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(A) Given $\int_0^1 f(\lambda x) d\lambda = a f(x)$.Let $\lambda x = t$,then $d\lambda = \frac{1}{x} dt$.Substituting these into the integral: $\frac{

Vedclass · Accepted Answer

$112$

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(A) Given $\int_0^1 f(\lambda x) d\lambda = a f(x)$.Let $\lambda x = t$,then $d\lambda = \frac{1}{x} dt$.Substituting these into the integral: $\frac{1}{x} \int_0^x f(t) dt = a f(x)$,which implies $\int_0^x f(t) dt = a x f(x)$.Differentiating both sides with respect to $x$ using the Fundamental Theorem of Calculus: $f(x) = a(f(x) + x f^{\prime}(x))$.Rearranging gives $(1 - a) f(x) = a x f^{\prime}(x)$,so $\frac{f^{\prime}(x)}{f(x)} = \frac{1 - a}{a} \cdot \frac{1}{x}$.Integrating both sides: $\ln|f(x)| = \frac{1 - a}{a} \ln x + C$.Since $f(1) = 1$,we get $C = 0$,so $f(x) = x^{\frac{1-a}{a}}$.Given $f(16) = \frac{1}{8}$,we have $16^{\frac{1-a}{a}} = 2^{-3}$.Since $16 = 2^4$,we have $2^{4 \cdot \frac{1-a}{a}} = 2^{-3}$,so $\frac{4(1-a)}{a} = -3$.$4 - 4a = -3a \Rightarrow a = 4$.Thus,$f(x) = x^{\frac{1-4}{4}} = x^{-3/4}$.Then $f^{\prime}(x) = -\frac{3}{4} x^{-7/4}$.$f^{\prime}\left(\frac{1}{16}\right) = -\frac{3}{4} \left(2^{-4}\right)^{-7/4} = -\frac{3}{4} \cdot 2^7 = -\frac{3}{4} \cdot 128 = -3 \cdot 32 = -96$.Therefore,$16 - f^{\prime}\left(\frac{1}{16}\right) = 16 - (-96) = 112$.

Let $f : (0, \infty) \rightarrow \mathbb{R}$ be a twice differentiable function. If for some $a \neq 0$,$\int_0^1 f(\lambda x) d\lambda = a f(x)$,$f(1) = 1$ and $f(16) = \frac{1}{8}$,then $16 - f^{\prime}\left(\frac{1}{16}\right)$ is equal to . . . . . .

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