A) \[\gamma =\frac{{{C}_{p}}}{{{C}_{v}}}=\frac{{{n}_{1}}{{C}_{{{p}_{1}}}}+{{n}_{2}}{{C}_{{{p}_{2}}}}}{{{n}_{1}}{{C}_{{{v}_{1}}}}+{{n}_{2}}{{C}_{{{v}_{2}}}}}\]
B) \[\therefore \]
C) \[\frac{3}{2}=\frac{2\left( \frac{5}{2}R \right)+n\left( \frac{7}{2}R \right)}{2\left( \frac{3}{2}R \right)+n\left( \frac{5}{2}R \right)}\]
D) All of these
Correct Answer: D
Solution :
We have, \[u={{e}^{x}}\sin x\]and \[v={{e}^{x}}\cos x\] On differentiating both sides w.r.t. x, we get \[\frac{du}{dx}={{e}^{x}}\sin x+{{e}^{x}}\cos x=u+v\] and \[\frac{dv}{dx}={{e}^{x}}\cos x-{{e}^{x}}\sin x=v-u\] \[\therefore \]\[v\frac{du}{dx}-u\frac{dv}{dx}=v(u+v)-u(v-u)={{u}^{2}}+{{v}^{2}}\] Now,\[\frac{{{d}^{2}}u}{d{{x}^{2}}}=\frac{du}{dx}+\frac{dv}{dx}=u+v+v-u=2v\]and\[\frac{{{d}^{2}}v}{d{{x}^{2}}}=\frac{dv}{dx}-\frac{du}{dx}\]\[=(v-u)-(v+u)=-2u\]
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