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Exemplar Solution Class 11th

Choose the correct answer from the given four options in each of the Exercises 30 to 59:

If $s i n θ + c o s e c θ = 2$ , then $s i n^{2} θ + c o s e c^{2} θ$ is equal to:

(a) 1

(b) 4

(c) 2

(d) None of these

11 Views|Posted a year ago

Asked by Alok Kumar Singh

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Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : s i n θ + c o s e c θ = 2 \\ S q u a r i n g b o t h s i d e s, w e g e t \\ {(s i n θ + c o s e c θ)}^{2} = {(2)}^{2} \\ \Rightarrow s i n^{2} θ + c o s e c^{2} θ + 2 s i n θ c o s e c θ = 4 \\ \Rightarrow s i n^{2} θ + c o s e c^{2} θ + 2 s i n θ \times \frac{1}{s i n θ} = 4 \\ \Rightarrow s i n^{2} θ + c o s e c^{2} θ + 2 = 4 \\ \Rightarrow s i n^{2} θ + c o s e c^{2} θ = 2 \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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Exemplar Solution Class 11th

Prove that, for all $n \in ℕ$ :

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

5 Views|Posted a year ago

Asked by Vishal Baghel

S

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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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Exemplar Solution Class 11th

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) $\{- 1, \frac{4}{3}\}$ <!-- [if gte msEquation 12]>-1,43

(d) $\{- 1, \frac{4}{3}\}$ <!-- [if gte msEquation 12]>-1,43

2 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by
Payal Gupta | Contributor-Level 10

a year ago

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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Exemplar Solution Class 11th

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 $n \in ℕ$ .

5 Views|Posted a year ago

Asked by Vishal Baghel

S

Answered by
Vishal Baghel | Contributor-Level 10

a year ago

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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Exemplar Solution Class 11th

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 $n \in ℕ$ .

6 Views|Posted a year ago

Asked by Vishal Baghel

S

Answered by
Vishal Baghel | Contributor-Level 10

a year ago

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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Exemplar Solution Class 11th

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 = $(- \infty, 2]$

2 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by
Payal Gupta | Contributor-Level 10

a year ago

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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Exemplar Solution Class 11th

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 $n \in ℕ$ .

4 Views|Posted a year ago

Asked by Vishal Baghel

S

Answered by
Vishal Baghel | Contributor-Level 10

a year ago

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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Exemplar Solution Class 11th

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) $? - (3, - 2)$

2 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by
Payal Gupta | Contributor-Level 10

a year ago

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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Exemplar Solution Class 11th

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 = $[0, \infty)$

3 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by
Payal Gupta | Contributor-Level 10

a year ago

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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Exemplar Solution Class 11th

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

4 Views|Posted a year ago

Asked by Vishal Baghel

S

Answered by
Vishal Baghel | Contributor-Level 10

a year ago

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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Choose the correct answer from the given four options in each of the Exercises 30 to 59:

If $s i n θ + c o s e c θ = 2$ , then $s i n^{2} θ + c o s e c^{2} θ$ is equal to:

(a) 1

(b) 4

(c) 2

(d) None of these

Prove that, for all $n \in ℕ$ :

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

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 $n \in ℕ$ .

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 $n \in ℕ$ .

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 $n \in ℕ$ .

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