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QJEEAdv24Mab4c0: JEE Advanced 2024 Mathematics - Class 12 JEEAdv24Mab4c0 Solution: Let be a continuously differentiable function on the interval…

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

lim⁡t→xt10f(x)−x10f(t)t9−x9=1\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 :

Multiple Choice Options

  • A 3111x−911x10\frac{31}{11 x}-\frac{9}{11} x^{10}
  • B 911x+1311x10\frac{9}{11 x}+\frac{13}{11} x^{10}
  • C −911x+3111x10\frac{-9}{11 x}+\frac{31}{11} x^{10}
  • D 1311x+911x10\frac{13}{11 x}+\frac{9}{11} x^{10}

Step-by-Step Verified Solution

amp;lim⁡1→xt10f(x)−x10f(t)t9−x9=1amp;⇒lim⁡1→x10t9f(x)−x10f′(t)9t8=1amp;⇒10xf(x)−x2f′(x)=9amp;⇒x2f′(x)=10xf(x)−9amp;⇒f′(x)=10f(x)x−9x2amp;⇒dydx−10xy=−9x2amp;⇒y⋅1x10=∫−9x2⋅1x10dxamp;⇒yx10=911x11+camp;∵f(1)=2⇒21=911+c⇒c=1311amp;∴f(x)=911x+1311x10\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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