Maths

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

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Contributor-Level 10

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

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

0 0

New answer posted

a year ago

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) $c o s 2 β$

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Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted

a year ago

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

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Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted

a year ago

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

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Contributor-Level 10

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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New answer posted

a year ago

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

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Contributor-Level 10

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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New answer posted

a year ago

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

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Contributor-Level 10

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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New answer posted

a year ago

The value of $s i n \frac{2 π}{18} + s i n \frac{5 π}{18} + s i n \frac{7 π}{18} + s i n \frac{4 π}{9}$ is given by:

(a) 1

(b) -1

(c) $c o s \frac{3 π}{9} + s i n \frac{π}{18}$

(d) $c o s \frac{2 π}{9} + s i n \frac{π}{18}$

0 Follower 2 Views

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n expression i s s i n \frac{π}{18} + s i n \frac{π}{9} + s i n \frac{2 π}{9} + s i n \frac{5 π}{18} \\ = (s i n \frac{5 π}{18} + s i n \frac{π}{18}) + (s i n \frac{2 π}{9} + s i n \frac{π}{9}) \\ = 2 s i n (\frac{\frac{5 π}{18} + \frac{π}{18}}{2}) . c o s (\frac{\frac{5 π}{18} - \frac{π}{18}}{2}) + 2 s i n (\frac{\frac{2 π}{9} + \frac{π}{9}}{2}) . c o s (\frac{\frac{2 π}{9} - \frac{π}{9}}{2}) \\ = 2 s i n \frac{π}{6} . c o s \frac{π}{9} + 2 s i n \frac{π}{6} . c o s \frac{π}{18} \\ = 2 * \frac{1}{2} c o s \frac{π}{9} + 2 * \frac{1}{2} c o s \frac{π}{18} = c o s \frac{π}{9} + c o s \frac{π}{18} \\ = s i n (\frac{π}{2} - \frac{π}{9}) + s i n (\frac{π}{2} - \frac{π}{18}) = s i n \frac{7 π}{18} + s i n \frac{8 π}{18} \\ = s i n \frac{7 π}{18} + s i n \frac{4 π}{9} . \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

0 0

New answer posted

a year ago

If $\overset{?}{? z} + 1 ? = z + 2 (1 + i)$ , then find $z$ .

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Contributor-Level 10

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

$\begin{array}{l} G i v e n t h a t : | z + 1 | = z + 2 (1 + i) \\ L e t z = x + i y \\ S o, | x + i y + 1 | = (x + i y) + 2 (1 + i) \\ \Rightarrow | (x + 1) + i y | = x + i y + 2 + 2 i \\ \Rightarrow | (x + 1) + i y | = (x + 2) + (y + 2) i \\ \Rightarrow \sqrt[]{{(x + 1)}^{2} + y^{2}} = (x + 2) + (y + 2) i [? | x + i y | = \sqrt[]{x^{2} + y^{2}}] \\ S q u a r i n g b o t h s i d e s \\ {(x + 1)}^{2} + y^{2} = {(x + 2)}^{2} + {(y + 2)}^{2} . i^{2} + 2 (x + 2) (y + 2) i \\ \Rightarrow x^{2} + 1 + 2 x + y^{2} = x^{2} + 4 + 4 x - y^{2} - 4 y - 4 + 2 (x + 2) (y + 2) i \\ C o m p a r i n g t h e r e a l a n d i m a g i n a r y p a r t s, w e g e t \\ x^{2} + 1 + 2 x + y^{2} = x^{2} + 4 x - y^{2} - 4 y a n d 2 (x + 2) (y + 2) = 0 \\ \Rightarrow 2 y^{2} - 2 x + 4 y + 1 = 0 \dots (i) \\ a n d (x + 2) (y + 2) = 0 \dots (i i) \\ x + 2 = 0 o r y + 2 = 0 \\ ∴ x = - 2 o r y = - 2 \\ N o w p u t x = - 2 i n e q n . (i) \\ \Rightarrow 2 y^{2} - 2 * (- 2) + 4 y + 1 = 0 \\ \Rightarrow 2 y^{2} + 4 + 4 y + 1 = 0 \\ \Rightarrow 2 y^{2} + 4 y + 5 = 0 \\ b^{2} - 4 a c = {(4)}^{2} - 4 * 2 * 5 = 16 - 40 = - 24 < 0 n o r e a l r o o t s . \\ N o w p u t y = - 2 i n e q n . (i) \\ \Rightarrow 2 {(- 2)}^{2} - 2 x + 4 (- 2) + 1 = 0 \\ \Rightarrow 8 - 2 x - 8 + 1 = \end{array}$

0 0

New answer posted

a year ago

Number of solutions of the equation $t a n x + s e c x = 2 c o s x$ lying in the interval $[0, 2 π]$ is:

(a) 0

(b) 1

(c) 2

(d) 3

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Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : t a n x + s e c x = 2 c o s x \\ \Rightarrow \frac{s i n x}{c o s x} + \frac{1}{c o s x} = 2 c o s x \\ \Rightarrow 1 + s i n x = 2 c o s^{2} x \Rightarrow 2 c o s^{2} x - s i n x - 1 = 0 \\ \Rightarrow 2 (1 - si n^{2} x) - s i n x - 1 = 0 \Rightarrow 2 - 2 si n^{2} x - s i n x - 1 = 0 \\ \Rightarrow - 2 si n^{2} x - s i n x + 1 = 0 \Rightarrow 2 si n^{2} x + s i n x - 1 = 0 \\ Since t h e e q u a t i o n i s q u a d r a t i c e q u a t i o n i n s i n x . S o i t w i l l h a v e 2 s o l u t i o n s . \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

0 0

New answer posted

a year ago

If $s i n θ = - \frac{2}{5}$ and $θ$ lies in the third quadrant, then the value of $c o s θ$ is:

(a) $- \frac{1}{5}$

(b) $- \frac{\sqrt[]{21}}{5}$

(c) $\frac{1}{5}$

(d) $\frac{\sqrt[]{21}}{5}$

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