A) \[P=2,Q=-3,R=4\]
B) \[P=-5,Q=2,R=3\]
C) \[P=5,Q=-2,R=3\]
D) \[P=5,Q=-6,R=3\]
Correct Answer: D
Solution :
Given function \[f(x)=P{{e}^{2x}}+Q{{e}^{x}}+Rx\] ?(i) Given conditions \[f(0)=-1,\] \[f'(\log 2)=31\] and \[\int_{0}^{\log \,4}{[f(x)-Rx]\,dx}=\frac{39}{2}\] differentiate equation (i) \[f'(x)=2P{{e}^{2x}}+Q{{e}^{x}}+R\] ...(ii) Put \[x=\log \,2\] in equation (ii) \[f'(\log \,2)=2P{{e}^{2\log \,2}}+Q{{e}^{\log \,2}}+R\] \[31=8P+2Q+R\] ...(iii) and, put \[x=0\]in equation (i) \[f(0)=P{{e}^{2\times 0}}+Q{{e}^{0}}+R.0\] \[=P+Q-1=P+Q\] \[\Rightarrow \,\,P=-1-Q\] ...(iv) Thus \[\int\limits_{0}^{\log \,\,4}{[f(x)-Rx]dx=\frac{39}{2}}\] \[\Rightarrow \,\,\int\limits_{0}^{\log \,\,4}{[P{{e}^{2x}}+Q{{e}^{x}}+Rx-Rx]dx=\frac{39}{2}}\] \[\Rightarrow \,\,\int\limits_{0}^{\log \,4}{[P{{e}^{2x}}+Q{{e}^{x}}]dx=\frac{39}{2}}\] \[\Rightarrow \,\,\left[ \frac{P{{e}^{2x}}}{2}+Q{{e}^{x}} \right]_{0}^{\log \,4}=\frac{39}{2}\] \[\Rightarrow \,\,\frac{P}{2}\times 16+4Q-\frac{P}{2}-Q=\frac{39}{2}\] \[\Rightarrow \,\,\frac{15P}{2}+3Q=\frac{39}{2}\] ?..(v) From (iv) and (v), we get \[\frac{15P}{2}+3(-1-P)=\frac{39}{2}\] \[\Rightarrow \,\,\frac{9P}{2}=\frac{45}{2}\Rightarrow P=5\] And \[Q=-1-P=-1-5=-6\] and from equation (iii) \[31=8\times 5+2\times -6+R\] \[31=40-12+R\] \[\therefore \,P=5;\,Q=-6,R=3\]
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