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7.Binomial Theorem
medium
The number of positive integers $k$ such that the constant term in the binomial expansion of $\left(2 x^{3}+\frac{3}{x^{k}}\right)^{12}, x \neq 0$ is $2^{8} \cdot \ell$, where $\ell$ is an odd integer, is......
A
$20$
B
$9$
C
$2$
D
$70$
(JEE MAIN-2022)
Solution
$\left(2 x^{3}+\frac{3}{x^{k}}\right)^{12}$
$t _{ r +1}={ }^{12} C _{ r }\left(2 x ^{3}\right)^{ r }\left(\frac{3}{ x ^{ k }}\right)^{12- r }$
$x ^{3 r -(12- r ) k } \rightarrow constant$
$\therefore 3 r -12 k + rk =0$
$\Rightarrow k =\frac{3 r }{12- r }$
$\therefore$ possible values of $r$ are $3,6,8,9,10$ and corresponding values of $k$ are $1,3,6,9,15$
Now ${ }^{12} C _{ r }=220,924,495,220,66$
$\therefore$ possible values of $k$ for which we will get $2^{8}$ are $3. 6$
Standard 11
Mathematics
