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7.Binomial Theorem
hard
જો $\left(x^{\frac{2}{3}}+\frac{\alpha}{x^3}\right)^{22}$ ના વિસ્તરણમાં $x$ વગર નું પદ $7315 $ હોય, તો $|\alpha|=...............$
A
$2$
B
$1$
C
$4$
D
$6$
(JEE MAIN-2023)
Solution
$T _{ r +1}={ }^{22} C _{ r } \cdot\left( x ^{\frac{2}{3}}\right)^{22- r } \cdot(\alpha)^{ r }, x ^{-3 r }$
$={ }^{22} C _{ r } \cdot x ^{\frac{44}{3}-\frac{2 r }{3}-3 r }(\alpha)^{ r }$
$\frac{44}{3}=\frac{11 r }{3}$
$r =4$
${ }^{22} C _4 \cdot \alpha^4=7315$
$\frac{22 \times 21 \times 20 \times 19}{24} \cdot \alpha^4=7315$
$\alpha=1$
Standard 11
Mathematics
