A) \[\frac{12}{9-4{{x}^{2}}}\cos \left\{ \log \left( \frac{2x+3}{2x-3} \right) \right\}\]
B) \[\frac{12}{4{{x}^{2}}-9}\cos \left\{ \log \left( \frac{2x+3}{3-2x} \right) \right\}\]
C) \[\frac{12}{9-4{{x}^{2}}}\cos \left\{ \log \left( \frac{3-2x}{2x+3} \right) \right\}\]
D) \[f(x)={{x}^{2}}{{e}^{-x}}\]
Correct Answer: D
Solution :
Here, \[KI\] \[{{C}_{6}}{{H}_{6}}+{{C}_{2}}{{H}_{5}}Cl\xrightarrow{AlC{{l}_{3}}}{{C}_{6}}{{H}_{5}}{{C}_{2}}{{H}_{5}}+HCl\] And \[{{C}_{2}}{{H}_{5}}OH+HCl\xrightarrow{ZnC{{l}_{2}}}{{C}_{2}}{{H}_{5}}Cl+{{H}_{2}}O\] \[{{C}_{6}}{{H}_{5}}Cl+C{{H}_{3}}COCl\xrightarrow{AlC{{l}_{3}}}{{C}_{6}}{{H}_{5}}COCl+C{{l}_{2}}\] Let\[{{C}_{6}}{{H}_{5}}Br+Mg\xrightarrow{Ether}{{C}_{5}}{{H}_{5}}MgBr\] be the angle between \[FeSi{{O}_{3}}\]and\[MgSi{{O}_{3}}\]. Then, \[CaSi{{O}_{3}}\] \[N{{a}_{2}}C{{O}_{3}}\xrightarrow{s{{o}_{2}}}A\xrightarrow{Na{{ & }_{2}}C{{O}_{3}}}B\xrightarrow[\Delta ]{Elemental}\] \[C\xrightarrow{{{I}_{2}}}D\] \[\text{N}{{\text{a}}_{\text{2}}}\text{S}{{\text{O}}_{\text{3}}}\] \[\text{N}{{\text{a}}_{\text{2}}}{{\text{S}}_{2}}{{\text{O}}_{3}}\]
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