JEE Advance - Mathematics (2024 - Paper 1 Online - No. 1)

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

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

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

$\frac{31}{11 x}-\frac{9}{11} x^{10}$
$\frac{9}{11 x}+\frac{13}{11} x^{10}$
$\frac{-9}{11 x}+\frac{31}{11} x^{10}$
$\frac{13}{11 x}+\frac{9}{11} x^{10}$

Explanation

$$ \begin{aligned} & \lim _{1 \rightarrow x} \frac{t^{10} f(x)-x^{10} f(t)}{t^9-x^9}=1 \\\\ & \Rightarrow \lim _{1 \rightarrow x} \frac{10 t^9 f(x)-x^{10} f^{\prime}(t)}{9 t^8}=1 \\\\ & \Rightarrow 10 x f(x)-x^2 f^{\prime}(x)=9 \\\\ & \Rightarrow x^2 f^{\prime}(x)=10 x f(x)-9 \\\\ & \Rightarrow f^{\prime}(x)=\frac{10 f(x)}{x}-\frac{9}{x^2} \\\\ & \Rightarrow \frac{d y}{d x}-\frac{10}{x} y=-\frac{9}{x^2} \\\\ & \Rightarrow y \cdot \frac{1}{x^{10}}=\int-\frac{9}{x^2} \cdot \frac{1}{x^{10}} d x \\\\ & \Rightarrow \frac{y}{x^{10}}=\frac{9}{11 x^{11}}+c \\\\ & \because f(1)=2 \Rightarrow \frac{2}{1}=\frac{9}{11}+c \Rightarrow c=\frac{13}{11} \\\\ & \therefore f(x)=\frac{9}{11 x}+\frac{13}{11} x^{10} \end{aligned} $$

$\Rightarrow$ Option (B) is correct.

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