Let f : (0, ) R be a twice differentiable function. If for some a 0, _0^1 f( x) d = a f(x) , f(1) = 1 and f(16) = 18 , then 16 - f' ( 116 ) is equal to ______.

_0^1 f( x) d =af(x) \\ x=t \\ ~d =1x dt \\ 1x _0^x f(t) d t=a f(x) \\ _0^x f(t) d t=axf(x) \\ f(x)=a(x f^(x)+f(x)) \\ (1-a) f(x)=a x f^(x

Let f 0 fty arrow R be a twice differentiable function If fo: (Mathematics)

$\int_0^1 f(\lambda x) d \lambda=\mathrm{af}(x) \\$

$\lambda \mathrm{x}=\mathrm{t} \\$

$\mathrm{~d} \lambda=\frac{1}{x} \mathrm{dt} \\$

$\frac{1}{x} \int_0^x f(t) d t=a f(x) \\$

$\int_0^x f(t) d t=\operatorname{axf}(x) \\$

$f(x)=a\left(x f^{\prime}(x)+f(x)\right) \\$

$(1-a) f(x)=a \cdot x f^{\prime}(x) \\$

$\frac{f^{\prime}(x)}{f(x)}=\frac{(1-a)}{a} \frac{1}{x} \\$

$\ell \mathrm{nf}(\mathrm{x})=\frac{1-\mathrm{a}}{\mathrm{a}} \ell \mathrm{nx}+\mathrm{c} \\$

$\mathrm{x}=1, \mathrm{f}(1)=1 \Rightarrow \mathrm{c}=0 \\$

$\mathrm{x}=16, \mathrm{f}(16)=\frac{1}{8} \\$

$\frac{1}{8}=(16)^{\frac{1-a}{a}} \Rightarrow-3=\frac{4-4 a}{a} \Rightarrow a=4 \\$

$f(x)=x^{-\frac{3}{4}} \\$

$f^{\prime}(x)=-\frac{3}{4} x^{-\frac{7}{4}} \\$

$\therefore 16-\mathrm{f}^{\prime}\left(\frac{1}{16}\right) \\$

$=16-\left(-\frac{3}{4}\left(2^{-4}\right)^{-7 / 4}\right) \\$

$=16+96=112$