A) \[\left( 16,\frac{251}{3} \right)\]
B) \[\left( 14,\frac{251}{3} \right)\]
C) \[\left( 14,\frac{272}{3} \right)\]
D) \[\left( 16,\frac{272}{3} \right)\]
Correct Answer: D
Solution :
\[1{{(1-2x)}^{18}}+ax\,{{(1-2x)}^{18}}+b{{x}^{2}}\,{{(1-2x)}^{18}}\] | ||
Coefficient of \[{{x}^{3}}:\] \[{{(-2)}^{3}}{{\,}^{18}}{{C}_{3}}+a\,{{(-2)}^{2}}{}^{18}{{C}_{2}}+b(-2){}^{18}{{C}_{1}}=0\] | ||
\[\frac{4\times (17\times 16)}{(3\times 2)}-2a.\frac{17}{2}+b=0\] | ? (i) | |
Coefficient of \[{{x}^{4}}\]: \[{{(-2)}^{4}}{}^{18}{{C}_{4}}+a{{(-2)}^{3}}{}^{18}{{C}_{3}}+b{{(-2)}^{2}}{}^{18}{{C}_{2}}=0\] | ||
\[(4\times 20)-2a.\frac{16}{3}+b=0\] | ? (ii) | |
From equation (i) and (ii), we get, \[4\left( \frac{17\times 8}{3}-20 \right)+2a\left( \frac{16}{3}-\frac{17}{2} \right)=0\] |
\[4\left( \frac{17\times 8-60}{3} \right)+\frac{2a(-19)}{6}=0\] |
\[a=\frac{4\times 76\times 6}{3\times 2\times 19}\] |
\[\Rightarrow \] b = 16 |
\[\Rightarrow \] \[b=\frac{2\times 16\times 6}{3}=80=\frac{272}{3}.\] |
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