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Question

If x1=a+3b+a3ba+3ba3b then, which one of the following statements is true?


A

3bx2+9b2ax=0

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B

3bx2+3b2ax=0

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C

3bx2+3b6ax=0

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D

5bx2+3b2ax=0

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Solution

The correct option is B

3bx2+3b2ax=0


x1=a+3b+a3ba+3ba3b
Applying componendo and dividendo
x+1x1=a+3b+a3b+a+3ba3ba+3b+a3ba+3b+a3b
= 2a+3b2a3b
= a+3ba3b
Squaring on both sides, we get:
(x+1)2(x1)2=(a+3b)2(a3b)2
= (x+1)2(x1)2=a+3ba3b
Again, applying componendo and dividendo
(x+1)2+(x1)2(x+1)2(x1)2=a+3b+a3ba+3ba+3b
2(x2+1)4x=2a6b
=(x2+1)2x=a3b
=3bx2+3b=2ax
= 3bx2+3b2ax=0


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