A) \[6\]
B) \[9\]
C) \[12\]
D) \[24\]
Correct Answer: C
Solution :
\[\therefore \]\[{{(1+x)}^{m}}{{(1-x)}^{n}}\] \[=\left[ 1+mx+\frac{m(m-1){{x}^{2}}}{2!}+.. \right]\] \[\times \left[ 1-nx+\frac{n(n-1){{x}^{2}}}{2!}-... \right]\] \[=1+(m-n)x+\] \[\left( \frac{{{n}^{2}}-n}{n}-mn+\frac{{{m}^{2}}-m}{2} \right){{x}^{2}}+...\] Since, the coefficient of \[x\] and \[{{x}^{2}}\] are \[3\] and \[-6\] respectively. \[\therefore \] \[m-n=3\] and \[\frac{{{n}^{2}}-n}{2}-mn+\frac{{{m}^{2}}-m}{2}=-6\] \[\Rightarrow \]\[\frac{(m-3)(m-4)}{2}-m(m-3)+\frac{{{m}^{2}}-m}{2}=-6\] \[\Rightarrow \] \[{{m}^{2}}-7m+12-2{{m}^{2}}+6m\] \[+{{m}^{2}}-m+12=0\] \[\Rightarrow \] \[-2m+24=0\] \[\Rightarrow \] \[m=12\]
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