This is a Test of Mathematics Solution Subjective 144 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.
Also visit: I.S.I. & C.M.I. Entrance Course of Cheenta
Suppose $ f(x)$ is a real valued differentiable function defined on $ [1,\infty)$ with $ f(1)=1$. Suppose moreover $ f(x)$ satisfies
$ f'(x) = \frac {1}{x^2+f^2(x)}$
Show that $ f(x) \leq 1+\frac{\pi}{4}$ for every $ x \geq 1$
As the question doesn't require us to find an exact solution rather just an upper bound, we can easily find it by manipulating the given statement after establishing certain properties of $ f(x)$.
We see that $ f'(x)>0$ for all $ x$ which means $ f(x)$ is an increasing function.
As the domain is $ [1,\infty)$ we can say that $ f(x)\geq f(1) $ for all $ x$.
$ => f^2(x)\geq f^2(1)$
$ => x^2+f^2(x)\geq x^2+f^2(1)$ (as $ x^2> 0$)
$ => \frac{1}{x^2+f^2(x)}\leq \frac{1}{x^2+f^2(1)}$
$ => f'(x)\leq \frac{1}{x^2+f^2(1)}$
Integrating both sides from 1 to $ x$
$ => \int_{1}^{x}f'(x)\leq \int_{1}^{x}\frac{1}{x^2+f^2(1)}\leq \int_{1}^{\infty}\frac{1}{x^2+f^2(1)}$
As $ f(1)=1$ we have,
$ => \int_{1}^{x}f'(x)\leq \int_{1}^{\infty}\frac{1}{x^2+1}$
$ => f(x) - f(1) \leq tan^{-1}\infty - tan^{-1}1$
$ => f(x) - f(1) \leq \frac{\pi}{2} - \frac{\pi}{4}$
Substituting $ f(x) = 1$
$ => f(x) \leq 1+\frac{\pi}{4}$
Hence Proved.
This is a Test of Mathematics Solution Subjective 144 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.
Also visit: I.S.I. & C.M.I. Entrance Course of Cheenta
Suppose $ f(x)$ is a real valued differentiable function defined on $ [1,\infty)$ with $ f(1)=1$. Suppose moreover $ f(x)$ satisfies
$ f'(x) = \frac {1}{x^2+f^2(x)}$
Show that $ f(x) \leq 1+\frac{\pi}{4}$ for every $ x \geq 1$
As the question doesn't require us to find an exact solution rather just an upper bound, we can easily find it by manipulating the given statement after establishing certain properties of $ f(x)$.
We see that $ f'(x)>0$ for all $ x$ which means $ f(x)$ is an increasing function.
As the domain is $ [1,\infty)$ we can say that $ f(x)\geq f(1) $ for all $ x$.
$ => f^2(x)\geq f^2(1)$
$ => x^2+f^2(x)\geq x^2+f^2(1)$ (as $ x^2> 0$)
$ => \frac{1}{x^2+f^2(x)}\leq \frac{1}{x^2+f^2(1)}$
$ => f'(x)\leq \frac{1}{x^2+f^2(1)}$
Integrating both sides from 1 to $ x$
$ => \int_{1}^{x}f'(x)\leq \int_{1}^{x}\frac{1}{x^2+f^2(1)}\leq \int_{1}^{\infty}\frac{1}{x^2+f^2(1)}$
As $ f(1)=1$ we have,
$ => \int_{1}^{x}f'(x)\leq \int_{1}^{\infty}\frac{1}{x^2+1}$
$ => f(x) - f(1) \leq tan^{-1}\infty - tan^{-1}1$
$ => f(x) - f(1) \leq \frac{\pi}{2} - \frac{\pi}{4}$
Substituting $ f(x) = 1$
$ => f(x) \leq 1+\frac{\pi}{4}$
Hence Proved.
