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

Mathematics · Limits, Continuity and Differentiability

JEE Advanced 2024 — Paper 1 — Question 1

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

Option A:
$\frac{31}{11 \mathrm{x}}-\frac{9}{11} \mathrm{x}^{10}$
Option B:
$\frac{9}{11 \mathrm{x}}+\frac{31}{11} \mathrm{x}^{10}$
Option C:
$\frac{-9}{11 \mathrm{x}}+\frac{31}{11} \mathrm{x}^{10}$
Option D:
$\frac{13}{11 \mathrm{x}}+\frac{9}{11} \mathrm{x}^{10}$
Correct

Answer: D

Step-by-step solution

\ell = \frac{1}{4v} \sqrt{\frac{\gamma \text{RT}}{\text{M}}}

Calculations for $\frac{1}{4v} \sqrt{\frac{\gamma \text{RT}}{\text{M}}}$ for gases mentioned in options A, B, C and D, work out to be 0.459 m, 0.363 m 0.340 m & 0.348 m respectively. As $\ell = (0.350 \pm 0.005)\,\text{m}$ ; Hence correct option is D.

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2024
Paper: Paper 1
Subject: Mathematics
Chapter: Limits, Continuity and Differentiability
Topic: Application of L'Hospital rule, series expansion.