MathematicsQ51–51 of 91 questions
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51
MathematicsDifficultMCQJEE Main · 2016
यदि ${\left( {{x^{\frac{1}{3}}} + \frac{1}{{2{x^{\frac{1}{3}}}}}} \right)^{18}}, (x > 0)$ के विस्तार में $x^{-2}$ और $x^{-4}$ के गुणांक क्रमशः $m$ और $n$ हैं,तो $\frac{m}{n}$ का मान ज्ञात कीजिए।
A
$27$
B
$182$
C
$\frac{5}{4}$
D
$\frac{4}{5}$
Solution
(B) विस्तार का सामान्य पद $T_{r+1} = ^{18}C_{r} (x^{1/3})^{18-r} (\frac{1}{2x^{1/3}})^r$ है।
इसे सरल करने पर,$T_{r+1} = ^{18}C_{r} \cdot \frac{1}{2^r} \cdot x^{6 - 2r/3}$ प्राप्त होता है।
$x^{-2}$ के गुणांक के लिए,$6 - \frac{2r}{3} = -2$ रखने पर,$\frac{2r}{3} = 8$,अतः $r = 12$ प्राप्त होता है।
अतः,$m = ^{18}C_{12} \cdot \frac{1}{2^{12}}$.
$x^{-4}$ के गुणांक के लिए,$6 - \frac{2r}{3} = -4$ रखने पर,$\frac{2r}{3} = 10$,अतः $r = 15$ प्राप्त होता है।
अतः,$n = ^{18}C_{15} \cdot \frac{1}{2^{15}}$.
अब,$\frac{m}{n} = \frac{^{18}C_{12} \cdot 2^{-12}}{^{18}C_{15} \cdot 2^{-15}} = \frac{^{18}C_{12}}{^{18}C_{15}} \cdot 2^3 = 182$.
इसे सरल करने पर,$T_{r+1} = ^{18}C_{r} \cdot \frac{1}{2^r} \cdot x^{6 - 2r/3}$ प्राप्त होता है।
$x^{-2}$ के गुणांक के लिए,$6 - \frac{2r}{3} = -2$ रखने पर,$\frac{2r}{3} = 8$,अतः $r = 12$ प्राप्त होता है।
अतः,$m = ^{18}C_{12} \cdot \frac{1}{2^{12}}$.
$x^{-4}$ के गुणांक के लिए,$6 - \frac{2r}{3} = -4$ रखने पर,$\frac{2r}{3} = 10$,अतः $r = 15$ प्राप्त होता है।
अतः,$n = ^{18}C_{15} \cdot \frac{1}{2^{15}}$.
अब,$\frac{m}{n} = \frac{^{18}C_{12} \cdot 2^{-12}}{^{18}C_{15} \cdot 2^{-15}} = \frac{^{18}C_{12}}{^{18}C_{15}} \cdot 2^3 = 182$.
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