JEE Maths#5

$\begin{aligned} f(x) & = ae^{2x} + be^x + cx \\ f(0) & = - 1 & \small \color{#3D99F6} ... \text{given} \\ \implies a + b & = -1 & ...(1) \end{aligned}$

$\begin{aligned} f'(x) & = 2ae^{2x} + be^x + c \\ f'(\ln 2) & = 31 & \small \color{#3D99F6} ... \text{given} \\ \implies 8a + 2b + c & = 31 & ...(2) \end{aligned}$

$\begin{aligned} \int_0^{\ln 4} (f(x) - cx) \ dx & = \frac {39}2 & \small \color{#3D99F6} ... \text{given} \\ \int_0^{\ln 4} (ae^{2x}+be^x) \ dx & = \frac {39}2 \\ \frac {ae^{2x}}2 + be^x \ \bigg|_0^{\ln 4} & = \frac {39}2 \\ 8a+4b-\frac a2 - b & = \frac {39}2 \\ \implies 15a + 6b & = 39 &...(3) \end{aligned}$

$\begin{aligned} (3)-6(1): \quad (15-6)a & = 39+6 \\ \implies a & = 5 \end{aligned}$

$\begin{aligned} (1): \quad 5+b & = -1\\ \implies b & = -6 \end{aligned}$

$\begin{aligned} (2): \quad 8(5)+2(-6)+c & = 31 \\ \implies c & = 3 \end{aligned}$

Therefore, the answer is $\boxed{a=5}$ .

$\begin{aligned} & f(0) = -1 \\ \implies & ae^0 + be^0 + c(0) = -1 \\ \implies & a+b = -1 \qquad \color{#3D99F6} (*) \end{aligned}$

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$\begin{aligned} & f'(x) = 2ae^{2x} + be^x + c \\ \implies & f'(\ln 2) = 2ae^{2 \ln 2} + be^{\ln 2} + c = 31 \\ \implies & f'(\ln 2) = 8a + 2b + c = 31 \qquad \color{#3D99F6} (**) \end{aligned}$

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$\begin{aligned} & \displaystyle \int_0^{\ln 4} \left( f(x) - cx \right) \,dx = \dfrac{39}{2} \\ \implies & \int_0^{\ln 4} \left( ae^{2x} + be^x \right) \,dx = \dfrac{39}{2} \\ \implies & \dfrac{a}{2} \left[ e^{2x} \right] {\huge |}_0^{\ln 4} + b \left[ e^x \right] {\huge |}_0^{\ln 4} = \dfrac{39}{2} \\ \implies & \dfrac{15a}{2} + 3b = \dfrac{39}{2} \\ \implies & \dfrac{5a}{2} + b = \dfrac{13}{2} \qquad \color{#3D99F6} (***) \end{aligned}$

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Solving the system of three simultaneous linear equations $\color{#3D99F6} (*)$ , $\color{#3D99F6} (**)$ and $\color{#3D99F6} (***)$ , we obtain

$\boxed{a = 5} \\ b = -6 \\ c = 3$