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# Test of Mathematics Solution Subjective 144 - Finding a Function's Upper Bound

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

## Problem

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 $x \geq 1$

## Solution

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.

## Important Resources:

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

## Problem

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 $x \geq 1$

## Solution

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.