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4 months ago

Class 11th Maths NCERT Exemplar Solutions Class 11th Chapter Five

The real value of $α$ for which the expression $\frac{1 - i s i n α}{1 + i s i n α}$ is purely real is:

(a) $(n + 1) + \frac{π}{2}$

(b) $\frac{(2 n + 1) π}{2}$

(c) $n π$

(d) None of these, where $n \in ?$

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Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} L e t z = \frac{1 - i s i n α}{1 + 2 i s i n α} = \frac{(1 - i s i n α) (1 - 2 i s i n α)}{(1 + 2 i s i n α) (1 - 2 i s i n α)} \\ = \frac{1 - 2 i s i n α - i s i n α + 2 i^{2} s i n^{2} α}{{(1)}^{2} - {(2 i s i n α)}^{2}} \\ = \frac{1 - 3 i s i n α - 2 s i n^{2} α}{1 - 4 i^{2} s i n^{2} α} = \frac{(1 - 2 s i n^{2} α) - 3 i s i n α}{1 + 4 s i n^{2} α} \\ = \frac{1 - 2 s i n^{2} α}{1 + 4 s i n^{2} α} - \frac{3 s i n α}{1 + 4 s i n^{2} α} . i \\ Since, z i s p u r e l y r e a l, t h e n \\ \frac{3 s i n α}{1 + 4 s i n^{2} α} = 0 \Rightarrow s i n α = 0 \\ S o, α = n π, n \in N . \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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Class 11th Exemplar Solution

$s i n x + i c o s 2 x$ and $c o s x - i s i n 2 x$ are conjugate to each other for:

(a) $x = n π$

(b) $x = n + \frac{π}{2} + \frac{π}{2}$

(c) $x = 0$

(d) no value of $x$

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Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} L e t z = s i n x + i c o s 2 x \\ \bar{z} = s i n x - i c o s 2 x \\ B u t w e a r e g i v e n t h a t \bar{z} = c o s x - i s i n 2 x \\ ∴ s i n x - i c o s 2 x = c o s x - i s i n 2 x \\ C o m p a r i n g t h e r e a l a n d i m a g i n a r y p a r t s, w e g e t \\ s i n x = c o s x a n d c o s 2 x = s i n 2 x \\ \Rightarrow t a n x = 1 a n d t a n 2 x = 1 \\ \Rightarrow t a n x = t a n \frac{π}{4} a n d t a n 2 x = t a n \frac{π}{4} \\ ∴ x = n π + \frac{π}{4}, n \in I a n d 2 x = n π + \frac{π}{4} \\ \Rightarrow x = 2 x \Rightarrow 2 x - x = 0 \Rightarrow x = 0 \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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4 months ago

Class 11th Exemplar Solution

Where does $z$ lie, if $\frac{z - 5 i}{z + 5 i} = \frac{1}{5} ?$

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Payal Gupta

Contributor-Level 10

This is Other Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : | \frac{z - 5 i}{z + 5 i} | = 1 \\ L e t z = x + y i \\ ∴ | \frac{x + y i - 5 i}{x + y i + 5 i} | = 1 \Rightarrow | \frac{x + (y - 5) i}{x + (y + 5) i} | = 1 \\ \Rightarrow | x + (y - 5) i | = | x + (y + 5) i | \\ \Rightarrow x^{2} + {(y - 5)}^{2} = x^{2} + {(y + 5)}^{2} \\ \Rightarrow {(y - 5)}^{2} = {(y + 5)}^{2} \\ \Rightarrow y^{2} + 25 - 10 y = y^{2} + 25 + 10 y \\ \Rightarrow 20 y = 0 \Rightarrow y = 0 \\ H e n c e, z l i e s o n x - a x i s i . e ., r e a l a x i s . \end{array}$

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4 months ago

Class 11th Maths NCERT Exemplar Solutions Class 11th Chapter Five

Find the principal argument of ${(1 + i \sqrt[]{3})}^{2}$ .

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Payal Gupta

Contributor-Level 10

This is Other Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : {(1 + i \sqrt[]{3})}^{2} = 1 + i^{2} . 3 + 2 \sqrt[]{3} i \\ = 1 - 3 + 2 \sqrt[]{3} i = - 2 + 2 \sqrt[]{3} i \\ t a n α = | \frac{2 \sqrt[]{3}}{- 2} | [? t a n α = | \frac{I m g (z)}{R e (z)} |] \\ \Rightarrow t a n α = | - \sqrt[]{3} | = \sqrt[]{3} \\ \Rightarrow t a n α = t a n \frac{π}{3} ∴ α = \frac{π}{3} \\ N o w R e (z) < 0 a n d i m a g e (z) > 0 \\ ∴ a r g (z) = π - α = π - \frac{π}{3} = \frac{2 π}{3} \\ H e n c e, t h e p r i n c i p l e a r g = \frac{2 π}{3} . \end{array}$

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4 months ago

Class 11th Physics

4.22 $\hat{i}$ and $\hat{j}$ are unit vectors along x- and y- axis respectively. What is the magnitude and direction of the vectors $\hat{i}$ + $\hat{j}$ , and $\hat{i}$ - $\hat{j}$ ? What are the components of a vector A= 2 $\hat{i}$ + 3 $\hat{j}$ along the directions of $\hat{i}$ + $\hat{j}$ , and $\hat{i}$ - $\hat{j}$ ? [You may use graphical method]

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Vishal Baghel

Contributor-Level 10

4.22

Let us consider a vector $\overset{?}{P}$ . The equation can be written as

Px = Py = 1 $|\overset{?}{P}|$ = $\sqrt (P_{x}^{2} + P_{y}^{2})$ $|\overset{?}{P}|$ = $\sqrt (1^{2} + 1^{2}) |\overset{?}{P}|$ = $\sqrt{2}$ …….(i)

So the magnitude of vector $\overset{?}{i}$ + $\overset{?}{j}$ = $\sqrt{2}$

Let $θ$ be the angle made by vector $\overset{?}{P}$ , with the x axis as given in the above figure

$\tan ? θ$ = $(P x / P y) θ$ = $\tan^{- 1} ? (1 / 1)$ , $θ$ = 45 $°$ with the x axis

Let $\overset{?}{Q}$ = $\overset{?}{i}$ - $\overset{?}{j}$

$\overset{?}{Q_{x}} \overset{?}{i}$ – $\overset{?}{Q_{y}}$ $\overset{?}{j}$ = ( $\overset{?}{i}$ – $\overset{?}{j})$

$\overset{?}{Q_{x}}$ = $\overset{?}{Q_{y}}$ = 1

$|\overset{?}{Q}|$ = $\sqrt{{Q_{x}}^{2} + {Q_{y}}^{2}}$ = $\sqrt 2$

Hence $|\overset{?}{Q}|$ = $\sqrt 2$ . Therefore the magnitude of ( $\overset{?}{i}$ + $\overset{?}{j})$ = $\sqrt 2$

Let $θ$ be the angle made

...more

4.22

Let us consider a vector $\overset{?}{P}$ . The equation can be written as

Px = Py = 1 $|\overset{?}{P}|$ = $\sqrt (P_{x}^{2} + P_{y}^{2})$ $|\overset{?}{P}|$ = $\sqrt (1^{2} + 1^{2}) |\overset{?}{P}|$ = $\sqrt{2}$ …….(i)

So the magnitude of vector $\overset{?}{i}$ + $\overset{?}{j}$ = $\sqrt{2}$

Let $θ$ be the angle made by vector $\overset{?}{P}$ , with the x axis as given in the above figure

$\tan ? θ$ = $(P x / P y) θ$ = $\tan^{- 1} ? (1 / 1)$ , $θ$ = 45 $°$ with the x axis

Let $\overset{?}{Q}$ = $\overset{?}{i}$ - $\overset{?}{j}$

$\overset{?}{Q_{x}} \overset{?}{i}$ – $\overset{?}{Q_{y}}$ $\overset{?}{j}$ = ( $\overset{?}{i}$ – $\overset{?}{j})$

$\overset{?}{Q_{x}}$ = $\overset{?}{Q_{y}}$ = 1

$|\overset{?}{Q}|$ = $\sqrt{{Q_{x}}^{2} + {Q_{y}}^{2}}$ = $\sqrt 2$

Hence $|\overset{?}{Q}|$ = $\sqrt 2$ . Therefore the magnitude of ( $\overset{?}{i}$ + $\overset{?}{j})$ = $\sqrt 2$

Let $θ$ be the angle made by vector $\overset{?}{Q}$ , with the x axis as given in the above figure

$\tan ? θ$ = $(Q x / Q y) θ$ = ${- \tan}^{- 1} ? (1 / 1)$ , $θ$ = - 45 $°$ with the x axis

It is given that $\overset{?}{A}$ = 2 $\overset{?}{i}$ + 3 $\overset{?}{j}$ . We can write Ax $\overset{?}{i}$ + Ay $\overset{?}{j}$ = 2 $\overset{?}{i}$ + 3 $\overset{?}{j}$

Comparing the components on both sides, we get

Ax =2 and Ay = 3

$|\overset{?}{A}|$ = $\sqrt{{(2}^{2}} + 3^{2}$ ) = $\sqrt 13$

Let $\overset{?}{A_{x}}$ make an angle $θ$ with the x axis, then

$\tan ? θ$ = (Ax / Ay), $θ$ = $\tan^{- 1} (? 3 / 2)$ = 56.31 $°$

So the angle between (2 $\overset{?}{i}$ + 3 $\overset{?}{j}$ ) and ( $\overset{?}{i}$ + $\overset{?}{j}$ )

$θ'$ = 56.31 – 45 = 11.31 $°$

Component of vector $\overset{?}{A}$ , along the direction of $\overset{?}{P}$ , making an angle $θ$

= ( A cos $θ'$ ) $\overset{?}{P}$

= ( A cos 11.31) $\frac{(\overset{?}{i} + \overset{?}{j})}{\sqrt{2}}$

= $\sqrt{13}$ $* \frac{0.9806}{\sqrt{2}}$ ( $\overset{?}{i}$ + $\overset{?}{j}$ )

= 2.5 $*$ $\sqrt{2}$

= $\frac{5}{\sqrt{2}}$

Let $θ'$ be the angle between the vectors (2 $\overset{?}{i}$ + 3 $\overset{?}{j}$ ) and ( $\overset{?}{i}$ - $\overset{?}{j})$

$θ " = 45 + 56.31 = 101.31 °$

Component of vector $\overset{?}{A}$ , along the direction of $\overset{?}{Q}$ , making an angle $θ$

= (A cos $θ "$ ) $\frac{(\overset{?}{i} - \overset{?}{j})}{\sqrt{2}}$

= $\sqrt{13}$ cos ( $101.31 °) \frac{(\overset{?}{i} - \overset{?}{j})}{\sqrt{2}}$

= -0.5( $\overset{?}{i}$ - $\overset{?}{j})$

= $- \frac{1}{\sqrt{2}}$

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New answer posted

4 months ago

Class 11th Maths NCERT Exemplar Solutions Class 11th Chapter Five

Find $| (1 + i) \frac{(2 + i)}{(3 + i)} |$

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Payal Gupta

Contributor-Level 10

This is Other Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : | (1 + i) \frac{(2 + i)}{(3 + i)} | \\ | (1 + i) \frac{(2 + i)}{(3 + i)} * \frac{3 - i}{3 - i} | = | (1 + i) . \frac{6 - 2 i + 3 i - i^{2}}{9 - i^{2}} | \\ = | \frac{(1 + i) . (7 + i)}{9 + 1} | = | \frac{7 + i + 7 i + i^{2}}{10} | = | \frac{7 + 8 i - 1}{10} | \\ = | \frac{6 + 8 i}{10} | = | \frac{3}{5} + \frac{4}{5} i | = \sqrt[]{{(\frac{3}{5})}^{2} + {(\frac{4}{5})}^{2}} = \sqrt[]{\frac{9}{25} + \frac{16}{25}} = \sqrt[]{\frac{25}{25}} = 1 \\ H e n c e, | (1 + i) \frac{(2 + i)}{(3 + i)} | = 1 \end{array}$

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New answer posted

4 months ago

Class 11th Maths NCERT Exemplar Solutions Class 11th Chapter Five

If $? z_{1} ? = ? z_{2} ? = 1$ , is it necessary that $z_{1} = z_{2}$ ?

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Payal Gupta

Contributor-Level 10

This is Other Questions as classified in NCERT Exemplar

$\begin{array}{l} L e t z_{1} = x_{1} + y_{1} i a n d z_{2} = x_{2} + y_{2} i \\ ∴ | x_{1} + y_{1} i | = | x_{2} + y_{2} i | \\ \Rightarrow \sqrt[]{x_{1}^{2} + y_{1}^{2}} = \sqrt[]{x_{2}^{2} + y_{2}^{2}} \\ \Rightarrow x_{1}^{2} + y_{1}^{2} = x_{2}^{2} + y_{2}^{2} \Rightarrow x_{1}^{2} = x_{2}^{2} a n d y_{1}^{2} = y_{2}^{2} \\ \Rightarrow x_{1} = \pm x_{2} a n d y_{1} = \pm y_{2} \\ S o, z_{1} = x_{1} + y_{1} i a n d z_{2} = \pm x_{2} \pm y_{2} i \\ ∴ z_{1} \neq z_{2} \\ H e n c e, i t i s n o t n e c e s s a r y t h a t z_{1} = z_{2} . \end{array}$

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New answer posted

4 months ago

Class 11th Maths NCERT Exemplar Solutions Class 11th Chapter Five

If $? z_{1} ? = ? z_{2} ? = 1$ , is it necessary that $z_{1} = z_{2}$ ?

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Payal Gupta

Contributor-Level 10

This is Other Questions as classified in NCERT Exemplar

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New answer posted

4 months ago

Class 11th Maths NCERT Exemplar Solutions Class 11th Chapter Five

What is the conjugate of $\frac{2 - i}{(1 - 2 i)}$ z

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Payal Gupta

Contributor-Level 10

This is Other Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t z = \frac{2 - i}{{(1 - 2 i)}^{2}} = \frac{2 - i}{1 + 4 i^{2} - 4 i} = \frac{2 - i}{1 - 4 - 4 i} \\ = \frac{2 - i}{- 3 - 4 i} = \frac{2 - i}{- 3 - 4 i} * \frac{- 3 + 4 i}{- 3 + 4 i} \\ = \frac{- 6 + 8 i + 3 i - 4 i^{2}}{{(- 3)}^{2} - {(4 i)}^{2}} = \frac{- 6 + 11 i + 4}{9 - 16 i^{2}} \\ = \frac{- 2 + 11 i}{9 + 16} = \frac{- 2}{25} + \frac{11}{25} i \\ ∴ \bar{z} = \frac{- 2}{25} - \frac{11}{25} i \\ H e n c e, \bar{z} = \frac{- 2}{25} - \frac{11}{25} i \end{array}$

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Class 11th Exemplar Solution

Find $z$ if $? z ? = 4$ and $(z) = \frac{5 π}{6}$ .

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Payal Gupta

Contributor-Level 10

This is Other Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : | z | = 4 a n d a r g (z) = \frac{5 π}{6} \Rightarrow θ = \frac{5 π}{6} \\ | z | = 4 \Rightarrow r = 4 \\ S o P o l a r f o r m o f z = r [c o s θ + i s i n θ] \\ = 4 [c o s \frac{5 π}{6} + i s i n \frac{5 π}{6}] \\ = 4 [c o s (π - \frac{π}{6}) + i s i n (π - \frac{π}{6})] \\ = 4 [- c o s \frac{π}{6} + i s i n \frac{π}{6}] = 4 [\frac{\sqrt[]{3}}{2} + i . \frac{1}{2}] \\ = - 2 \sqrt[]{3} + 2 i \\ H e n c e, z = - 2 \sqrt[]{3} + 2 i . \end{array}$

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