Let $f(\mathrm{x})$ be a continuously differentiable function on the interval $(0, \infty)$ such that $f(1)=2$ and $\lim _{\mathrm{t} \rightarrow \mathrm{x}} \frac{\mathrm{t}^{10} f(\mathrm{x})-\mathrm{x}^{10} f(\mathrm{t})}{\mathrm{t}^{9}-\mathrm{x}^{9}}=1$ for each $\mathrm{x}>0$. Then, for all $\mathrm{x}>0, f(\mathrm{x})$ is equal to :

(B) $\frac{9}{11 \mathrm{x}}+\frac{13}{11} \mathrm{x}^{10}$

(A) $\frac{31}{11 \mathrm{x}}-\frac{9}{11} \mathrm{x}^{10}$

(C) $\frac{-9}{11 \mathrm{x}}+\frac{31}{11} \mathrm{x}^{10}$

(D) $\frac{13}{11 \mathrm{x}}+\frac{9}{11} \mathrm{x}^{10}$

JEE Advanced · 2024 · 26 May 2024 · Paper 1 · Differential Equations

Let

f (x)

be a continuously differentiable function on the interval

(0, \infty)

such that

f (1) = 2

and

lim_{t \to x} \frac{t ^{10} f ( x ) - x ^{10} f ( t )}{t ^{9} - x ^{9}} = 1

for each

x > 0

. Then, for all

x > 0, f (x)

is equal to :

Answer: (B)

Apply L'Hôpital's rule to the given limit:

t \to x lim \frac{10 t ^{9} f ( x ) - x ^{10} f ^{'} ( t )}{9 t ^{8}} = 1

Substituting

t = x

10 x f (x) - x^{2} f^{'} (x) = 9

Rearranging:

f^{'} (x) - \frac{10}{x} f (x) = - \frac{9}{x ^{2}}

This is a first-order linear ODE. The integrating factor is

e^{- 10 l n x} = x^{- 10}

. Multiplying:

\frac{d}{d x} (\frac{f ( x )}{x ^{10}}) = - \frac{9}{x ^{12}}

Integrating:

\frac{f ( x )}{x ^{10}} = \frac{9}{11 x ^{11}} + c

Using

f (1) = 2

2 = \frac{9}{11} + c \Rightarrow c = \frac{13}{11}

Therefore:

f (x) = \frac{9}{11 x} + \frac{13}{11} x^{10}

JEE Advanced · 2024 · 26 May 2024 · Paper 1 · Single correct option