for each $x>0$. Then, for all $x>0, f(x)$ is equal to
$\Rightarrow \lim _{t \rightarrow x} \frac{10 t^9 f(x)-x^{10} f^{\prime}(t)}{9 t^8}=1$
$\Rightarrow 10 x f(x)-x^2 f^{\prime}(x)=9$
$\Rightarrow x^2 f^{\prime}(x)=10 x f(x)-9$
$\Rightarrow f^{\prime}(x)=\frac{10 f(x)}{x}-\frac{9}{x^2}$
$\Rightarrow \frac{d y}{d x}-\frac{10}{x} y=-\frac{9}{x^2}$
$\Rightarrow y \cdot \frac{1}{x^{10}}=\int-\frac{9}{x^2} \cdot \frac{1}{x^{10}} d x$
$\Rightarrow \frac{y}{x^{10}}=\frac{9}{11 x^{11}}+c$ $. . . . (1)$
$\because f(1)=2 \Rightarrow \frac{2}{1}=\frac{9}{11}+c \Rightarrow c=\frac{13}{11}$
$\therefore f(x)=\frac{9}{11 x}+\frac{13}{11} x^{10}$
$\Rightarrow$ Option $(B)$ is correct.
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(The inverse trigonometric functions take the principal values)