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Question

Let a=^i+4^k,b=5^i+2^k and c=3^i+^k. If a=xb+yc, then the value of (x,y) is

A
(1,1)
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B
(2,2)
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C
(1,2)
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D
(2,1)
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Solution

The correct option is C (1,2)
Given :
a=xb+yc
^i+4^k=x(5^i+2^k)+y(3^i+^k)
Equating the cofficient of like vectors, we get
5x3y=1 (i)2x+y=4 (ii)
Solving (i) and (ii), we get
x=1,y=2

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