JEE Main 2022 (Online) 28th June Morning ShiftQuestion 1
The number of positive integers $k$ such that the constant term in the binomial expansion of $\left(2x^{3}+\frac{3}{x^{k}}\right)^{12}$, $x\neq0$ is $2^{8}\cdot\ell$, where $\ell$ is an odd integer, is $\_\_\_\_\_\_$ .
Binomial Theorem · Coefficient of a Given Power / Term Independent of x
Answer$2$
Solution
Concept. List all $k$ that give a constant term, then count the factors of $2$ in each. Formulas used. (1) $T_{r+1}={}^{n}C_r\,a^{n-r}b^r$Step 1 - Condition. $T_{r+1}={}^{12}C_r\,2^{12-r}\,3^r\,x^{36-3r-kr}$ is constant when $r(k+3)=36$. Step 2 - The possible $k$. $k+3\ge4$ must divide $36$: $$k+3\in\{4,6,9,12,18,36\},\qquad k=1,3,6,9,15,33,\qquad r=9,6,4,3,2,1.$$ Step 3 - Power of $2$ in ${}^{12}C_r\,2^{12-r}$ ($3^r$ is odd): $r=9$: ${}^{12}C_9=220=2^2\cdot55$, so $2+3=5$. $r=6$: ${}^{12}C_6=924=2^2\cdot231$, so $2+6=8$. $r=4$: ${}^{12}C_4=495$ is odd, so $0+8=8$. $r=3$: $220$, so $2+9=11$. $r=2$: $66$, so $1+10=11$. $r=1$: $12$, so $2+11=13$. Step 4 - Count. The power is exactly $8$ for $k=3$ and $k=6$: two values. Answer: 2