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(B) Given $f^{\prime\prime}(x) = f(x)$. Multiplying both sides by $f^{\prime}(x)$,we get $f^{\prime}(x)f^{\prime\prime}(x) = f(x)f^{\prime}(x)$.Integr

Vedclass · Accepted Answer

$36$

Vedclass · Answer

(B) Given $f^{\prime\prime}(x) = f(x)$. Multiplying both sides by $f^{\prime}(x)$,we get $f^{\prime}(x)f^{\prime\prime}(x) = f(x)f^{\prime}(x)$.Integrating both sides with respect to $x$,we have $\frac{1}{2}(f^{\prime}(x))^2 = \frac{1}{2}(f(x))^2 + C$.Using the initial conditions $f(0) = 0$ and $f^{\prime}(0) = 3$,we get $\frac{1}{2}(3)^2 = \frac{1}{2}(0)^2 + C$,so $C = \frac{9}{2}$.Thus,$(f^{\prime}(x))^2 = (f(x))^2 + 9$. Since $f^{\prime}(x) \geq 0$,we have $f^{\prime}(x) = \sqrt{(f(x))^2 + 9}$.Separating variables,$\int \frac{df}{\sqrt{f^2 + 9}} = \int dx$,which gives $\ln|f(x) + \sqrt{(f(x))^2 + 9}| = x + C_1$.Using $f(0) = 0$,we get $\ln|0 + \sqrt{0 + 9}| = 0 + C_1$,so $C_1 = \ln 3$.Therefore,$f(x) + \sqrt{(f(x))^2 + 9} = 3e^x$.Let $y = f(x)$. Then $\sqrt{y^2 + 9} = 3e^x - y$. Squaring both sides,$y^2 + 9 = 9e^{2x} - 6ye^x + y^2$,which simplifies to $6ye^x = 9e^{2x} - 9$.So,$f(x) = \frac{9(e^{2x} - 1)}{6e^x} = \frac{3}{2}(e^x - e^{-x}) = 3\sinh(x)$.At $x = \ln 3$,$f(\ln 3) = \frac{3}{2}(3 - \frac{1}{3}) = \frac{3}{2}(\frac{8}{3}) = 4$.Thus,$9f(\ln 3) = 9 	imes 4 = 36$.

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a thrice differentiable odd function satisfying $f^{\prime}(x) \geq 0$,$f^{\prime\prime}(x) = f(x)$,$f(0) = 0$,and $f^{\prime}(0) = 3$. Then $9f(\log_e 3)$ is equal to . . . . . . .

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