A) \[\sqrt{{{a}^{2}}+{{b}^{2}}}\]
B) \[\frac{1}{\sqrt{ab}}\]
C) \[ab\]
D) \[a+b\]
Correct Answer: A
Solution :
Given, \[\frac{d}{dx}({{e}^{ax}}\,x\cos \,bx)=r{{e}^{ax}}\cos (bx+\alpha ),\] then r=? Let \[y={{e}^{ax}}.\cos \,bx\] \[\frac{dy}{dx}=a{{e}^{ax}}.\cos bx-b{{e}^{ax}}.\sin bx\] \[\frac{dy}{dx}={{e}^{ax}}(a\,\cos \,bx-b\,\sin bx)\] Let \[\left. \begin{matrix} a=r\,\cos \,\alpha \\ b=r\,\sin \,\alpha \\ \end{matrix} \right\}\] …..(i) They, \[\frac{dy}{dx}={{e}^{ax}}.r(\cos \,bx.cos\alpha -sinbx.sin\alpha )\] \[\frac{dy}{dx}={{e}^{ax}}.r\cos (bx+\alpha )\] ….(ii) Where, \[\tan \alpha =\frac{b}{a}\Rightarrow \alpha ={{\tan }^{-1}}\left( \frac{b}{a} \right)\] and \[{{r}^{2}}({{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha )={{a}^{2}}+{{b}^{2}}\] \[{{r}^{2}}={{a}^{2}}+{{b}^{2}}\] \[r=\sqrt{{{a}^{2}}+{{b}^{2}}}\]
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