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Exemplar Solution Class 11th

$1 + 2 + 2^{2} + \dots + 2^{n} = 2^{n + 1} - 1$ , 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:

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

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Exemplar Solution Class 11th

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}.

2 Views|Posted a year ago

Asked by Payal Gupta

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Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

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

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Exemplar Solution Class 11th

Find the general solution of the equation

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



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

3 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

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

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Exemplar Solution Class 11th

$2 + 4 + 6 + \dots + 2 n = n^{2} + n$ , 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:

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

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Exemplar Solution Class 11th

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

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

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

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Exemplar Solution Class 11th

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

6 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:

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

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Exemplar Solution Class 11th

Find the general solution of the equation $s i n x - 3 s i n 2 x + s i n 3 x = c o s x - 3 c o s 2 x + c o s 3 x .$

4 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

$\begin{array}{l} G i v e n t h a t : s i n x - 3 s i n 2 x + s i n 3 x = c o s x - 3 c o s 2 x + c o s 3 x \\ \Rightarrow (s i n 3 x + s i n x) - 3 s i n 2 x = (c o s 3 x + c o s x) - 3 c o s 2 x \\ \Rightarrow 2 s i n (\frac{3 x + x}{2}) . c o s (\frac{3 x - x}{2}) - 3 s i n 2 x = 2 c o s (\frac{3 x + x}{2}) . c o s (\frac{3 x - x}{2}) - 3 c o s 2 x \\ \Rightarrow 2 s i n 2 x . c o s x - 3 s i n 2 x = 2 c o s 2 x . c o s x - 3 c o s 2 x \\ \Rightarrow 2 s i n 2 x . c o s x - 2 c o s 2 x . c o s x = 3 s i n 2 x - 3 c o s 2 x \\ \Rightarrow 2 c o s x (s i n 2 x - c o s 2 x) - 3 (s i n 2 x - c o s 2 x) = 0 \\ \Rightarrow (s i n 2 x - c o s 2 x) (2 c o s x - 3) = 0 \\ \Rightarrow s i n 2 x - c o s 2 x = 0 a n d 2 c o s x - 3 \neq 0 [? - 1 \leq c o s x \leq 1] \\ \Rightarrow \frac{s i n 2 x}{c o s 2 x} - 1 = 0 \Rightarrow t a n 2 x = 1 \\ \Rightarrow t a n 2 x = t a n \frac{π}{4} \Rightarrow 2 x = n π + \frac{π}{4} ∴ x = \frac{n π}{2} + \frac{π}{8} \\ H e n c e, t h e g e n e r a l s o l u t i o n o f t h e e q u a t i o n i s \\ x = \frac{n π}{2} + \frac{π}{8}, n \in Z . \end{array}$

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Exemplar Solution Class 11th

$2^{n} < (n + 2)!$ , 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:

$\begin{array}{l} L e t P (n) : 2 n < (n + 2)! f o r a l l k \in N . \\ S t e p 1 : P (1) : 2.1 < (1 + 2)! \Rightarrow 2 < 3! \Rightarrow 2 < 6 w h i c h i s t r u e f o r P (1) . \\ S t e p 2 : P (k) : 2 k < (k + 2)! L e t i t b e t r u e f o r P (k) \\ S t e p 3 : P (k + 1) : 2 (k + 1) < (k + 1 + 2)! \\ F r o m S t e p 2, w e g e t \\ 2 k < (k + 2)! \\ \Rightarrow 2 k + 2 < (k + 2)! + 2 \\ \Rightarrow 2 (k + 1) < (k + 2)! + 2 \\ Also, (k + 2)! + 2 < (k + 3)! \\ ∴ 2 (k + 1) < (k + 3)! \\ \Rightarrow 2 (k + 1) < (k + 2 + 1)! w h i c h i s t r u e f o r P (k + 1) \\ H e n c e, P (k + 1) i s t r u e w h e n e v e r P (k) i s t r u e . \end{array}$

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Exemplar Solution Class 11th

$n^{2} < 2^{n}$ , for all natural numbers $n \geq 5$ .

3 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:

$\begin{array}{l} L e t P (n) : n^{2} < 2^{n} f o r a l l n a t u r a l n u m b e r s, n \geq 5 \\ S t e p 1 : P (5) : 1^{5} < 2^{5} \Rightarrow 1 < 32 w h i c h i s t r u e f o r P (5) . \\ S t e p 2 : P (k) : k^{2} < 2^{k} . L e t i t b e t r u e f o r k \in N . \\ S t e p 3 : P (k + 1) : {(k + 1)}^{2} < 2^{k + 1} \\ F r o m S t e p 2, w e g e t \\ k^{2} < 2^{k} \\ k^{2} + 2 k + 1 < 2^{k} + 2 k + 1 \\ {(k + 1)}^{2} < 2^{k} + 2 k + 1 \dots (i) \\ Since (2 k + 1) < 2^{k} \\ S o k^{2} + 2 k + 1 < 2^{k} + 2^{k} \\ k^{2} + 2 k + 1 < 2 . 2^{k} \\ k^{2} + 2 k + 1 < 2^{k + 1} (i i) \\ F r o m e q n . (i) a n d (i i), w e g e t {(k + 1)}^{2} < 2^{k + 1} . \\ H e n c e, P (k + 1) i s t r u e w h e n e v e r P (k) i s t r u e f o r k \in N, n \geq 5 . \end{array}$

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Exemplar Solution Class 11th

If $f (x) = a x + b$ , where  $a$ and  $b$ are integers, $f (- 1) = - 5$  and $f (3) = 3$ , then  $a$ and $b$ are equal to

(a) $a = - 3, b = - 1$  

(b) $a = 2, b = - 3$  

(c) $a = 0, b = 2$  

(d) $a = 2, b = 3$

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

$\begin{array}{l} G i v e n t h a t : f (x) = a x + b \\ \Rightarrow f (- 1) = a (- 1) + b \\ \Rightarrow - 5 = - a + b \\ \Rightarrow a - b = 5 \dots (i) \\ \Rightarrow f (3) = 3 a + b \\ \Rightarrow 3 = 3 a + b \\ \Rightarrow 3 a + b = 3 \dots (i i) \\ O n s o l v i n g e q n . (i) a n d (i i), w e g e t a = 2, b = - 3 \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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

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

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

Find the general solution of the equation $s i n x - 3 s i n 2 x + s i n 3 x = c o s x - 3 c o s 2 x + c o s 3 x .$

$2^{n} < (n + 2)!$ , for all natural numbers $n$ .

$n^{2} < 2^{n}$ , for all natural numbers $n \geq 5$ .