Question

The composite function $f_1\left(f_2\right(f_3(\dots ..(f_n\left(x\right))\dots .)$ is a decreasing function and 'r' functions out of total 'n' functions are decreasing functions while the rest are increasing. The maximum value of $r(n-r)$ is

• A. $\frac{n^2-4}{4}$ when n is of the form $4k$
• B. $\frac{n^2}{4}$ when n is an even number
• C. $\frac{n^2-1}{4}$ when n is an odd number
• D. $\frac{n^2}{4}$ when n is of the form $4k+2$

Answer 1

Answer: (A), (C), (D)

The correct options are
A $\frac{n^2-4}{4}$ when n is of the form $4k$
C $\frac{n^2-1}{4}$ when n is an odd number
D $\frac{n^2}{4}$ when n is of the form $4k+2$

Let $g\left(x\right)=f_1\left(f_2\right(f_3(\dots ..(f_n\left(x\right))\dots )$
$g\left(x\right)$ is a decreasing function.
Hence its derivative will be negative.
Now,
$g^{\prime }\left(x\right)=f_1^{\prime }\left(f_2\right(f_3(...f_n(x\left)\right)...)\times f_2^{\prime }(f_3(...f_n(x\left)\right)...)\times ...f_{n-1}^{\prime }(f\left(x\right))\times f_n^{\prime }(x)$
Since $g^{\prime }\left(x\right)<0$ there must be odd number of negative terms out of the $n$ terms in g'(x).
It is given that $r$ functions are decreasing. Hence, $r$ must be odd.
Now, we wish to maximise $r(n-r)$
Let $p\left(r\right)=rn-r^2$, for $0<r\le n$
$\Rightarrow p^{\prime }\left(r\right)=n-2r$
$\Rightarrow p^{\prime \prime }\left(r\right)=-2$
Hence maximum value of $p\left(r\right)$ occurs at $r=\frac{n}{2}$
In our case $r$ is an odd integer. Hence, let us consider different cases.
Case I - $n$ is an even integer.
For $p\left(r\right)$ to be maximum , $r=\frac{n}{2}$
For $r$ to be an odd integer, we will have to consider sub cases.

Case I - (a) $n=4m+2$ , where $m$ is an integer
In this case, $r=\frac{n}{2}$ will be an odd integer.
Hence, $p\left(r\right)=\frac{n^2}{4}$.
Case I - (b) $n=4m$ , where $m$ is an integer
In this case, $\frac{n}{2}$ will not be an odd integer.
Hence let us consider the integer before $\frac{n}{2}$ and after it.
i.e. $r=\frac{n}{2}-1$ and $r=\frac{n}{2}+1$
In both the cases, maximum value of $p\left(r\right)=\frac{n^2-4}{4}$.
Case II - $n$ is an odd number
In this case let us consider the integers between which $\frac{n}{2}$ lies.
i.e. $\frac{n-1}{2}$ and $\frac{n+1}{2}$
In both the cases, the maximum value of $p\left(r\right)=\frac{n^2-1}{4}$