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Question

A real value of x will satisfy the equation 34ix3+4ix=αiβ(α,βR) , if

A
α2β2=1
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B
α2β2=1
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C
α2+β2=1
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D
α2β2=2
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Solution

The correct option is C α2+β2=1
Given 34ix3+4ix=αiβ(α,βR)

(34ix)2(3+4ix)(34ix)=αiβ(α,βR)

916x224xi9+16x2=αiβ(α,βR)

comparing we get

α=(916x2)(9+16x2) and β=(24)(9+16x2)

Also,

α2+β2=(916x2)2+242(9+16x2)2(9+16x2)2=1

Hence x satisfies the above equation.

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