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The numbers P, Q and R for which the function $f (x) = P e^{2 x} + Q e^{x} + R x$ satisfies $f (0) = - 1$
$f^{'} (\log 2) = 31$ and $\int_{0}^{\log 4} [f (x) - R x] d x = 39 / 2$ are given by

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P = 2, Q = – 3, R = 4

P = – 5, Q = 2, R = 3

P = 5, Q = – 2, R = 3

P= 5, Q = – 6, R = 3

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detailed solution

Correct option is D

We have f′(x)=2Pe2x+Qex+R, so that 31=f′(log⁡2)=8P+2Q+RAlso, −1=f(0)=P+Q. Further,392=∫0log⁡4 [f(x)−Rx]dx=∫0log⁡4 Pe2x+Qexdx=P2e2x+Qex0log⁡4=P2×16+4Q−P2−Q=15P2+3QSolving the above equations, we get P=5,Q=−6 and R=3

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The numbers P, Q and R for which the function f(x)=Pe2x+Qex+Rx satisfies f(0)=–1 f′(log⁡2)=31 and ∫0log⁡4 [f(x)−Rx]dx=39/2 are given by

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The numbers P, Q and R for which the function $f (x) = P e^{2 x} + Q e^{x} + R x$ satisfies $f (0) = - 1$
$f^{'} (\log 2) = 31$ and $\int_{0}^{\log 4} [f (x) - R x] d x = 39 / 2$ are given by