Question

In a triangle $ABC$, $D,E,F$ are taken on the sides $BC,CA,AB$ respectively. If $\frac{BD}{DC}=\frac{CE}{EA}=\frac{AF}{FB}=n$ and $\text{Area}\left(△DEF\right)=f\left(n\right)\times \text{Area}\left(△ABC\right)$, then the correct statements is/are

• A. $f\left(n\right)$ is monotonic function.
• B. $f\left(n\right)$ is increasing in $[1,\infty )$
• C. $\underset{0}{\overset{1}{\int }}f\left(n\right)dn=\frac{5}{2}-3\ln 2$
• D. $\underset{0}{\overset{1}{\int }}f\left(n\right)dn=-\frac{1}{2}+3\ln 2$

Answer 1

Answer: (B), (C)

The correct options are
B $f\left(n\right)$ is increasing in $[1,\infty )$
C $\underset{0}{\overset{1}{\int }}f\left(n\right)dn=\frac{5}{2}-3\ln 2$

Assuming $A$ to be origin and the position vectors of $B$ and $C$ be $\overrightarrow{b}$ and $\overrightarrow{c}$ respectively.

Position vectors of
$D:\frac{n\overrightarrow{c}+\overrightarrow{b}}{n+1}E:\frac{\overrightarrow{c}}{n+1}F:\frac{n\overrightarrow{b}}{n+1}$
Now,
$\overrightarrow{FD}=\overrightarrow{AD}-\overrightarrow{AF}=\frac{n\overrightarrow{c}+(1-n)\overrightarrow{b}}{n+1}\overrightarrow{FE}=\overrightarrow{AE}-\overrightarrow{AF}=\frac{\overrightarrow{c}-n\overrightarrow{b}}{n+1}$
We know that,
$\text{Area}\left(△ABC\right)=\frac{1}{2}|\overrightarrow{b}\times \overrightarrow{c}|$
So,
$\text{Area}\left(△DEF\right)=\frac{1}{2}|\overrightarrow{FD}\times \overrightarrow{FE}|=\frac{1}{2(n+1{)}^2}\left|\right(n\overrightarrow{c}+(1-n)\overrightarrow{b})\times (\overrightarrow{c}-n\overrightarrow{b}\left)\right|=\frac{1}{2(n+1{)}^2}\left|n^2\right(\overrightarrow{b}\times \overrightarrow{c})+(n-1)\overrightarrow{c}\times \overrightarrow{b}|=\frac{n^2-n+1}{(n+1{)}^2}\times \frac{1}{2}\left|\right(\overrightarrow{b}\times \overrightarrow{c}\left)\right|$
Therefore,
$f\left(n\right)=\frac{n^2-n+1}{(n+1{)}^2}\Rightarrow f\left(n\right)=1-\frac{3n}{(n+1{)}^2}\Rightarrow f^{\prime }\left(n\right)=-3\left[\frac{(n+1{)}^2-2n(n+1)}{(n+1{)}^4}\right]\Rightarrow f^{\prime }\left(n\right)=\frac{3(n-1)}{(n+1{)}^3}$
So, $f\left(n\right)$ is increasing when $n\ge 1$
Now let
$I=\underset{0}{\overset{1}{\int }}f\left(n\right)dn\Rightarrow I=\underset{0}{\overset{1}{\int }}1-\frac{3n}{(n+1{)}^2}dn\Rightarrow I=1-3\underset{0}{\overset{1}{\int }}\left[\frac{n}{(n+1{)}^2}\right]dn\Rightarrow I=1-3\underset{0}{\overset{1}{\int }}[\frac{1}{(n+1)}-\frac{1}{(n+1{)}^2}]dn\Rightarrow I=1-3{\left[\ln \right(n+1)+\frac{1}{(n+1)}]}_0^1\Rightarrow I=1-3\ln 2+\frac{3}{2}\Rightarrow \underset{0}{\overset{1}{\int }}f\left(n\right)dn=\frac{5}{2}-3\ln 2$