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JEE Advanced 2016 · Differential Equations

Let f:(0,∞)→Rf:(0,\infty)\to\mathbb{R} be a differentiable function such that f′(x)=2−f(x)xf'(x)=2-\frac{f(x)}{x} for all x∈(0,∞)x\in(0,\infty) and f(1)≠1f(1)\neq 1. Then

Answers: A. lim⁡x→0+f′(1x)=1\lim_{x\to 0^{+}} f'\left(\frac{1}{x}\right)=1

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Written solution

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

f′(x)=2−f(x)x,x∈(0,∞),f(1)≠1f'(x)=2-\dfrac{f(x)}{x}, \quad x\in(0,\infty), \quad f(1)\neq 1

To find: which statement(s) is/are correct\text{To find: which statement(s) is/are correct}

f′(x)+1xf(x)=2,IF=xf'(x)+\dfrac{1}{x}f(x)=2, \quad \text{IF}=x

ddx(xf(x))=2x\dfrac{d}{dx}\big(x f(x)\big) = 2x

xf(x)=x2+C  ⇒  f(x)=x+Cx,C≠0x f(x) = x^2 + C \;\Rightarrow\; f(x) = x + \dfrac{C}{x}, \quad C\neq 0

f′(x)=1−Cx2  ⇒  f′ ⁣(1x)=1−Cx2→1f'(x) = 1-\dfrac{C}{x^2} \;\Rightarrow\; f'\!\left(\dfrac{1}{x}\right) = 1-Cx^2 \to 1

⇒(1) is correct\Rightarrow (1) \text{ is correct}

x f ⁣(1x)=x(1x+Cx)=1+Cx2→1≠2x\,f\!\left(\dfrac{1}{x}\right) = x\left(\dfrac{1}{x}+Cx\right) = 1+Cx^2 \to 1 \neq 2

⇒(2) is incorrect\Rightarrow (2) \text{ is incorrect}

x2f′(x)=x2(1−Cx2)=x2−C→−C≠0x^2 f'(x) = x^2\left(1-\dfrac{C}{x^2}\right) = x^2 - C \to -C \neq 0

⇒(3) is incorrect\Rightarrow (3) \text{ is incorrect}

f(x)=x+Cx→±∞ as x→0+  ⇒  ∣f∣≤2 failsf(x)=x+\dfrac{C}{x}\to\pm\infty \text{ as } x\to 0^{+} \;\Rightarrow\; |f|\le 2 \text{ fails}

⇒(4) is incorrect\Rightarrow (4) \text{ is incorrect}

Only statement (1) is correct\boxed{\text{Only statement (1) is correct}}

Answer: (1)\text{Answer: (1)}

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