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New answer posted•a year ago

Exemplar Solution Class 11th

Write the complex number $z = \frac{1 - i}{c o s \frac{π}{3} + i s i n \frac{π}{3}}$  in polar form.

3 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

$\begin{array}{l} G i v e n t h a t : \\ z = \frac{1 - i}{\frac{1}{2} + i \frac{\sqrt[]{3}}{2}} = \frac{2 - 2 i}{1 + i \sqrt[]{3}} = \frac{2 - 2 i}{1 + i \sqrt[]{3}} \times \frac{1 - i \sqrt[]{3}}{1 - i \sqrt[]{3}} \\ \Rightarrow z = \frac{2 - 2 \sqrt[]{3} i - 2 i + 2 \sqrt[]{3} i^{2}}{{(1)}^{2} - {(i \sqrt[]{3})}^{2}} = \frac{2 - 2 \sqrt[]{3} i - 2 i - 2 \sqrt[]{3}}{1 - 3 i^{2}} \\ = \frac{(2 - 2 \sqrt[]{3}) - (2 + 2 \sqrt[]{3}) i}{4} = \frac{1 - \sqrt[]{3}}{2} - \frac{1 + \sqrt[]{3}}{2} i \\ r = \sqrt[]{{(\frac{1 - \sqrt[]{3}}{2})}^{2} + {(- \frac{1 + \sqrt[]{3}}{2})}^{2}} = \sqrt[]{\frac{1 + 3 - 2 \sqrt[]{3}}{4} + \frac{1 + 3 + 2 \sqrt[]{3}}{4}} \\ = \sqrt[]{\frac{4 - 2 \sqrt[]{3} + 4 + 2 \sqrt[]{3}}{4}} = \sqrt[]{\frac{8}{4}} = \sqrt[]{2} \\ S o r = \sqrt[]{2} \\ N o w a r g (z) = t a n^{- 1} \frac{y}{x} \\ \Rightarrow θ = t a n^{- 1} \frac{- (\frac{1 + \sqrt[]{3}}{2})}{(\frac{1 - \sqrt[]{3}}{2})} = t a n^{- 1} [- (\frac{1 + \sqrt[]{3}}{1 - \sqrt[]{3}})] = t a n^{- 1} \frac{\sqrt[]{3} + 1}{\sqrt[]{3} - 1} \\ \Rightarrow θ = t a n^{- 1} [t a n (\frac{π}{4} + \frac{π}{6})] [? t a n (\frac{π}{4} + \frac{π}{6}) = \frac{t a n \frac{π}{4} + t a n \frac{π}{6}}{1 - t a n \frac{π}{4} t a n \frac{π}{6}}] \\ \Rightarrow θ = \frac{5 π}{12} \\ H e n c e, t h e p o l a r i s z = \sqrt[]{2} [c o s (\frac{5 π}{12}) + i s i n (\frac{5 π}{12})] . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

If $c o s e c x = 1 + c o t x$ , then $x = 2 n π, 2 n π + \frac{π}{2}$ .

5 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

$\begin{array}{l} G i v e n t h a t : c o s e c x = 1 + c o t x \\ \Rightarrow \frac{1}{s i n x} = 1 + \frac{c o s x}{s i n x} \\ \Rightarrow \frac{1}{s i n x} = \frac{s i n x + c o s x}{s i n x} \\ \Rightarrow s i n x + c o s x = 1 \\ \Rightarrow \frac{1}{\sqrt[]{2}} s i n x + \frac{1}{\sqrt[]{2}} c o s x = \frac{1}{\sqrt[]{2}} \\ \Rightarrow s i n \frac{π}{4} s i n x + c o s \frac{π}{4} c o s x = \frac{1}{\sqrt[]{2}} \\ \Rightarrow c o s (x - \frac{π}{4}) = \frac{1}{\sqrt[]{2}} \\ \Rightarrow c o s (x - \frac{π}{4}) = c o s \frac{π}{4} \\ \Rightarrow x = 2 n π + \frac{π}{4} + \frac{π}{4} \Rightarrow x = 2 n π + \frac{π}{2} \\ o r x = 2 n π + \frac{π}{4} - \frac{π}{4} \Rightarrow x = 2 n π \\ H e n c e, t h e s t a t e m e n t i s' T r u e' . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

Find the complex number satisfying the equation $z + \sqrt 2 ? z + 1 ? + i = 0$ .

3 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

$\begin{array}{l} G i v e n t h a t : z + \sqrt[]{2} | (z + 1) | + i = 0 \\ L e t z = x + y i \\ ∴ (x + y i) + \sqrt[]{2} | (x + y i + 1) | + i = 0 \\ \Rightarrow x + (y + 1) i + \sqrt[]{2} | (x + 1) + y i | = 0 \\ \Rightarrow x + (y + 1) i + \sqrt[]{2} \sqrt[]{{(x + 1)}^{2} + y^{2}} = 0 \\ \Rightarrow x + (y + 1) i + \sqrt[]{2} \sqrt[]{x^{2} + 2 x + 1 + y^{2}} = 0 + 0 i \\ \Rightarrow x + \sqrt[]{2} \sqrt[]{x^{2} + 2 x + 1 + y^{2}} = 0, y + 1 = 0 \\ \Rightarrow x = - \sqrt[]{2} \sqrt[]{x^{2} + 2 x + 1 + y^{2}} a n d y = - 1 \\ \Rightarrow x^{2} = 2 (x^{2} + 2 x + 1 + y^{2}) \\ \Rightarrow x^{2} = 2 x^{2} + 4 x + 2 + 2 y^{2} \\ \Rightarrow x^{2} + 4 x + 2 + 2 y^{2} = 0 \\ \Rightarrow x^{2} + 4 x + 2 + 2 {(- 1)}^{2} = 0 [? y = - 1] \\ \Rightarrow x^{2} + 4 x + 4 = 0 \\ \Rightarrow {(x + 2)}^{2} = 0 \\ \Rightarrow x + 2 = 0 \Rightarrow x = - 2 \\ H e n c e, z = x + y i = - 2 - i . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

One value of $θ$ which satisfies the equation $s i n^{4} θ - 2 s i n^{2} θ - 1 = 0$ lies between $0$ and $2 π$ .

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 True or False Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n e q u a t i o n i s \\ s i n^{4} θ - 2 s i n^{2} θ - 1 = 0 \\ s i n^{2} θ = \frac{- (- 2) \pm \sqrt[]{{(- 2)}^{2} - 4 \times 1 \times - 1}}{2 \times 1} \\ = \frac{2 \pm \sqrt[]{4 + 4}}{2} = \frac{2 \pm \sqrt[]{8}}{2} = \frac{2 \pm 2 \sqrt[]{2}}{2} = 1 \pm \sqrt[]{2} \\ ∴ s i n^{2} θ = (1 + \sqrt[]{2}) o r (1 - \sqrt[]{2}) \Rightarrow - 1 \leq s i n θ \leq 1 \\ \Rightarrow s i n^{2} θ \leq 1 b u t s i n^{2} θ = (1 + \sqrt[]{2}) o r (1 - \sqrt[]{2}) \\ W h i c h i s n o t p o s s i b l e . \\ H e n c e, t h e s t a t e m e n t i s' F a l s e' . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

Solve the system of equations $Re (z^{2}) = 0, z = 2$ .

2 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

$\begin{array}{l} G i v e n t h a t : R e (z^{2}) = 0 a n d | z | = 2 \\ L e t z = x + y i \\ ∴ | z | = \sqrt[]{x^{2} + y^{2}} \\ \Rightarrow \sqrt[]{x^{2} + y^{2}} = 2 \Rightarrow x^{2} + y^{2} = 4 \dots (i) \\ Since z = x + y i \\ z^{2} = x^{2} + y^{2} i^{2} + 2 x y i \Rightarrow z^{2} = x^{2} - y^{2} + 2 x y i \\ ∴ R e (z^{2}) = x^{2} - y^{2} \\ \Rightarrow x^{2} - y^{2} = 0 \dots (i i) \\ F r o m e q n . (i) a n d (i i), w e g e t \\ x^{2} + y^{2} = 4 \Rightarrow 2 x^{2} = 4 \Rightarrow x^{2} = 2 \Rightarrow x = \pm \sqrt[]{2} a n d y = \pm \sqrt[]{2} \\ H e n c e, z = \pm \sqrt[]{2} \pm i \sqrt[]{2}, - \sqrt[]{2} \pm i \sqrt[]{2} . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

. $c o s \frac{π}{15} \cdot c o s \frac{4 π}{15} \cdot c o s \frac{8 π}{15} \cdot c o s \frac{16 π}{15} = \frac{1}{16}$

2 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

$\begin{array}{l} L . H . S . c o s \frac{2 π}{15} . c o s \frac{4 π}{15} . c o s \frac{8 π}{15} . c o s \frac{16 π}{15} \\ = c o s 2 4^{0} . c o s 4 8^{0} . c o s 9 6^{0} . c o s 19 2^{0} \\ = \frac{1}{16 s i n 2 4^{0}} [(2 s i n 2 4^{0} c o s 2 4^{0}) (2 c o s 4 8^{0}) (2 c o s 9 6^{0}) (2 c o s 19 2^{0})] \\ = \frac{1}{16 s i n 2 4^{0}} [s i n 4 8^{0} . 2 c o s 4 8^{0} (2 c o s 9 6^{0}) (2 c o s 19 2^{0})] \\ = \frac{1}{16 s i n 2 4^{0}} [2 s i n 4 8^{0} c o s 4 8^{0} (2 c o s 9 6^{0}) (2 c o s 19 2^{0})] \\ = \frac{1}{16 s i n 2 4^{0}} [s i n 9 6^{0} (2 c o s 9 6^{0}) (2 c o s 19 2^{0})] \\ = \frac{1}{16 s i n 2 4^{0}} [2 s i n 9 6^{0} c o s 9 6^{0} (2 c o s 19 2^{0})] \\ = \frac{1}{16 s i n 2 4^{0}} [s i n 19 2^{0} (2 c o s 19 2^{0})] \\ = \frac{1}{16 s i n 2 4^{0}} [2 s i n 19 2^{0} c o s 19 2^{0}] \\ = \frac{1}{16 s i n 2 4^{0}} s i n 38 4^{0} = \frac{1}{16 s i n 2 4^{0}} s i n (36 0^{0} + 2 4^{0}) \\ = \frac{1}{16 s i n 2 4^{0}} \times s i n 2 4^{0} [? s i n (36 0^{0} + θ) = s i n θ] \\ = \frac{1}{16} R . H . S . \\ H e n c e, t h e s t a t e m e n t i s' T r u e' . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

$s i n 1 0^{\circ}$ is greater than $c o s 1 0^{\circ}$ .

4 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

$\begin{array}{l} I f s i n 1 0^{0} > c o s 1 0^{0} \\ \Rightarrow s i n 1 0^{0} > c o s (9 0^{0} - 8 0^{0}) \\ \Rightarrow s i n 1 0^{0} > s i n 8 0^{0} w h i c h i s n o t p o s s i b l e . \\ H e n c e, t h e s t a t e m e n t i s' F a l s e' . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

The equality $s i n A + s i n 2 A + s i n 3 A = 3$ holds for some real value of $A$ .

4 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

$\begin{array}{l} G i v e n t h a t : s i n A + s i n 2 A + s i n 3 A = 3 \\ Since t h e maximum v a l u e o f s i n A i s 1 b u t f o r s i n 2 A a n d s i n 3 A i t i s n o t e q u a l t o 1 . \\ S o i t i s n o t p o s s i b l e . \\ H e n c e, t h e s t a t e m e n t i s' F a l s e' . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

If $t a n A = \frac{1 - c o s B}{s i n B}$ , then $t a n 2 A = t a n B$ .

2 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

$\begin{array}{l} G i v e n t h a t : t a n A = \frac{1 - c o s B}{s i n B} = \frac{2 s i n^{2} B / 2}{2 s i n B / 2 c o s B / 2} = t a n \frac{B}{2} \\ t a n 2 A = \frac{2 t a n A}{1 - t a n^{2} A} = \frac{2 t a n B / 2}{1 - t a n^{2} B / 2} \\ ∴ t a n 2 A = t a n B \\ H e n c e, t h e s t a t e m e n t i s' T r u e' . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

The maximum distance of a point on the graph of the function $y = 3 s i n x + c o s x$ from the $x$ -axis is ___.

2 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

$\begin{array}{l} G i v e n t h a t : y = \sqrt[]{3} s i n x + c o s x \dots (i) \\ ∴ T h e maximum distance f r o m a point o n t h e g r a p h o f e q n . (i) f r o m x - a x i s \\ = \sqrt[]{{(\sqrt[]{3})}^{2} + {(1)}^{2}} = \sqrt[]{3 + 1} = 2 . \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s 2 . \end{array}$

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