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7.Binomial Theorem
hard
જો $\left(a x-\frac{1}{b x^2}\right)^{13}$ માં $x^7$ નો સહગુણક અને $\left(a x+\frac{1}{b x^2}\right)^{13}$ માં $x^{-5}$ નો સહગુણક સરખા હોય,તો $a^4 b^4=.........$
A
$44$
B
$22$
C
$11$
D
$33$
(JEE MAIN-2023)
Solution
$T_{r+1}={ }^{13} C_r(a x)^{13-r}\left(-\frac{1}{b x^2}\right)^r$
$={ }^{13} C_r(a)^{13-r}\left(-\frac{1}{b}\right)^r x^{13-3 r}$
$1 3 – 3 r = 7 \Rightarrow r=2$
Coefficient of $x^7={ }^{13} C_2(a)^{11} \cdot \frac{1}{b^2}$
In the other expansion $T_{r+1}={ }^{13} C_r(a x)^{13-r}\left(\frac{1}{b x^2}\right)^r$
$13-3 r=-5 \Rightarrow r=6$
Coefficient of $x^{-5}={ }^{13} C_6(a)^7 \cdot \frac{1}{b^6}$
${ }^{13} C_2 \frac{a^{11}}{b^2}={ }^{13} C_6 \frac{a^7}{b^6}$
$a^4 b^4=\frac{{ }^{13} C_6}{{ }^{13} C_2}=22$
Standard 11
Mathematics
