JEE Main 2022 · 28 Jun · Shift 1 · Mathematics Q28

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MathematicsInteger+4 / -1Question 28 of 90

The number of positive integers $k$ such that the constant term in the binomial expansion of $\left(2x^3+\dfrac3{x^k}\right)^{12}$ , $x\ne0$ is $2^8\cdot l$ , where $l$ is an odd integer, is ______.

Your answer

Answer and worked solution

Official answer: 2

Constant term. $3(12-r)=kr$ , so $r=\frac{36}{k+3}$ : $k=1,3,6,9,15,33$ with $r=9,6,4,3,2,1$ .

Power of 2 in ${}^{12}C_r\,2^{12-r}3^r$ : $r=9$ : $2+3=5$ ; $r=6$ : $2+6=8$ ; $r=4$ : $0+8=8$ ; $r=3$ : 11; $r=2$ : 11; $r=1$ : 13.

Two values ( $k=3,6$ ).

Answer: 2

Topics:General and middle terms Number systems

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JEE Main 2022 · 28 Jun · Shift 1, Mathematics question 28