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7.Binomial Theorem
hard
ધારોકે $[t]$ એ મહત્તમ પૂર્ણાક $\leq t$ દર્શાવે છે.જો $\left(3 x^2-\frac{1}{2 x^5}\right)^7$ નાં વિસ્તરણમાં અયળ પદ $\alpha$ હોય, તો $[\alpha]=...........$
A
$1274$
B
$1275$
C
$1273$
D
$1272$
(JEE MAIN-2023)
Solution
$\left(3 x ^2-\frac{1}{2 x ^5}\right)^7$
$T _{ r +1}={ }^7 C _{ r }\left(3 x ^2\right)^{7- r }\left(-\frac{1}{2 x ^5}\right)^{ r }$
$14-2 r -5 r =14-7 r =0$
$\therefore r =2$
$\therefore T _3={ }^7 C _2 \cdot 3^5\left(-\frac{1}{2}\right)^2=\frac{21 \times 243}{4}=1275.75$
$\therefore[\alpha]=1275$
Standard 11
Mathematics
