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The number of positive integers k such that the constant term in the binomial expansion of 2x3+3xk12,x0 is 28·l, where l is an odd integer, is ______.
MathematicsBinomial TheoremJEE MainJEE Main 2022 (28 Jun Shift 1)
Solution:
2218 Upvotes Verified Answer
The correct answer is: 2

For 2x3+3xk12

Tr+1=Cr122x312-r·3xkr=Cr12·212-r·3r·x36-3r-kr

For the term to be independent of x,

36-3r-kr=0 r=36k+3 

Here k can be 0,1,3,6,9 for r to be a whole number.

But for Cr12·212-r·3r to be in the form of 28·l

Only k=3, 6 satisfies the above relation.

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