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A real value of x satisfies the equation (3-4ix3+4ix) = α − iβ (α, β ∈ R) if α2 + β2 = ______.

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प्रश्न

A real value of x satisfies the equation `((3 - 4ix)/(3 + 4ix))` = α − iβ (α, β ∈ R) if α2 + β2 = ______.

विकल्प

  • 1

  • –1

  • 2

  • –2

MCQ
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उत्तर

A real value of x satisfies the equation `((3 - 4ix)/(3 + 4ix))` = α − iβ (α, β ∈ R) if α2 + β2 = 1.

Explanation:

Given that: `((3 - 4ix)/(3 + 4ix))` = α − iβ

⇒ `((3 - 4ix)/(3 + 4ix) xx (3 - 4ix)/(3 - 4ix))` = α − iβ

⇒ `(9 - 12ix - 12ix + 16i^2 x^2)/(9 - 16i^2 x^2)` = α − iβ 

⇒ `(9 - 24ix - 16x^2)/(9 + 16x^2)` = α − iβ 

⇒ `(9 - 16x^2)/(9 + 16x^2) - (24x)/(9 + 16x^2) i` = α − iβ   .....(i)

⇒ `(9 - 16x^2)/(9 + 16x^2) + (24x)/(9 + 16x^2) i` = α + iβ

Multiplying equation (i) and (ii) we get

⇒ `((9 - 16x^2)/(9 + 16x^2))^2 + ((24x)/(9 + 16x^2))^2` = α2 + β2

⇒ `((9 - 16x^2)^2 + (24x)^2)/(9 + 16x^2)^2` = α2 + β2

⇒ `(81 + 256x^4 - 288x^2 + 576x^2)/(9 + 16x^2)^2` = α2 + β2

⇒ `(81 + 256x^4 + 288x^2)/(9 + 16x^2)^2` = α2 + β2

⇒ `(9 + 16x^2)^2/(9 + 16x^2)^2` = α2 + β2

So, = α2 + β= 1

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अध्याय 5: Complex Numbers and Quadratic Equations - Exercise [पृष्ठ ९६]

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एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 11
अध्याय 5 Complex Numbers and Quadratic Equations
Exercise | Q 40 | पृष्ठ ९६

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