Question

Let $f:[0,1]\to [0,\infty )$ be a differentiable function with decreasing first derivative in its domain and $f\left(0\right)=0$. If $f^{\prime }\left(x\right)>0$ for all $x\in [0,1],$ then

• A. $\underset{0}{\overset{1}{\int }}\frac{dx}{\left(f\right(x){)}^2+1}<\frac{{\tan }^{-1}f\left(1\right)}{f^{\prime }\left(1\right)}$
• B. $\underset{0}{\overset{1}{\int }}\frac{dx}{\left(f\right(x){)}^2+1}<\frac{f\left(1\right)}{f^{\prime }\left(1\right)}$
• C. $\underset{0}{\overset{1}{\int }}\frac{dx}{\left(f\right(x){)}^2+1}>\frac{{\tan }^{-1}f\left(1\right)}{f^{\prime }\left(1\right)}$
• D. $\underset{0}{\overset{1}{\int }}\frac{dx}{\left(f\right(x){)}^2+1}=\frac{f\left(1\right)}{f^{\prime }\left(1\right)}$ for some $x\in [0,1]$

Answer 1

Answer: (A), (B)

$\underset{0}{\overset{1}{\int }}\frac{dx}{\left(f\right(x){)}^2+1}<\frac{f\left(1\right)}{f^{\prime }\left(1\right)}$

$f^{\prime }$ is decreasing in its domain.
$\Rightarrow f^{\prime }\left(x\right)\ge f^{\prime }\left(1\right)\forall x\in [0,1]$
$\therefore \frac{f^{\prime }\left(1\right)}{\left(f\right(x){)}^2+1}\le \frac{f^{\prime }\left(x\right)}{\left(f\right(x){)}^2+1}$
On integrating,
$f^{\prime }\left(1\right)\underset{0}{\overset{1}{\int }}\frac{dx}{\left(f\right(x){)}^2+1}\le {\tan }^{-1}\left(f\right(1\left)\right)-{\tan }^{-1}\left(f\right(0\left)\right)$
$\Rightarrow \underset{0}{\overset{1}{\int }}\frac{dx}{\left(f\right(x){)}^2+1}\le \frac{{\tan }^{-1}\left(f\right(1\left)\right)}{f^{\prime }\left(1\right)}\le \frac{f\left(1\right)}{f^{\prime }\left(1\right)}[\because {\tan }^{-1}x\le x\forall x\ge 0]$

For rightmost equality to hold, ${\tan }^{-1}f\left(1\right)=f\left(1\right)$
$\therefore f\left(1\right)=0$, then $\underset{0}{\overset{1}{\int }}\frac{dx}{\left(f\right(x){)}^2+1}=0$
which is not possible as this is strictly positive function.
Hence, $\underset{0}{\overset{1}{\int }}\frac{dx}{\left(f\right(x){)}^2+1}<\frac{f\left(1\right)}{f^{\prime }\left(1\right)}$