Question

Help me

Answer 1

If f(x) be a positive, continuous and differentiable on the interval (a,b). If $$lim_{x\rightarrow a^+}{f(x)=1} \ and \ lim_{x\rightarrow b^-}f(x)=3^{1/4}. Also f'(x)\ge f^3(x)+1/f(x), than$$

$$a.b-a\ge \frac{\pi}{4}$$

$$b. b-a\le\frac{\pi}{4}$$

$$c.b-a\le\frac{\pi}{24}$$

d. none of these.

@hartnn

Answer 2

have u tried solving the given DE ?

Answer 3

nah, :(

Answer 4

say f(x) = y :

\[\large \begin{align}y' &\ge y^3 + 1/y \\~\\\dfrac{dy}{dx} &\ge \dfrac{y^4+1}{y}\end{align}\]

Answer 5

since y=f(x) is positive in the region we are interested, you can do separate variables by "cross multiplying"

Answer 6

say f(x) = y :

\[\large \begin{align}y' &\ge y^3 + 1/y \\~\\\dfrac{dy}{dx} &\ge \dfrac{y^4+1}{y}\\~\\ \dfrac{y}{y^4+1}dy&\ge dx\end{align}\]

Answer 7

integrate both sides

Answer 8

rest should be easy

Answer 9

here $$f(x)=\sqrt{tan2x}$$

Answer 10

\[\large \begin{align}y' &\ge y^3 + 1/y \\~\\
\dfrac{dy}{dx} &\ge \dfrac{y^4+1}{y}\\~\\
\dfrac{y}{y^4+1}dy&\ge dx\\~\\
\int_a^b\dfrac{y}{y^4+1}dy&\ge \int_a^b dx\\~\\
\frac{1}{2} \arctan(y^2) \Bigg|_a^b&\ge x\Bigg|_a^b\\~\\
\frac{1}{2}( \arctan((3^{1/4})^2) - \arctan(1^2)) \Bigg|_a^b&\ge b-a\\~\\
\frac{1}{2}( \arctan(\sqrt{3}) - \arctan(1))&\ge b-a\\~\\
\frac{1}{2}( \frac{\pi}{3} - \frac{\pi}{4} ) &\ge b-a\\~\\

\end{align}\]