Definite Integration Question 350
Question: The numbers P, Q and $ R $ for which the function $ f(x)=Pe^{2x}+Qe^{x}+Rx $ satisfies the conditions $ f(0)=-1, $ $ {f}’(\log 2)=31 $ and $ \int_0^{\log 4}{[f(x)-Rx]dx=\frac{39}{2}} $ are given by
Options:
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 $
Show Answer
Answer:
Correct Answer: D
Solution:
We have $ f’(x)=2Pe^{2x}+Qe^{x}+R $
therefore $ f’(\log 2)=8P+2Q+R $
Also, $ -1=f(0)=P+Q $
$ \frac{39}{2}=\int_0^{\log 4}{[f(x)-Rx]dx=\int_0^{\log 4}{(Pe^{2x}}+Q}e^{x})dx $
$ \Rightarrow $ $ \frac{15P}{2}+3Q=\frac{39}{2} $
Now
On solving the above equations, we get $ P=5,Q=-6 $ and $ R=3 $ .
BETA
