JEE Advanced 2024 — Paper 1 — MATH-I solution

JEE Advanced2024Paper 1MATH-I

Let $f (x)$ be a continuously differentiable function on the interval $(0, \infty)$ such that $f (1) = 2$ and

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

for each $x > 0$ . Then, for all $x > 0$ , $f (x)$ is equal to

Solution

Applying L'Hopital's rule to the limit (differentiating numerator and denominator with respect to $t$ ):

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

Substituting $t = x$ gives $10 x^{9} f (x) - x^{10} f^{'} (x) = 9 x^{8}$ , which rearranges to the linear ODE

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

The integrating factor is $e^{- \int \frac{10}{x} d x} = x^{- 10}$ . Therefore

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

Integrating, $\frac{f ( x )}{x ^{10}} = \frac{9}{11 x ^{11}} + C$ . Using $f (1) = 2$ gives $C = \frac{13}{11}$ .

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

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