Question

Let $f$ be a continuous and differentiable function in $(a,b)$, $\underset{x\to a^{+}}{lim}f\left(x\right)\to \infty$ and $\underset{x\to b^{-}}{lim}f\left(x\right)\to -\infty$. If $f^{\prime }\left(x\right)+f^2\left(x\right)\ge -1$ for $a<x<b$, then

• A. $b-a\le \pi$
• B. $b-a\ge \pi$
• C. $b-a=\pi$
• D. None of these

Answer 1

The correct option is B $b-a\ge \pi$

We have $f^{\prime }\left(x\right)+f^2\left(x\right)\ge -1$

$\Rightarrow \frac{f^{\prime }\left(x\right)}{1+f^2\left(x\right)}+1\ge 0,$ for $x\in (a,b)$ ...(1)

$\Rightarrow h\left(x\right)={\tan }^{-1}f\left(x\right)+x,$ is a non-decreasing function in the interval $(a,b)$

$\Rightarrow \underset{x\to a}{lim}h\left(x\right)\le \underset{x\to a}{lim}h\left(x\right)$

$\Rightarrow \underset{x\to a^{+}}{lim}({\tan }^{-1}f\left(x\right)+x)\le \underset{x\to b^{-}}{lim}({\tan }^{-1}f\left(x\right)+x)$

$\Rightarrow \frac{\pi }{2}+a\le -\frac{\pi }{2}+b$

Hence $b-a\ge \pi$