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(B) Given the determinant condition: $f(x)f^{\prime \prime}(x) - (f^{\prime}(x))^2 = 0$.This can be rewritten as $\frac{f^{\prime \prime}(x)}{f^{\prim

Vedclass · Accepted Answer

$(6, 9)$

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(B) Given the determinant condition: $f(x)f^{\prime \prime}(x) - (f^{\prime}(x))^2 = 0$.This can be rewritten as $\frac{f^{\prime \prime}(x)}{f^{\prime}(x)} = \frac{f^{\prime}(x)}{f(x)}$.Integrating both sides with respect to $x$,we get $\ln(f^{\prime}(x)) = \ln(f(x)) + \ln(c)$,which implies $f^{\prime}(x) = c f(x)$.This is a first-order linear differential equation: $\frac{f^{\prime}(x)}{f(x)} = c$.Integrating again,$\ln(f(x)) = cx + k_1$,so $f(x) = k e^{cx}$.Using the initial conditions:$f(0) = 1 \implies k e^0 = 1 \implies k = 1$.$f^{\prime}(x) = c e^{cx}$,and $f^{\prime}(0) = 2 \implies c e^0 = 2 \implies c = 2$.Thus,$f(x) = e^{2x}$.Calculating $f(1) = e^2 \approx 7.389$.Since $6 < 7.389 < 9$,the value lies in the interval $(6, 9)$.

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