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7.Binomial Theorem
medium
${\left( {\frac{a}{x} + bx} \right)^{12}}$ ના વિસ્તરણમાં $x^{-10}$ સહગુણક મેળવો.
A
$12{a^{11}}$
B
$12{b^{11}}a$
C
$12{a^{11}}b$
D
$12{a^{11}}{b^{11}}$
Solution
(c) ${\left[ {\exp \frac{a}{x}} \right]^{12 – r}} + {[\exp bx]^r} = – 10$
$ \Rightarrow – 12 + r + r = – 10 $
$\Rightarrow r = 1$
Then coefficient of ${x^{ – 10}}$ is $^{12}{C_1}{(a)^{11}}{(b)^1} = 12{a^{11}}b$.
Standard 11
Mathematics
