Coefficient of x^{-5} in binomial expansion of {left( {frac{{x + 1}}{{{x^{frac{2}{3}}}

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7.Binomial Theorem

hard

The coefficient of $x^{-5}$ in the binomial expansion of ${\left( {\frac{{x + 1}}{{{x^{\frac{2}{3}}} - {x^{\frac{1}{3}}} + 1}} - \frac{{x - 1}}{{x - {x^{\frac{1}{2}}}}}} \right)^{10}}$ where $x \ne 0, 1$ , is

$1$

$4$

$-4$

$-1$

(JEE MAIN-2017)

Solution

${\left[ {\frac{{\left( {{x^{1/3}} + 1\left( {{x^{2/3}}} \right) + {x^{1/3}} + 1} \right)}}{{\left( {{x^{2/3}} – {x^{\sqrt 3 }} + 1} \right)}} – \frac{{(\sqrt x – 1(\sqrt x ) + 1)}}{{\sqrt x (\sqrt x – 1)}}} \right]^{10}}$

$ = {({x^{1/3}} + 1 – 1 – 1/{x^{1/2}})^{10}}$

$ = {({x^{1/3}} – 1/{x^{1/2}})^{10}}$

${T_{r + 1}}{ = ^{10}}{C_r}{x^{\frac{{20 – 5r}}{6}}}$

for $r = 10$

${T_{11}}{ = ^{10}}{C_{10}}{x^{ – 5}}$

${\text { Coefficient of } x^{-5}=^{10} C_{10}(1)(-1)^{10}=1}$

Standard 11

Mathematics

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hard

(JEE MAIN-2023)

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Solution

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