Q. If $ y=f\left( \frac{2x+3}{3-2x} \right) $ and $ f(x)=\sin (\log x) $ , then the angle between $ \frac{dy}{dx} $ and $ \frac{12}{9-4{{x}^{2}}}\cos \left\{ \log \left( \frac{2x+3}{3-2x} \right) \right\} $ is

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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}} $

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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