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Q.

If the value of the third term in the expansion of x+xlog10⁡x5is 106  , then x may have value(s)

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a

1, 10

b

10,10−5/2

c

10,10−3/2

d

102,10−3/2

answer is B.

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Detailed Solution

T3=5C2x5−2xlog10⁡x2=106⇒x2log10⁡x+3=105⇒t(2t+3)=5 where t=log10⁡x⇒t=−5/2,1
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