A) (35, 20)
B) (45, 35)
C) (35, 45)
D) (20, 45)
Correct Answer: C
Solution :
Given, \[{{(1-y)}^{m}}{{(1+y)}^{n}}=1+{{a}_{1}}y+{{a}_{2}}{{y}^{2}}+.....\] Since,\[{{(1-y)}^{m}}{{(1+y)}^{n}}={{(}^{m}}{{C}_{0}}{{-}^{m}}{{C}_{1}}y{{+}^{m}}{{C}_{2}}{{y}^{2}}-....)\] \[\times {{(}^{n}}{{C}_{0}}{{+}^{n}}{{C}_{1}}y{{+}^{n}}{{C}_{2}}{{y}^{2}}+....)\] \[\therefore \]\[{{a}_{1}}=\]Coefficient of y in \[{{(1-y)}^{m}}{{(1+y)}^{n}}\] \[{{=}^{n}}{{C}_{1}}{{-}^{m}}{{C}_{1}}=10\] \[\Rightarrow \]\[n-m=10\] \[\Rightarrow \]\[n=m+10\] ...(i) and\[{{a}_{2}}=\]Coefficient of\[{{y}^{2}}\]in \[{{(1-y)}^{m}}{{(1+y)}^{n}}\] \[{{=}^{n}}{{C}_{2}}{{-}^{m}}{{C}_{1}}{{.}^{n}}{{C}_{1}}{{+}^{m}}{{C}_{2}}\] \[\therefore \]\[^{n}{{C}_{2}}{{-}^{m}}{{C}_{1}}{{.}^{n}}{{C}_{1}}{{+}^{m}}{{C}_{2}}=10\] \[\Rightarrow \]\[\frac{n(n-1)}{2}+\frac{m(m-1)}{2}-mn=10\] \[\Rightarrow \]\[\frac{(10-m)(9+m)}{2}+\frac{m(m-1)}{2}-m(10+m)=10\] [from Eq. (i)] \[\Rightarrow \]\[45+\frac{19m}{2}+\frac{{{m}^{2}}}{2}+\frac{{{m}^{2}}}{2}-\frac{m}{2}-10m-{{m}^{2}}=10\] \[\Rightarrow \]\[45-m=10\Rightarrow m=35\] \[\therefore \] \[n=45\] [from Eq. (i)] Hence, \[(m,\text{ }n)=(35,\text{ }45)\] Alternate Solution \[{{(1-y)}^{m}}{{(1+y)}^{n}}=1+{{a}_{1}}y+{{a}_{2}}{{y}^{2}}+a{{ & }_{3}}{{y}^{3}}+.....\] On differentiating w.r.t. y. we get \[-m{{(1-y)}^{m-1}}{{(1+y)}^{n}}+{{(1-y)}^{m}}n{{(1+y)}^{n-1}}\] \[={{a}_{1}}+2{{a}_{2}}y+3{{a}_{3}}{{y}^{3}}+...\] ... (i) Now, putting\[y=0\]in Eq. (i), we get \[-m+n={{a}_{1}}=10\] ...(ii) where, \[{{a}_{1}}=10\] Again, differentiating Eq. (i), we get \[-m+[(m-1){{(1-y)}^{m-2}}{{(1+y)}^{2}}{{(1-y)}^{m-1}}n{{(1+y)}^{n-1}}]\] \[+n[-m{{(1-y)}^{m-1}}{{(1+y)}^{n-1}}+{{(1-y)}^{m}}(n-1){{(1+y)}^{n-2}}]\] \[=2{{a}_{2}}+6{{a}_{3}}+y+...\] ...(iii) Now, putting y = 0 in Eq. (iii), we get \[-m[-(m-1)+n]+n[-m+(n-1)]=2{{a}_{2}}=20\] \[\Rightarrow \]\[m(m-1)-mn-mn+n(n-1)=20\] \[\Rightarrow \]\[{{m}^{2}}+{{n}^{2}}-m-n-2mn=20\] \[\Rightarrow \]\[{{(m-n)}^{2}}-(m+n)=20\] \[\Rightarrow \]\[100-(m+n)=20\] \[(\because -m+n=10)\] \[\Rightarrow \] \[m+n=80\] ...(iv) On solving Eqs. (ii) and (iv), we get \[m=35\]and \[n=45\]
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