- Home
- Standard 11
- Mathematics
यदि $\left(a x^2+\frac{1}{2 b x}\right)^{11}$ में $x^7$ तथा in $\left(a x-\frac{1}{3 b x^2}\right)^{11}$ में $\mathrm{x}^{-7}$ के गुणांक बराबर हैं, तो
$64 ab =243$
$729 ab =32$
$243 ab =64$
$32 ab =729$
Solution
$\left(a x^2+\frac{1}{2 b x}\right)^{11}$
$T _{ r +1}={ }^{11} C _{ r }\left( ax ^2\right)^{11- r } \cdot\left(\frac{1}{2 bx }\right)^{ r }$
$={ }^{11} C _{ r } a ^{11- r } \cdot\left(\frac{1}{2 b }\right)^{ I } \cdot x ^{22-2 r – r }={ }^{11} C _{ r } a^{11- r } \cdot\left(\frac{1}{2 b }\right)^{ I } \cdot x ^{22-3 r }$
$\therefore 22-3 r=7$
$3 r =15$
$r=5$
Again $\left(a x-\frac{1}{3 b x^2}\right)^{11}$
$T _{ r +1}={ }^{11} C _{ r }( ax )^{11- r }\left(-\frac{1}{3 b x^2}\right)^{ r }$
$={ }^{11} C _{ r } a ^{11- r } \cdot\left(\frac{-1}{3 b }\right)^{ r } \cdot x ^{11- r -2 r }$ $\therefore 11-3 r=-7$
$3 r =18$
$r=6$
$\text { Now, } \frac{{ }^{11} C _5 a ^6}{32 b^5}=\frac{{ }^{11} C _6 \cdot a ^5}{3^6 \cdot b ^6}$
$729 ab =32$
