If in the expansion of (1x+xtanx)5, the ratio of 4th term to the 2nd term is 227π4, then the value of x can be
A
−π6
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B
−π3
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C
π3
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D
π12
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Solution
The correct option is Dπ3 From the above given expression by applying binomial theorem, we get. T4=5C3xtan3x =10xtan3x T2=5C1x−3tanx =5x−3tanx Therefore T4T2 =10xtan3x5x−3tanx =2x4tan2x =2π427 x4tan2x=π427 Taking root on both sides, we get x2tanx=π23√3 =√3π29 Therefore x2=π29 ...(i) and tanx=√3...(ii) Both Eq(i) and Eq (ii) give x=π3