SATHEE: Differential Equations 2 Question 21

Differential Equations 2 Question 21

22. Let $f : (0, \infty) \to R$ be a differentiable function such that $f^{'} (x) = 2 - \frac{f (x)}{x}$ for all $x \in (0, \infty)$ and $f (1) \neq 1$ . Then

(a) $lim_{x \to 0 +} f^{'} \frac{1}{x} = 1$

(b) $lim_{x \to 0 +} x f \frac{1}{x} = 2$

(d) $| f (x) | \leq 2$ for all $x \in (0, 2)$

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Answer:

Correct Answer: 22. (a)

Solution:

Here, $f^{'} (x) = 2 - \frac{f (x)}{x}$

or $\frac{d y}{d x} + \frac{y}{x} = 2$ [i.e. linear differential equation in $y$ ]

Integrating Factor, $I F = e^{\int \frac{1}{x} d x} = e^{\log x} = x$

$∴$ Required solution is $y \cdot (I F) = \int Q (I F) d x + C$

$\begin{aligned} \Rightarrow & y (x) & = \int 2 (x) d x + C \\ \Rightarrow & y x & = x^{2} + C \\ ∴ & y & = x + \frac{C}{x} [∵ C \neq 0, as f (1) \neq 1] \end{aligned}$

(a) $lim_{x \to 0^{+}} f^{'} \frac{1}{x} = lim_{x \to 0^{+}} (1 - C x^{2}) = 1$

$∴$ Option (a) is correct.

(b) $lim_{x \to 0^{+}} x f \frac{1}{x} = lim_{x \to 0^{+}} (1 + C x^{2}) = 1$

$∴$ Option (b) is incorrect.

$∴$ Option (c) is incorrect.

(d) $f (x) = x + \frac{C}{x}, C \neq 0$

For $C > 0, lim_{x \to 0^{+}} f (x) = \infty$

$∴$ Function is not bounded in $(0, 2)$ .

$∴$ Option (d) is incorrect.

Differential Equations 2 Question 21

22. Let $f : (0, \infty) \to R$ be a differentiable function such that $f^{'} (x) = 2 - \frac{f (x)}{x}$ for all $x \in (0, \infty)$ and $f (1) \neq 1$ . Then

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Differential Equations 2 Question 21

22. Let f:(0,∞)→R be a differentiable function such that f′(x)=2−f(x)x for all x∈(0,∞) and f(1)≠1. Then

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22. Let $f : (0, \infty) \to R$ be a differentiable function such that $f^{'} (x) = 2 - \frac{f (x)}{x}$ for all $x \in (0, \infty)$ and $f (1) \neq 1$ . Then