Suppose f is a real valued differentiable function defined on [1, ∞) with f(1)=1 such that f satisfies f prime(x)=(1/x2+f2(x)) then value of f(x) can not exceed.

Question

Suppose f is a real valued differentiable function defined on [1, ∞) with f(1)=1 such that f satisfies f prime(x)=(1/x2+f2(x)) then value of f(x) can not exceed. (A)  ln 2 (B) 1+(π/4) (C) 1+ ln tan 1 (D) (π/4)

Tardigrade Admin · Accepted Answer

f prime( x )=(1/ x 2+ f 2( x ))>0 ⇒ f ( x ) is an increasing function ∀ x >1 Now f prime(x)=(1/x2+f2(x)) ≤ (1/x2+1) ∀ x ≥ 1 .∫ limits1x f prime(x) d x ≤ ∫ limits1x (1/(x2+1.)) d x f(x)-1 ≤ tan -1 x-(π/4) displaystyle lim x arrow ∞ f(x)-1 ≤ π / 2-(π/4) ⇒ displaystyle lim x arrow ∞ f(x) ≤ 1+(π/4)

Q. Suppose $f$ is a real valued differentiable function defined on $[1, \infty)$ with $f (1) = 1$ such that $f$ satisfies $f^{'} (x) = \frac{1}{x ^{2} + f ^{2} ( x )}$ then value of $f (x)$ can not exceed.

$ln 2$

$1 + \frac{π}{4}$

$1 + ln tan 1$

$\frac{π}{4}$

Q. Suppose f is a real valued differentiable function defined on [1,∞) with f(1)=1 such that f satisfies f′(x)=x2+f2(x)1​ then value of f(x) can not exceed.

ln2

1+4π​

1+lntan1

4π​

f′(x)=x2+f2(x)1​>0 ⇒f(x) is an increasing function ∀x>1 Now f′(x)=x2+f2(x)1​≤x2+11​∀x≥1 1∫x​f′(x)dx≤1∫x​(x2+11​)dx f(x)−1≤tan−1x−4π​ x→∞lim​f(x)−1≤π/2−4π​ ⇒x→∞lim​f(x)≤1+4π​

Q. Suppose $f$ is a real valued differentiable function defined on $[1, \infty)$ with $f (1) = 1$ such that $f$ satisfies $f^{'} (x) = \frac{1}{x ^{2} + f ^{2} ( x )}$ then value of $f (x)$ can not exceed.

$ln 2$

$1 + \frac{π}{4}$

$1 + ln tan 1$

$\frac{π}{4}$