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New answer posted•a year ago

Exemplar Solution Class 11th

If $\frac{1 + i}{1 - i} = x + i y$ , then find the value of $x + y$ .

2 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

$\begin{array}{l} W e h a v e \frac{{(1 + i)}^{2}}{2 - i} = x + i y \\ \Rightarrow \frac{1 + i^{2} + 2 i}{2 - i} = x + i y \Rightarrow \frac{1 - 1 + 2 i}{2 - i} = x + i y \\ \Rightarrow \frac{2 i}{2 - i} = x + i y \Rightarrow \frac{2 i}{2 - i} \times \frac{2 + i}{2 + i} = x + i y \\ \Rightarrow \frac{4 i + 2 i^{2}}{4 - i^{2}} = x + i y \Rightarrow \frac{4 i - 2}{4 + 1} = x + i y [? i^{2} = - 1] \\ \Rightarrow \frac{- 2 + 4 i}{5} = x + i y \Rightarrow \frac{- 2}{5} + \frac{4}{5} i = x + i y \\ C o m p a r i n g t h e r e a l a n d i m a g i n a r y p a r t s, w e g e t x = \frac{- 2}{5}, y = \frac{4}{5} . \\ H e n c e, x + y = \frac{- 2}{5} + \frac{4}{5} = \frac{2}{5} . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

If $\frac{{(1 + i)}^{3}}{1 - i^{3}} - \frac{{(1 - i)}^{3}}{1 + i^{3}} = x + i y$ , then find $(x, y)$ .

2 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

$\begin{array}{l} W e h a v e {(\frac{1 + i}{1 - i})}^{3} - {(\frac{1 - i}{1 + i})}^{3} = x + i y \\ \Rightarrow {[\frac{1 + i}{1 - i} \times \frac{1 + i}{1 + i}]}^{3} - {[\frac{1 - i}{1 + i} \times \frac{1 - i}{1 - i}]}^{3} = x + i y \\ \Rightarrow {[\frac{1 + i^{2} + 2 i}{1 - i^{2}}]}^{3} - {[\frac{1 + i^{2} - 2 i}{1 - i^{2}}]}^{3} = x + i y \\ \Rightarrow {[\frac{1 - 1 + 2 i}{1 + 1}]}^{3} - {[\frac{1 - 1 - 2 i}{1 + 1}]}^{3} = x + i y \\ \Rightarrow {[\frac{2 i}{2}]}^{3} - {[\frac{2 i}{2}]}^{3} = x + i y \\ \Rightarrow {(i)}^{3} - {(- i)}^{3} = x + i y \Rightarrow i^{2} . i + i^{2} . i = x + i y \\ \Rightarrow - i - i = x + i y \Rightarrow 0 - 2 i = x + i y \\ C o m p a r i n g t h e r e a l a n d i m a g i n a r y p a r t s, w e g e t x = 0, y = - 2 . \\ H e n c e, (x, y) = (0, - 2) . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

The value of tan 3A -tan2A-tanA is equal to

(a) tan3Atan2AtanA

(b) -tan3Atan2AtanA

(c) tanAtan2A-tan2Atan3A-tan3AtanA

(d) None of these

7 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n expression i s t a n 3 A - t a n 2 A - t a n A \\ t a n 3 A = t a n (2 A + A) \\ \Rightarrow t a n 3 A = \frac{t a n 2 A + t a n A}{1 - t a n 2 A . t a n A} \\ \Rightarrow t a n 3 A (1 - t a n 2 A . t a n A) = t a n 2 A + t a n A \\ \Rightarrow t a n 3 A - t a n 3 A t a n 2 A t a n A = t a n 2 A + t a n A \\ \Rightarrow t a n 3 A - t a n 2 A - t a n A = t a n 3 A t a n 2 A t a n A \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

Evaluate $\sum_{n = 1} {(1 + i)}^{3}$ , where $n \in ?$ .

3 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

$\begin{array}{l} W e h a v e \sum_{n = 1}^{13} (i^{n} + i^{n + 1}) \\ = (i + i^{2}) + (i^{2} + i^{3}) + (i^{3} + i^{4}) + (i^{4} + i^{5}) + (i^{5} + i^{6}) + (i^{6} + i^{7}) + (i^{7} + i^{8}) + (i^{8} + i^{9}) + (i^{9} + i^{10}) \\ (i^{10} + i^{11}) + (i^{11} + i^{12}) + (i^{12} + i^{13}) + (i^{13} + i^{14}) \\ = i + 2 (i^{2} + i^{3} + i^{4} + i^{5} + i^{6} + i^{7} + i^{8} + i^{9} + i^{10} + i^{11} + i^{12} + i^{13}) + i^{14} \\ = i + 2 [- 1 - i + 1 + i - 1 - i + 1 + i - 1 - i + 1 + i] + (- 1) \\ = i + 2 (0) - 1 \Rightarrow - 1 + i \\ H e n c e, \sum_{n = 1}^{13} (i^{n} + i^{n + 1}) = - 1 + i . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

For a positive integer $n$ , find the value of ${(1 - i)}^{n} \frac{1}{1 - i^{n}}$ .

6 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

$\begin{array}{l} W e h a v e {(1 - i)}^{n} {(1 - \frac{1}{i})}^{n} \\ = {[(1 - i) (1 - \frac{1}{i})]}^{n} = {[(1 - i) (1 - \frac{1}{i} \times \frac{i}{i})]}^{n} = {[(1 - i) (1 - \frac{i}{i^{2}})]}^{n} \\ = {[(1 - i) (1 + i)]}^{n} [? i^{2} = - 1] \\ = {[1 - i^{2}]}^{n} = {[1 + 1]}^{n} = 2^{n} \\ H e n c e, {(1 - i)}^{n} {(1 - \frac{1}{i})}^{n} = 2^{n} . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

The minimum value of $3 c o s x + 4 s i n x + 8$ is:

(a) 5

(b) 9

(c) 7

(d) 3

3 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n expression i s 3 c o s x + 4 s i n x + 8 \\ L e t y = 3 c o s x + 4 s i n x + 8 \\ \Rightarrow y - 8 = 3 c o s x + 4 s i n x \\ M i n i m u m v a l u e o f y - 8 = - \sqrt[]{{(3)}^{2} + {(4)}^{2}} \\ y - 8 = - \sqrt[]{9 + 16} = - 5 \\ y = 8 - 5 = 3 \\ S o, t h e m i n i m u m v a l u e o f t h e g i v e n expression i s 3 . \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

If $t a n α = \frac{m}{m + 1}$ and $t a n β = \frac{1}{m + 2}$ , then $α + β$ is equal to:

(a) $\frac{π}{2}$  

(b) $\frac{π}{3}$  

(c) $\frac{π}{6}$  

(d) $\frac{π}{4}$

3 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t t a n α = \frac{m}{m + 1} a n d t a n β = \frac{1}{2 m + 1} \\ t a n (α + β) = \frac{t a n α + t a n β}{1 - t a n α . t a n β} = \frac{\frac{m}{m + 1} + \frac{1}{2 m + 1}}{1 - \frac{m}{m + 1} \times \frac{1}{2 m + 1}} \\ = \frac{\frac{2 m^{2} + m + m + 1}{(m + 1) (2 m + 1)}}{\frac{(m + 1) (2 m + 1) - m}{(m + 1) (2 m + 1)}} = \frac{2 m^{2} + 2 m + 1}{2 m^{2} + 2 m + m + 1 - m} \\ = \frac{2 m^{2} + 2 m + 1}{2 m^{2} + 2 m + 1} = 1 \\ t a n (α + β) = t a n \frac{π}{4} ∴ α + β = \frac{π}{4} \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

Which of the following is correct?

(a) $s i n 1^{\circ} > s i n 1$

(b) $s i n 1^{\circ} < s i n 1$

(c) $s i n 1^{\circ} = s i n 1$

(d) $s i n 1^{\circ} = \frac{π}{1 8^{0}}$

Hint 1 radian = $\frac{π}{18 0^{0}} = 5 7^{0} 3 0^{'} a p p r o x$

5 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} W e k n o w t h a t i f θ i n c r e a s e s t h e n t h e v a l u e o f s i n θ a l s o i n c r e a s e s . \\ S o, s i n 1^{0} < s i n 1 [? 1 r a d i a n = \frac{π}{180} s i n 1] \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

The value of $t a n 7 5^{\circ} - c o t 7 5^{\circ}$ is equal to:

(a) $2 \sqrt[]{3}$

(b) $2 + \sqrt[]{3}$

(c) $2 - \sqrt[]{3}$

(d) 1

3 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n expression is t a n 7 5^{0} - c o t 7 5^{0} \\ t a n 7 5^{0} - c o t 7 5^{0} = t a n 7 5^{0} - c o t {(90 - 15)}^{0} \\ = t a n 7 5^{0} - t a n 1 5^{0} = \frac{s i n 7 5^{0}}{c o s 7 5^{0}} - \frac{s i n 1 5^{0}}{c o s 1 5^{0}} \\ = \frac{s i n 7 5^{0} c o s 1 5^{0} - c o s 7 5^{0} s i n 1 5^{0}}{c o s 7 5^{0} c o s 1 5^{0}} \\ = \frac{s i n (7 5^{0} - 1 5^{0})}{\frac{1}{2} \times 2 c o s 7 5^{0} c o s 1 5^{0}} \\ = \frac{2 s i n 6 0^{0}}{c o s (7 5^{0} + 1 5^{0}) + c o s (7 5^{0} - 1 5^{0})} \\ = \frac{2 \times \frac{\sqrt[]{3}}{2}}{c o s 9 0^{0} c o s 6 0^{0}} = \frac{\sqrt[]{3}}{0 + \frac{1}{2}} = 2 \sqrt[]{3} \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$