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Exemplar Solution Class 11th

If  $z_{1} = 1$ ( $z_{1} \neq - 1$ ) and $z_{2} = \frac{z_{1} - 1}{z_{1} + 1}$ , then show that the real part of  $z_{2}$ is zero.

7 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

$\begin{array}{l} L e t z_{1} = x + y i \\ | z_{1} | = \sqrt[]{x^{2} + y^{2}} = 1 [G i v e n t h a t | z_{1} | = 1] \\ \Rightarrow x^{2} + y^{2} = 1 \dots (i) \\ N o w z_{2} = \frac{z_{1} - 1}{z_{1} + 1} = \frac{x + y i - 1}{x + y i + 1} \\ = \frac{(x - 1) + y i}{(x + 1) + y i} = \frac{(x - 1) + y i}{(x + 1) + y i} \times \frac{x + 1 - y i}{x + 1 - y i} \\ = \frac{(x - 1) (x + 1) - y (x - 1) i + y (x + 1) i - y^{2} i^{2}}{{(x + 1)}^{2} - y^{2} i^{2}} \\ = \frac{x^{2} - 1 + y i (x + 1 - x + 1) + y^{2}}{x^{2} + 1 + 2 x + y^{2}} \\ = \frac{(x^{2} + y^{2} - 1) + 2 y i}{x^{2} + y^{2} + 2 x + 1} \\ = \frac{(1 - 1)}{x^{2} + y^{2} + 2 x + 1} + \frac{2 y}{x^{2} + y^{2} + 2 x + 1} i \\ = 0 + \frac{2 y}{x^{2} + y^{2} + 2 x + 1} i \\ H e n c e, t h e r e a l p a r t o f z_{2} i s 0 . \end{array}$

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Exemplar Solution Class 11th

If $c o s θ = \frac{x}{x + 1}$ , then $θ$ is:

(a)   An acute angle

(b)   A right angle

(c)   An obtuse angle

(d)   Not possible

2 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : c o s θ = x + \frac{1}{x} \Rightarrow c o s θ = \frac{x^{2} + 1}{x} \\ \Rightarrow x^{2} + 1 = x c o s θ \Rightarrow x^{2} - x c o s θ + 1 = 0 \\ F o r r e a l v a l u e o f x, b^{2} - 4 a c \geq 0 \\ \Rightarrow {(- c o s θ)}^{2} - 4 \times 1 \times 1 \geq 0 \\ \Rightarrow c o s^{2} θ - 4 \geq 0 \Rightarrow c o s^{2} θ \geq 4 \\ \Rightarrow c o s θ \geq \pm 2 [- 1 \leq c o s θ \leq 1] \\ S o, t h e v a l u e o f θ i s n o t p o s s i b l e . \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

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Exemplar Solution Class 11th

$z_{1}$ and  $z_{2}$ are two complex numbers such that $? z_{1} ? = ? z_{2} ? = 1$  and $arg (z_{1}) + arg (z_{2}) = π$ , then show that $z_{1} = - \frac{}{z_{2}}$ .

2 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

$\begin{array}{l} L e t z_{1} = r_{1} (c o s θ_{1} + i s i n θ_{1}) a n d z_{2} = r_{2} (c o s θ_{2} + i s i n θ_{2}) a r e p o l a r f o r m o f t w o c o m p l e x \\ n u m b e r s z_{1} a n d z_{2} . \\ G i v e n t h a t : | z_{1} | = | z_{2} | \Rightarrow r_{1} = r_{2} \\ a n d a r g (z_{1}) + a r g (z_{2}) = π \\ \Rightarrow θ_{1} + θ_{2} = π \\ \Rightarrow θ_{1} = π - θ_{2} \\ N o w z_{1} = r_{1} [c o s (π - θ_{2}) + i s i n (π - θ_{2})] \\ \Rightarrow z_{1} = r_{1} [- c o s θ_{2} + i s i n θ_{2}] \\ \Rightarrow z_{1} = - r_{1} [c o s θ_{2} - i s i n θ_{2}] \dots (i) \\ \Rightarrow z_{2} = r_{2} [c o s θ_{2} + i s i n θ_{2}] \\ \Rightarrow {\bar{z}}_{2} = r_{1} [c o s θ_{2} - i s i n θ_{2}] [? r_{1} = r_{2}] \dots (i i) \\ F r o m e q n . (i) a n d (i i) w e g e t, \\ z_{1} = - {\bar{z}}_{2} . H e n c e p r o v e d . \end{array}$

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Exemplar Solution Class 11th

If $t a n θ = \frac{a}{b}$ , then $b c o s 2 θ + a s i n 2 θ$ is equal to:

(a) a 

(b) b 

(c) $\frac{b}{a}$  

(d) None

2 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : t a n θ = \frac{a}{b} \\ b c o s 2 θ + a s i n 2 θ = b [\frac{1 - t a n^{2} θ}{1 + t a n^{2} θ}] + a [\frac{2 t a n θ}{1 + t a n^{2} θ}] \\ = b [\frac{1 - \frac{a^{2}}{b^{2}}}{1 + \frac{a^{2}}{b^{2}}}] + a [\frac{\frac{2 a}{b}}{1 + \frac{a^{2}}{b^{2}}}] \\ = b [\frac{b^{2} - a^{2}}{b^{2} + a^{2}}] + [\frac{\frac{2 a^{2}}{b}}{\frac{b^{2} + a^{2}}{b^{2}}}] \\ = \frac{b^{3} - a^{2} b}{b^{2} + a^{2}} + \frac{2 a^{2} b}{b^{2} + a^{2}} = \frac{b^{3} - a^{2} b + 2 a^{2} b}{b^{2} + a^{2}} \\ = \frac{b^{3} + a^{2} b}{b^{2} + a^{2}} = \frac{b (b^{2} + a^{2})}{b^{2} + a^{2}} = b \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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Exemplar Solution Class 11th

If $\frac{z - 1}{z + 1}$ is a purely imaginary number ( $z \neq - 1$ ), then find the value of $? z ?$ .

2 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

$\begin{array}{l} G i v e n t h a t : \frac{z - 1}{z + 1} i s p u r e l y i m a g i n a r y n u m b e r . \\ L e t z = x + y i \\ ∴ \frac{x + y i - 1}{x + y i + 1} = \frac{(x - 1) + i y}{(x + 1) + i y} = \frac{(x - 1) + i y}{(x + 1) + i y} \times \frac{(x + 1) - i y}{(x + 1) - i y} \\ \Rightarrow \frac{(x - 1) (x + 1) - i y (x - 1) + (x + 1) i y - i^{2} y^{2}}{{(x + 1)}^{2} - i^{2} y^{2}} \\ \Rightarrow \frac{x^{2} - 1 + i y (x + 1 - x + 1) + y^{2}}{x^{2} + 1 + 2 x + y^{2}} = \frac{x^{2} + y^{2} - 1 + 2 y i}{x^{2} + y^{2} + 2 x + 1} \\ \Rightarrow \frac{x^{2} + y^{2} - 1}{x^{2} + y^{2} + 2 x + 1} + \frac{2 y}{x^{2} + y^{2} + 2 x + 1} i \\ Since, t h e n u m b e r i s p u r e l y i m a g i n a r y, t h e n r e a l p a r t = 0 \\ ∴ \frac{x^{2} + y^{2} - 1}{x^{2} + y^{2} + 2 x + 1} = 0 \\ \Rightarrow x^{2} + y^{2} - 1 = 0 \Rightarrow x^{2} + y^{2} = 1 \\ \Rightarrow \sqrt[]{x^{2} + y^{2}} = 1 ∴ | z | = 1 \end{array}$

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Exemplar Solution Class 11th

If $t a n α = \frac{1}{7}$ , $t a n β = \frac{1}{3}$ , then $c o s 2 α$ is equal to:

(a) $s i n 2 β$  

(b) $s i n 4 β$  

(c) $s i n 3 β$  

(d) $c o s 2 β$

2 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : t a n α = \frac{1}{7} a n d t a n β = \frac{1}{3} \\ c o s 2 α = \frac{1 - t a n^{2} α}{1 + t a n^{2} α} = \frac{1 - {(\frac{1}{7})}^{2}}{1 + {(\frac{1}{7})}^{2}} = \frac{1 - \frac{1}{49}}{1 + \frac{1}{49}} = \frac{48}{50} = \frac{24}{25} \\ t a n 2 β = \frac{2 t a n β}{1 - t a n^{2} β} = \frac{2 \times \frac{1}{3}}{1 - \frac{1}{9}} = \frac{\frac{2}{3}}{\frac{8}{9}} = \frac{2}{3} \times \frac{9}{8} = \frac{3}{4} \\ ∴ t a n 2 β = \frac{3}{4} \\ s i n 4 β = \frac{2 t a n 2 β}{1 + t a n^{2} 2 β} = \frac{2 \times \frac{3}{4}}{1 + {(\frac{3}{4})}^{2}} = \frac{\frac{3}{2}}{1 + \frac{9}{16}} = \frac{3}{2} \times \frac{16}{25} = \frac{24}{25} \\ c o s 2 α = s i n 4 β = \frac{24}{25} \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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Exemplar Solution Class 11th

The value of $c o s^{2} 4 8^{?} - s i n^{2} 1 2^{?}$ is:

(a) $\frac{\sqrt[]{5} + 1}{4}$

(b) $- \frac{\sqrt[]{5} + 1}{4}$

(c) $\frac{1}{2}$

(d) $- \frac{1}{2}$

Hint : Use of $c o s^{2} A - s i n^{2} B = c o s (A + B) C o s (A - B)$

6 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n expression i s c o s^{2} 4 8^{0} - s i n^{2} 1 2^{0} \\ c o s^{2} 4 8^{0} - s i n^{2} 1 2^{0} = c o s (4 8^{0} + 1 2^{0}) . c o s (4 8^{0} - 1 2^{0}) [? c o s^{2} A - s i n^{2} B = c o s (A + B) . c o s (A - B)] \\ = c o s 6 0^{0} . c o s 3 6^{0} = \frac{1}{2} \times \frac{\sqrt[]{5} + 1}{4} = \frac{\sqrt[]{5} + 1}{8} \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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Exemplar Solution Class 11th

Show that  $? \frac{z - 2}{z + 3} ? = 2$ represents a circle. Find its centre and radius.

4 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

$\begin{array}{l} G i v e n t h a t : | \frac{z - 2}{z - 3} | = 2 \\ L e t z = x + i y \\ ∴ | \frac{x + i y - 2}{x + i y - 3} | = 2 \\ \Rightarrow | \frac{(x - 2) + i y}{(x - 3) + i y} | = 2 \\ \Rightarrow \frac{| (x - 2) + i y |}{| (x - 3) + i y |} = 2 [? | \frac{a}{b} | = \frac{| a |}{| b |}] \\ \Rightarrow | (x - 2) + i y | = 2 | (x - 3) + i y | \\ \Rightarrow \sqrt[]{{(x - 2)}^{2} + y^{2}} = 2 \sqrt[]{{(x - 3)}^{2} + y^{2}} [? | x + i y | = \sqrt[]{x^{2} + y^{2}}] \\ S q u a r i n g b o t h s i d e s \\ {(x - 2)}^{2} + y^{2} = 4 [{(x - 3)}^{2} + y^{2}] \\ \Rightarrow x^{2} + 4 - 4 x + y^{2} = 4 [x^{2} + 9 - 6 x + y^{2}] \\ \Rightarrow x^{2} + y^{2} - 4 x + 4 = 4 x^{2} + 4 y^{2} - 24 x + 36 \\ \Rightarrow 3 x^{2} + 3 y^{2} - 20 x + 32 = 0 \\ \Rightarrow x^{2} + y^{2} - \frac{20}{3} x + \frac{32}{3} = 0 \\ H e r e, g = - \frac{10}{3}, f = 0 \\ r = \sqrt[]{g^{2} + f^{2} - c} = \sqrt[]{\frac{100}{9} - 0 - \frac{32}{3}} = \sqrt[]{\frac{4}{9}} = \frac{2}{3} \\ H e n c e, t h e r e q u i r e d e q u a t i o n o f a c i r c l e i s \\ x^{2} + y^{2} - \frac{20}{3} x + \frac{32}{3} = 0 \\ C e n t r e = (- g, - f) = (\frac{10}{3}, 0) a n d r = \frac{2}{3} . \end{array}$

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Exemplar Solution Class 11th

If $A$ lies in the second quadrant and $3 t a n A + 4 = 0$ , then the value of $2 c o t A - 5 c o s A + s i n A$ is equal to:

(a) $- \frac{53}{10}$  

(b) $- \frac{23}{10}$  

(c) $- \frac{37}{10}$  

(d) None

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 an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : 3 t a n A + 4 = 0, A l i e s i n second q u a d r a n t \\ ∴ t a n A = \frac{- 4}{3} \\ c o s A = \frac{- 3}{5} [A l i e s i n second q u a d r a n t] \\ a n d s i n A = \frac{4}{5} a n d c o t A = \frac{- 3}{4} \\ ∴ 2 c o t A - 5 c o s A + s i n A = 2 (\frac{- 3}{4}) - 5 (\frac{- 3}{5}) + \frac{4}{5} \\ = \frac{- 3}{2} + 3 + \frac{4}{5} = \frac{- 15 + 30 + 8}{10} = \frac{23}{10} \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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Exemplar Solution Class 11th

If $arg (z - 1) = arg (z + 3 i)$ , then find $x - 1 : y$ , where $z = x + i y$ .

3 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

$\begin{array}{l} G i v e n t h a t : a r g (z - 1) = a r g (z + 3 i) \\ \Rightarrow a r g (x + y i - 1) = a r g (x + y i + 3 i) \\ \Rightarrow a r g [(x - 1) + y i] = a r g [x + (y + 3) i] \\ \Rightarrow t a n^{- 1} \frac{y}{x - 1} = t a n^{- 1} \frac{y + 3}{x} \\ \Rightarrow \frac{y}{x - 1} = \frac{y + 3}{x} \\ \Rightarrow x y = (x - 1) (y + 3) \Rightarrow x y = x y + 3 x - y - 3 \\ \Rightarrow 3 x - y = 3 \Rightarrow 3 x - 3 = y \\ \Rightarrow 3 (x - 1) = y \Rightarrow \frac{(x - 1)}{y} = \frac{1}{3} \Rightarrow x - 1 : y = 1 : 3 \\ H e n c e, x - 1 : y = 1 : 3 . \end{array}$

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If $\frac{z - 1}{z + 1}$ is a purely imaginary number ( $z \neq - 1$ ), then find the value of $? z ?$ .

The value of $c o s^{2} 4 8^{?} - s i n^{2} 1 2^{?}$ is:

(a) $\frac{\sqrt[]{5} + 1}{4}$

(b) $- \frac{\sqrt[]{5} + 1}{4}$

(c) $\frac{1}{2}$

(d) $- \frac{1}{2}$

Hint : Use of $c o s^{2} A - s i n^{2} B = c o s (A + B) C o s (A - B)$

Show that  $? \frac{z - 2}{z + 3} ? = 2$ represents a circle. Find its centre and radius.

If $arg (z - 1) = arg (z + 3 i)$ , then find $x - 1 : y$ , where $z = x + i y$ .