x^{2} + 2 x + 1}{x^{2} - x - 6}$ is

(a) $? - {3, - 2}$

(b) $? - {- 3, 2}$

(c) $? - [3, - 2]$

(d) $? - (3, - 2)$

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : f (x) = \frac{x^{2} + 2 x + 1}{x^{2} - x - 6} \\ f (x) i s d e f i n e d i f x^{2} - x - 6 \neq 0 \\ \Rightarrow x^{2} - 3 x + 2 x - 6 \neq 0 \\ \Rightarrow (x - 3) (x + 2) \neq 0 \Rightarrow x \neq - 2, x \neq 3 \\ S o, t h e d o m a i n o f f (x) = R - {- 2, 3} \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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9 months ago

Dhenkanal Autonomous College M.Sc

How many seats are there in Dhenkanal Autonomous College for PG Chemistry?

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9 months ago

DTU - Delhi Technological University DTU - Delhi Technological University Placement

Does DTU actually offers good placements?

0 Follower 9 Views

Shoaib Mehdi

Contributor-Level 10

The highest and average package recorded in 2025 was 85 LPA and 15.75 LPA. 1766 offers were made in 2025.

Delhi Technological University has a decent record of placements for its BTech, BDes, MTec and MBA courses. As per the latest report, 2,053 job offers were rolled out by over 350 participating companies during the DTU placements 2024. More than 1,900 students from BTech, MTech, MBA, and other courses were placed in reputed companies. The key highlights related to Delhi Technological University placements 2024 are tabulated below:

Particulars	Placement Statistics (2024)
Highest package	INR 85.3
Average package	INR 15.45 LPA
No. of offers	2,053
Students placed	1,900+
Total recruiters	350+

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Two Exemplar Solution

The domain and range of real function $f$ defined by $f (x) = \sqrt[]{x - 1}$ is given by

(a) Domain = $(1, \infty)$ , Range = $(0, \infty)$

(b) Domain = $[1, \infty)$ , Range = $(0, \infty)$

(c) Domain = $[1, \infty)$ , Range = $[0, \infty)$

(d) Domain = $[1, \infty)$ , Range = $[0, \infty)$

0 Follower 3 Views

Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

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

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Four Exemplar Solution

$1 + 5 + 9 +...$
$+ (4 n - 3) = n (2 n - 1)$ , for all natural numbers $n$ .

0 Follower 4 Views

Vishal Baghel

Contributor-Level 10

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

$\begin{array}{l} L e t P (n) : 1 + 5 + 9 + \dots + (4 n - 3) = n (2 n - 1), \forall n \in N . \\ S t e p 1 : P (1) : 1 = 1 (2.1 - 1) = 1 w h i c h i s t r u e f o r P (1) . \\ S t e p 2 : P (k) : 1 + 5 + 9 + \dots + (4 k - 3) = k (2 k - 1) . L e t i t b e t r u e . \\ S t e p 3 : P (k + 1) : 1 + 5 + 9 + \dots + (4 k - 3) + (4 k + 1) \\ = k (2 k - 1) + (4 k + 1) = 2 k^{2} - k + 4 k + 1 \\ = 2 k^{2} + 3 k + 1 = 2 k^{2} + 2 k + k + 1 \\ = 2 k (k + 1) + 1 (k + 1) = (2 k + 1) (k + 1) \\ = (k + 1) (2 k + 2 - 1) = (k + 1) [2 (k + 1) - 1] \\ w h i c h i s t r u e f o r P (k + 1) . \\ H e n c e, P (k + 1) i s t r u e w h e n e v e r P (k) i s t r u e . \end{array}$

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Four Exemplar Solution

$1 + 2 + 2^{2} + \dots + 2^{n} = 2^{n + 1} - 1$ , for all natural numbers $n$ .

0 Follower 4 Views

Vishal Baghel

Contributor-Level 10

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

$\begin{array}{l} L e t P (n) : 1 + 2 + 2^{2} + \dots + 2^{n} = 2^{n + 1} - 1, \forall n \in N . \\ P (n) : 2^{0} + 2^{1} + 2^{2} + \dots + 2^{n} = 2^{n + 1} - 1 \\ S t e p 1 : P (1) : 2^{0} = 2^{0 + 1} - 1 = 2 - 1 = 2^{0} w h i c h i s t r u e . \\ S t e p 2 : P (k) : 2^{0} + 2^{1} + 2^{2} + \dots + 2^{k} = 2^{k + 1} - 1 . L e t i t b e t r u e . \\ S t e p 3 : P (k + 1) : 2^{0} + 2^{1} + 2^{2} + \dots + 2^{k} + 2^{k + 1} \\ = 2^{k + 1} - 1 + 2^{k + 1} = 2 . 2^{k + 1} - 1 = 2^{k + 2} - 1 \\ = 2^{(k + 1) + 1} - 1 w h i c h i s t r u e f o r P (k + 1) \\ H e n c e, P (k + 1) i s t r u e w h e n e v e r P (k) i s t r u e . \end{array}$

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9 months ago

Public Policy in India Political Science

Which is the best MA in Public Policy college in India?

0 Follower 10 Views

Tasbiya Khan

Contributor-Level 10

As per popularity basis, listed below are the top MA Public Policy colleges in India along with their tuition fees:

Top Colleges	Tuition Fee
St Xavier College, Mumbai	INR 1.5 lakh
Amity University Noida	INR 2.2 lakh
Christ University Bangalore	INR 4.8 lakh
Galgotias University	INR 92,000
TISS Mumbai	INR 68,000

Disclaimer: This information is sourced from official website and may vary.

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Two Exemplar Solution

The domain and range of the real function $f$ defined by $f (x) = \frac{x - 4}{x - 4}$ is given by

(a) Domain = $?$ , Range = {-1, 1}

(b) Domain = $? - {1}$ , Range = $? - {1}$

(c) Domain = $? - {4}$ , Range = {-1}

(d) Domain = $? - {- 4}$ , Range = {-1, 1}.

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : f (x) = \frac{4 - x}{x - 4} \\ W e k n o w t h a t f (x) i s d e f i n e d i f x - 4 \neq 0 \Rightarrow x \neq 4 \\ S o, t h e d o m a i n o f f (x) i s = R - {4} \\ L e t f (x) = y = \frac{4 - x}{x - 4} \\ \Rightarrow y x - 4 y = 4 - x \Rightarrow y x + x = 4 y + 4 \\ \Rightarrow x (y + 1) = 4 y + 4 \Rightarrow x = \frac{4 (1 + y)}{1 + y} \\ I f x i s r e a l n u m b e r, t h e n 1 + y \neq 0 \Rightarrow y \neq - 1 \\ ∴ R a n g e o f f (x) = R - {- 1} \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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9 months ago

Political Science Humanities & Social Sciences

How many MA in Public Policy colleges are there in India?

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Tasbiya Khan

Contributor-Level 10

About 45+ MA in Public Policy colleges are there in India. CUET PG, TISSNET, etc. are some of the most accepting entrance exams in the top MA colleges in India. Some of the popular colleges include St. Xavier's College, Mumbai, Amity University, Noida, Christ University, School of Humanities and Social Sciences, Jain Deemed to be University, Galgotias University, TISS - Tata Institute of Social Sciences, Mumbai, University of Mumbai [MU], and many others.

Of these, 20 colleges are privately owned and 17 colleges are owned by public/government organisations.

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Three Exemplar Solution

Find the general solution of the equation

$(\sqrt[]{3} - 1) c o s θ + (\sqrt[]{3} + 1) s i n θ = 2 .$

[Hint: Put $\sqrt[]{3} - 1 = r s i n α, \sqrt[]{3} + 1 = r c o s α w h i c h g i v e s$ . ] $t a n α = t a n (\frac{π}{4} - \frac{π}{6}) α = \frac{π}{12}]$

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} G i v e n t h a t : (\sqrt[]{3} - 1) c o s θ + (\sqrt[]{3} + 1) s i n θ = 2 \\ P u t \sqrt[]{3} - 1 = r s i n α, \sqrt[]{3} + 1 = r c o s α \\ S q u a r i n g a n d a d d i n g, w e g e t \\ r^{2} = 3 + 1 - 2 \sqrt[]{3} + 3 + 1 + 2 \sqrt[]{3} \\ r^{2} = 8 \Rightarrow r = \pm 2 \sqrt[]{2} \\ N o w t h e g i v e n e q u a t i o n c a n b e w r i t t e n a s \\ r s i n α c o s θ + r c o s α s i n θ = 2 \\ \Rightarrow r (s i n α c o s θ + c o s α s i n θ) = 2 \\ \Rightarrow 2 \sqrt[]{2} s i n (α + θ) = 2 \\ \Rightarrow s i n (α + θ) = \frac{2}{2 \sqrt[]{2}} = \frac{1}{\sqrt[]{2}} \\ \Rightarrow s i n (α + θ) = s i n \frac{π}{4} \\ \Rightarrow α + θ = n π + {(- 1)}^{n} . \frac{π}{4} \dots (i) \\ N o w \frac{r s i n α}{r c o s α} = \frac{\sqrt[]{3} - 1}{\sqrt[]{3} + 1} \\ \Rightarrow t a n α = \frac{t a n \frac{π}{3} - t a n \frac{π}{4}}{1 + t a n \frac{π}{4} . t a n \frac{π}{3}} \\ \Rightarrow t a n α = t a n (\frac{π}{3} - \frac{π}{4}) \\ \Rightarrow t a n α = t a n \frac{π}{12} ∴ α = \frac{π}{12} \\ P u t t i n g t h e v a l u e o f α i n e q n . (i) w e g e t \\ \frac{π}{12} + θ = n π + {(- 1)}^{n} . \frac{π}{4} \\ ∴ θ = n π + {(- 1)}^{n} . \frac{π}{4} - \frac{π}{12} \\ H e n c e, t h e g e n e r a l s o l u t i o n o f t h e e q u a t i o n i s \\ θ = n π + {(- 1)}^{n} . \frac{π}{4} - \frac{π}{12}, n \in Z . \end{array}$

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Four Exemplar Solution

$2 + 4 + 6 + \dots + 2 n = n^{2} + n$ , for all natural numbers $n$ .

0 Follower 4 Views

Vishal Baghel

Contributor-Level 10

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

$\begin{array}{l} L e t P (n) : 2 + 4 + 6 + \dots + 2 n = n^{2} + n, \forall n \in N . \\ S t e p 1 : P (1) : 2 = 1^{2} + 1 = 2 w h i c h i s t r u e f o r P (1) . \\ S t e p 2 : P (k) : 2 + 4 + 6 + \dots + 2 k = k^{2} + k L e t i t b e t r u e . \\ S t e p 3 : P (k + 1) : 2 + 4 + 6 + \dots + 2 k + 2 k + 2 \\ = k^{2} + k + 2 k + 2 = k^{2} + 3 k + 2 \\ = k^{2} + 2 k + k + 1 + 1 = {(k + 1)}^{2} + (k + 1) \\ W h i c h i s t r u e f o r P (k + 1) \\ H e n c e, P (k + 1) i s t r u e w h e n e v e r P (k) i s t r u e . \end{array}$

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Two Exemplar Solution

The domain of the function $f$ defined by $f (x) = \sqrt[]{4 - x} + \frac{2}{\sqrt[]{x^{2}} - 1}$ is equal to

(a) $(- \infty, - 1) \cup (1, 4]$

(b) $(- \infty, - 1] \cup (1, 4]$

(c) $(- \infty, - 1) \cup [1, 4]$

(d) $(- \infty, - 1) \cup [1, 4)$

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : f (x) = \sqrt[]{4 - x} + \frac{1}{\sqrt[]{x^{2} - 1}} \\ f (x) i s d e f i n e d i f 4 - x \geq 0 o r x^{2} - 1 > 0 \\ \Rightarrow - x \geq - 4 o r (x - 1) (x + 1) > 0 \\ \Rightarrow x \leq 4 o r x < - 1 a n d x > 1 \\ ∴ D o m a i n o f f (x) = (- \infty, - 1) \cup (1, 4] \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Four Exemplar Solution

$\frac{1}{1} + \frac{1}{2} + \dots + \frac{1}{n} < n$ , for all natural numbers $n \geq 2$ .

0 Follower 5 Views

Vishal Baghel

Contributor-Level 10

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

$\begin{array}{l} L e t P (n) : \sqrt[]{n} < \frac{1}{\sqrt[]{1}} + \frac{1}{\sqrt[]{2}} + \dots + \frac{1}{\sqrt[]{n}}, \forall n \geq 2 \\ S t e p 1 : P (2) : \sqrt[]{2} < \frac{1}{\sqrt[]{1}} + \frac{1}{\sqrt[]{2}} w h i c h i s t r u e . \\ S t e p 2 : P (k) : \sqrt[]{k} < \frac{1}{\sqrt[]{1}} + \frac{1}{\sqrt[]{2}} + \dots + \frac{1}{\sqrt[]{k}} L e t i t b e t r u e . \\ S t e p 3 : P (k + 1) : \sqrt[]{k + 1} < \frac{1}{\sqrt[]{1}} + \frac{1}{\sqrt[]{2}} + \dots + \frac{1}{\sqrt[]{k + 1}} \\ F r o m S t e p 2, w e g e t \\ \sqrt[]{k} < \frac{1}{\sqrt[]{1}} + \frac{1}{\sqrt[]{2}} + \dots + \frac{1}{\sqrt[]{k}} \\ \Rightarrow \sqrt[]{k} + \frac{1}{\sqrt[]{k + 1}} < \frac{1}{\sqrt[]{1}} + \frac{1}{\sqrt[]{2}} + \dots + \frac{1}{\sqrt[]{k}} + \frac{1}{\sqrt[]{k + 1}} \\ \Rightarrow \frac{\sqrt[]{k} . \sqrt[]{k + 1} + 1}{\sqrt[]{k + 1}} < \frac{1}{\sqrt[]{1}} + \frac{1}{\sqrt[]{2}} + \dots + \frac{1}{\sqrt[]{k}} + \frac{1}{\sqrt[]{k + 1}} \dots (i) \\ N o w i f \sqrt[]{k + 1} < \frac{\sqrt[]{k} . \sqrt[]{k + 1} + 1}{\sqrt[]{k + 1}} \\ \Rightarrow (k + 1) < \sqrt[]{k} . \sqrt[]{k + 1} + 1 \\ \Rightarrow k < \sqrt[]{k} . \sqrt[]{k + 1} \dots (i i) \\ F r o m e q n . (i) a n d (i i) w e g e t \\ \sqrt[]{k + 1} < \frac{1}{\sqrt[]{1}} + \frac{1}{\sqrt[]{2}} + \dots + \frac{1}{\sqrt[]{k + 1}} \\ H e n c e, P (k + 1) i s t r u e w h e n e v e r P (k) i s t r u e . \end{array}$

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