1 - i x - i-2 - i - 1 - 2i y - i-2 - i - 1 - x- y -

The correct option is D ( 7/3, 7/15 ) (1+i)x i2+i+(1+2i)y+i2 i=1 ⇒((1+i)x i)(2 i)(2+i)(2 i)+((1+2i)y+i)(2+i)(2 i)(2+i)=1 ⇒ #160;(x+ix i) ...

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Byju's Answer

Standard XII

Mathematics

Algebra of Limits

1 + i x - i/2...

Question

$\frac{(1 + i) x - i}{2 + i} + \frac{(1 + 2 i) y + i}{2 - i} = 1 \Rightarrow (x, y) =$

(\frac{7}{3}, \frac{- 7}{15})

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(\frac{7}{3}, \frac{7}{15})

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(\frac{7}{5}, \frac{- 7}{15})

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(\frac{7}{5}, \frac{7}{15})

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Solution

The correct option is D $(\frac{7}{3}, \frac{- 7}{15})$
$\frac{(1 + i) x - i}{2 + i} + \frac{(1 + 2 i) y + i}{2 - i} = 1$
$\Rightarrow \frac{((1 + i) x - i) (2 - i)}{(2 + i) (2 - i)} + \frac{((1 + 2 i) y + i) (2 + i)}{(2 - i) (2 + i)} = 1$
$\Rightarrow (x + i x - i) (2 - i) + (y + 2 i y + i) (2 + i) = 4 - (- 1)$
$\Rightarrow 2 x - i x + 2 i x + x - 2 i - 1 + 2 y + 4 i y + 2 i + i y - 2 y - 1 = 5$
$\Rightarrow 3 x + i x + 5 y - 2 = 5$
$\Rightarrow 3 x + i x + 5 i y = 7$
Equating real part
$3 x = 7$
$x = \frac{7}{3}$

Also, $x + 5 y = 0$
$\Rightarrow \frac{7}{3} + 5 y = 0$
$\Rightarrow \frac{7}{3} = - 5 y$
$\Rightarrow y = \frac{- 7}{15}$

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Similar questions

Q. The property under addition used in each of the following are
(i)

(\frac{- 3}{7}) + \frac{1}{9} = \frac{1}{9} + (\frac{- 3}{7})

(ii)

\frac{4}{9} + (\frac{7}{8} + \frac{1}{2}) = (\frac{4}{9} + \frac{7}{8}) + \frac{1}{2}

(iii)

8 + \frac{7}{10} = \frac{7}{10} + 8

(iv)

(\frac{- 7}{15}) + 0 = \frac{- 7}{15} = 0 + (\frac{- 7}{15})

(v)

\frac{2}{5} + (\frac{- 2}{5}) = 0

(i) Commutative
(ii) Associative
(iii) Commutative
(iv) Additive identity
(v) Additive inverse.

Name the property of multiplication illustrated by each of the following statements :

(i) $\frac{- 15}{8} \times \frac{- 12}{7} = \frac{- 12}{7} \times \frac{- 15}{8}$

(ii) $(\frac{- 2}{3} \times \frac{7}{9}) \times \frac{- 9}{5} = \frac{- 2}{3} \times (\frac{7}{9} \times \frac{- 9}{5})$

(iii) $\frac{- 3}{4} \times (\frac{- 3}{4} \times \frac{- 5}{6}) + (\frac{- 3}{4} \times \frac{7}{8})$

(iv) $\frac{- 16}{9} \times 1 = 1 \times \frac{- 16}{9} = \frac{- 16}{9}$

(v) $\frac{- 11}{15} \times \frac{15}{- 11} = \frac{15}{- 11} \times \frac{- 11}{15} = 1$

(vi) $\frac{- 7}{5} \times 0 = 0$

Re-arrange suitably and find the sum in each of the following :

(i) $\frac{11}{12} + \frac{- 17}{3} + \frac{11}{2} + \frac{- 25}{2}$

(ii) $\frac{- 6}{7} + \frac{- 5}{6} + \frac{- 4}{9} + \frac{- 15}{7}$

(iii) $\frac{3}{5} + \frac{7}{3} + \frac{9}{5} + \frac{- 13}{15} + \frac{- 7}{3}$

(iv) $\frac{4}{13} + \frac{- 5}{8} + \frac{- 8}{13} + \frac{9}{13}$

(v) $\frac{2}{3} + \frac{- 4}{5} + \frac{1}{3} + \frac{2}{5}$

(vi) $\frac{1}{8} + \frac{5}{12} + \frac{2}{7} + \frac{7}{12} + \frac{9}{7} + \frac{- 5}{16}$

Prove that:
(i) $\frac{1}{3 + \sqrt{7}} + \frac{1}{\sqrt{7} + \sqrt{5}} + \frac{1}{\sqrt{5} + \sqrt{3}} + \frac{1}{\sqrt{3} + 1} = 1$
(ii) $\frac{1}{1 + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + \frac{1}{\sqrt{4} + \sqrt{5}} + \frac{1}{\sqrt{5} + \sqrt{6}} + \frac{1}{\sqrt{6} + \sqrt{7}} + \frac{1}{\sqrt{7} + \sqrt{8}} + \frac{1}{\sqrt{8} + \sqrt{9}} = 2$

Without adding find the sum

(i) 1 + 3 + 5 + 7 + 9

(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19

(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23

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(1+i)x−i2+i+(1+2i)y+i2−i=1⇒(x,y)=

$\frac{(1 + i) x - i}{2 + i} + \frac{(1 + 2 i) y + i}{2 - i} = 1 \Rightarrow (x, y) =$