The number of positive integers k such that t | vedclass

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(C) The general term in the expansion of $\left(2x^3 + \frac{3}{x^k}\right)^{12}$ is given by $T_{r+1} = {}^{12}C_r (2x^3)^r \left(\frac{3}{x^k}\right

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$2$

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(C) The general term in the expansion of $\left(2x^3 + \frac{3}{x^k}\right)^{12}$ is given by $T_{r+1} = {}^{12}C_r (2x^3)^r \left(\frac{3}{x^k}\right)^{12-r}$.Simplifying the expression,we get $T_{r+1} = {}^{12}C_r \cdot 2^r \cdot 3^{12-r} \cdot x^{3r - k(12-r)}$.For the constant term,the exponent of $x$ must be zero: $3r - k(12-r) = 0$,which implies $k = \frac{3r}{12-r}$.Since $k$ is a positive integer and $0 \le r \le 12$,we test values of $r$ such that $12-r$ divides $3r$:If $r=3, k = \frac{9}{9} = 1$.If $r=6, k = \frac{18}{6} = 3$.If $r=8, k = \frac{24}{4} = 6$.If $r=9, k = \frac{27}{3} = 9$.If $r=10, k = \frac{30}{2} = 15$.The constant term is $C = {}^{12}C_r \cdot 2^r \cdot 3^{12-r}$. We require $C = 2^8 \cdot \ell$,where $\ell$ is odd.For $r=3: {}^{12}C_3 \cdot 2^3 \cdot 3^9 = 220 \cdot 2^3 \cdot 3^9 = (2^2 \cdot 5 \cdot 11) \cdot 2^3 \cdot 3^9 = 2^5 \cdot (5 \cdot 11 \cdot 3^9)$,which is not $2^8 \cdot \ell$.For $r=6: {}^{12}C_6 \cdot 2^6 \cdot 3^6 = 924 \cdot 2^6 \cdot 3^6 = (2^2 \cdot 3 \cdot 7 \cdot 11) \cdot 2^6 \cdot 3^6 = 2^8 \cdot (3^7 \cdot 7 \cdot 11)$,which is $2^8 \cdot \ell$ (where $\ell$ is odd).For $r=8: {}^{12}C_8 \cdot 2^8 \cdot 3^4 = 495 \cdot 2^8 \cdot 3^4 = (3^2 \cdot 5 \cdot 11) \cdot 2^8 \cdot 3^4 = 2^8 \cdot (3^6 \cdot 5 \cdot 11)$,which is $2^8 \cdot \ell$ (where $\ell$ is odd).For $r=9: {}^{12}C_9 \cdot 2^9 \cdot 3^3 = 220 \cdot 2^9 \cdot 3^3 = (2^2 \cdot 5 \cdot 11) \cdot 2^9 \cdot 3^3 = 2^{11} \cdot (5 \cdot 11 \cdot 3^3)$,not $2^8 \cdot \ell$.For $r=10: {}^{12}C_{10} \cdot 2^{10} \cdot 3^2 = 66 \cdot 2^{10} \cdot 3^2 = (2 \cdot 3 \cdot 11) \cdot 2^{10} \cdot 3^2 = 2^{11} \cdot (3^3 \cdot 11)$,not $2^8 \cdot \ell$.Thus,there are $2$ such values of $k$ ($k=3$ and $k=6$).

The number of positive integers $k$ such that the constant term in the binomial expansion of $\left(2x^3 + \frac{3}{x^k}\right)^{12}, x \neq 0$ is $2^8 \cdot \ell$,where $\ell$ is an odd integer,is:

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