7.Binomial Theorem
easy

${\left( {\sqrt {\frac{x}{3}}  + \frac{3}{{2{x^2}}}} \right)^{10}}$ के विस्तार में $x$ से स्वतंत्र पद होगा   

A

$3\over2$

B

$5\over4$

C

$5\over2$

D

इनमें से कोर्इ नहीं

(IIT-1965)

Solution

$(10 – r)\left( {\frac{1}{2}} \right) + r( – 2) = 0 \Rightarrow 5 – \frac{r}{2} – 2r = 0 \Rightarrow r = 2$

अत: $x$ से स्वतंत्र पद

${ = ^{10}}{C_2} \times {\left( {\frac{1}{3}} \right)^4}{\left( {\frac{3}{2}} \right)^2} = \frac{{10 \times 9}}{{2 \times 1}} \times \frac{1}{{3 \times 3 \times 2 \times 2}} = \frac{5}{4}$

Standard 11
Mathematics

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