A) \[C{{s}^{+}}\]
B) \[C{{l}^{-}}\]
C) \[9.2g{{N}_{2}}{{O}_{4}}\]
D) \[{{N}_{2}}{{O}_{4}}(g)2N{{O}_{2}}(g)\]
Correct Answer: A
Solution :
Let\[I=\int_{\sqrt{\ln 2}}^{\sqrt{\ln 3}}{\frac{x\sin {{x}^{2}}}{\sin {{x}^{2}}+\sin (\ln 6-{{x}^{2}})}dx}\] Putting \[{{x}^{2}}=t\]\[\Rightarrow \]\[2x\,dx=dt\] \[\therefore \]\[I=\frac{1}{2}\int_{\ln 2}^{\ln 3}{\frac{\sin t}{\sin t+\sin (\ln 6-t)}}dt\] Using\[\int_{a}^{b}{f(x)dx}=\int_{a}^{b}{f(a+b-x)dx,}\]we get \[I=\frac{1}{2}\int_{\ln 2}^{\ln 3}{\frac{\sin (\ln 6-t)}{\sin (\ln 6-t)+\sin t}dt}\] ?(ii) On adding Eqs.(i) and (ii), we get \[2I=\int_{\ln 2}^{\ln 3}{1dt}=[t]_{\ln 2}^{\ln 3}\] \[\Rightarrow \]\[2I=\frac{1}{2}(\ln 3-\ln 2)=\frac{1}{2}\ln \left( \frac{3}{2} \right)\] \[\Rightarrow \]\[I=\frac{1}{4}\ln \left( \frac{3}{2} \right)\]
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