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Question

If x=2+5i(where 1i=1) and 2(11!9!+13!7!)+15!5!=2ab! then x35x2+33x10=

A
a+b
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B
ba
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C
ab
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D
ab
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E
(ab)(a+b)
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Solution

The correct option is A a+b
10!2ab!=2[10C1+10C3]+10C5
10!2ab!=10C1+10C3+10C5+10C7+10C9=29
a=9,b=10
x=2+5i
(x2)2=25
x24x+29=0
x35x2+33x10=(x1)(x24x+29)+19
19=a+b

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