Answer
64.8k+ views
Hint: Use binomial theorem of series expansion.
Expanding both the terms of the above term we get:
\[
{\left( {1 + x} \right)^m}{\left( {1 - x} \right)^n} \\
= \left( {1 + {}^m{C_1}{x^1} + {}^m{C_2}{x^2} + {}^m{C_3}{x^3} + ........} \right)\left( {1 - {}^n{C_1}{x^1} + {}^n{C_2}{x^2} - {}^n{C_3}{x^3} + ........} \right) \\
= 1\left( {1 + {}^m{C_1}{x^1} + {}^m{C_2}{x^2} + {}^m{C_3}{x^3} + ........} \right) \\
- {}^n{C_1}{x^1}\left( {1 + {}^m{C_1}{x^1} + {}^m{C_2}{x^2} + {}^m{C_3}{x^3} + ........} \right) \\
+ {}^n{C_2}{x^2}\left( {1 + {}^m{C_1}{x^1} + {}^m{C_2}{x^2} + {}^m{C_3}{x^3} + ........} \right) \\
+ \cdots \cdots \cdots \\
\\
\]
Expanding the equation further to find the coefficients, we get:
\[ \Rightarrow 1 + x\left( {{}^m{C_1} - {}^n{C_1}} \right) + {x^2}\left( {{}^m{C_2} - {}^m{C_1}{}^n{C_1} + {}^n{C_2}} \right) + \cdots \cdots \]
Hence, the coefficient of the term containing $x$ is
$
\Rightarrow {}^m{C_1} - {}^n{C_1} = 3 \\
\Rightarrow m - n = 3 \\
$ ---- (1)
Similarly coefficient of the term containing ${x^2}$ is
$
\Rightarrow {}^m{C_2} - {}^m{C_1}{}^n{C_1} + {}^n{C_2} = - 6 \\
\Rightarrow m\left( {m - 1} \right) - 2mn + n\left( {n - 1} \right) = - 12 \\
\Rightarrow {m^2} - m - 2mn + {n^2} - n = - 12 \\
\Rightarrow {\left( {m - n} \right)^2} - \left( {m + n} \right) = - 12 \\
\Rightarrow 9 - \left( {m + n} \right) = - 12 \\
\Rightarrow m + n = 21 \\
$ ---- (2)
Adding equation 1 and 2 to find the value of m
$
\Rightarrow \left( {m + n} \right) + \left( {m - n} \right) = 21 + 3 \\
\Rightarrow 2m = 24 \\
\Rightarrow m = 12 \\
$
Hence option C is the correct answer.
Note: - In the above question binomial theorem of series expansion has been used in the very first step to expand the terms. We do not need to expand and multiply the whole series as we were concerned with only the first and second power of x so only two to three terms have been considered.
Expanding both the terms of the above term we get:
\[
{\left( {1 + x} \right)^m}{\left( {1 - x} \right)^n} \\
= \left( {1 + {}^m{C_1}{x^1} + {}^m{C_2}{x^2} + {}^m{C_3}{x^3} + ........} \right)\left( {1 - {}^n{C_1}{x^1} + {}^n{C_2}{x^2} - {}^n{C_3}{x^3} + ........} \right) \\
= 1\left( {1 + {}^m{C_1}{x^1} + {}^m{C_2}{x^2} + {}^m{C_3}{x^3} + ........} \right) \\
- {}^n{C_1}{x^1}\left( {1 + {}^m{C_1}{x^1} + {}^m{C_2}{x^2} + {}^m{C_3}{x^3} + ........} \right) \\
+ {}^n{C_2}{x^2}\left( {1 + {}^m{C_1}{x^1} + {}^m{C_2}{x^2} + {}^m{C_3}{x^3} + ........} \right) \\
+ \cdots \cdots \cdots \\
\\
\]
Expanding the equation further to find the coefficients, we get:
\[ \Rightarrow 1 + x\left( {{}^m{C_1} - {}^n{C_1}} \right) + {x^2}\left( {{}^m{C_2} - {}^m{C_1}{}^n{C_1} + {}^n{C_2}} \right) + \cdots \cdots \]
Hence, the coefficient of the term containing $x$ is
$
\Rightarrow {}^m{C_1} - {}^n{C_1} = 3 \\
\Rightarrow m - n = 3 \\
$ ---- (1)
Similarly coefficient of the term containing ${x^2}$ is
$
\Rightarrow {}^m{C_2} - {}^m{C_1}{}^n{C_1} + {}^n{C_2} = - 6 \\
\Rightarrow m\left( {m - 1} \right) - 2mn + n\left( {n - 1} \right) = - 12 \\
\Rightarrow {m^2} - m - 2mn + {n^2} - n = - 12 \\
\Rightarrow {\left( {m - n} \right)^2} - \left( {m + n} \right) = - 12 \\
\Rightarrow 9 - \left( {m + n} \right) = - 12 \\
\Rightarrow m + n = 21 \\
$ ---- (2)
Adding equation 1 and 2 to find the value of m
$
\Rightarrow \left( {m + n} \right) + \left( {m - n} \right) = 21 + 3 \\
\Rightarrow 2m = 24 \\
\Rightarrow m = 12 \\
$
Hence option C is the correct answer.
Note: - In the above question binomial theorem of series expansion has been used in the very first step to expand the terms. We do not need to expand and multiply the whole series as we were concerned with only the first and second power of x so only two to three terms have been considered.
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