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Let $f:[1,\infty) \rightarrow R$ is differentiable and satisfies $f'(x) = \ \frac{1}{x^2+f(x)^2}$ and $f(1) = 1$ Then

MORE THAN ONE ANSWER MAY BE CORRECT

  1. $\lim_{x\to \infty} f(x)$< 1+ $\frac{\pi}{4}$
  2. $\lim_{x\to \infty} f(x)$< 1+ $\frac{\pi}{3}$
  3. $f'(x)$< $\frac{1}{x^2+1}$
  4. $f(x)$ is monotonically decreasing

I was able to prove the 3rd option and then I directly integrated it on both sides like $\int f'(x) dx< \int \frac{dx}{1+x^2}$ is this the correct way to approach options 1. or 2.

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    in the limit an argument is missing! – 2017-01-15
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    @MyGlasses nope. Answer are 1 and 3. – 2017-01-15

1 Answers 1

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since we have $$\frac{1}{x^2+f(x)^2}=f'(x)$$ we get $$\frac{1}{x^2+f(x)^2}>0$$ thus $$f'(x)>0$$ further we have $$\frac{1}{x^2+(f(x)^2}<\frac{1}{1+x^2}$$ since we have $$1

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    so the function is monotonically increasing huh.? what about other options ? $f'(x)$?? – 2017-01-15
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    yes this is true – 2017-01-15
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    How to approach A) And B) Options.?? – 2017-01-15
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    i don't know which limit do you mean, please correct it. – 2017-01-15
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    I edited the question. – 2017-01-15