The number of positive integers k such that t | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(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:

Similar Questions

If the fourth term in the binomial expansion of $\left(\sqrt{\frac{1}{x^{1+\log _{10} x}}}+x^{\frac{1}{12}}\right)^{6}$ is equal to $200$,and $x > 1$,then the value of $x$ is

The $6^{th}$ term in the expansion of $\left( 2x^2 - \frac{1}{3x^2} \right)^{10}$ is

If $p$ and $q$ are real numbers such that the $7^{\text{th}}$ term in the expansion of $\left(\frac{5}{p^3} - \frac{3q}{7}\right)^8$ is $700$,then $49p^2 =$ (in $q^2$)

If $C_0, C_1, C_2, \ldots, C_8$ are the binomial coefficients in the expansion of $(1+x)^8$,then $\sum_{r=1}^8 r^3 \frac{C_r}{C_{r-1}} =$

If the term independent of $x$ in the expansion of $\left( x^{\frac{2}{3}} + \frac{\alpha}{x^3} \right)^{22}$ is $7315$,then $|\alpha|$ is equal to $...........$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module