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7.Binomial Theorem
hard
${\left( {\frac{{x + 1}}{{{x^{\frac{2}{3}}} - {x^{\frac{1}{3}}} + 1}} - \frac{{x - 1}}{{x - {x^{\frac{1}{2}}}}}} \right)^{10}}$ ના વિસ્તરણમાં $x^{-5}$ નો સહગુણક મેળવો. જ્યાં $x \ne 0, 1$
A
$1$
B
$4$
C
$-4$
D
$-1$
(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
