Let f:(1,) R be a function defined as f(x)=(x-1)/(x+1).

Question

Let $f:(1,\infty)	o R$ be a function defined as $f(x)=\frac{x-1}{x+1}$. Let ${f}^{i+1}(x)=f({f}^{i}(x)),i=1,2,\ldots ,25$, where ${f}^{1}(x)=f(x)$. If $g(x)+{f}^{26}(x)=0,x\in (1,\infty)$, then the area of the region bounded by the curves $y=g(x),2y=2x-3,y=0$ and $x=4$ is :

Accepted Answer

We first focus on the orbit periodicity of ${f}^{n}(x)$ :

${f}^{1}(x)=f(x)=\frac{x-1}{x+1};x\in (1,\infty)\ldots ..(i)$

${f}^{2}(x)=f{{f}^{1}(x)}=f{f(x)}=f(\frac{x-1}{x+1})=\frac{(\frac{x-1}{x+1})-1}{(\frac{x-1}{x+1})+1}=-\frac{2}{2x}=-\frac{1}{x};x\in (1,\infty)\ldots ..(ii)$

${f}^{3}(x)=f{{f}^{2}(x)}=f(-\frac{1}{x})=\frac{(-\frac{1}{x})-1}{(-\frac{1}{x})+1}=\frac{1+x}{1-x};x\in (1,\infty)\ldots ..(iii)$

${f}^{4}(x)=f{{f}^{3}(x)}=f(\frac{1+x}{1-x})=\frac{(\frac{1+x}{1-x})-1}{(\frac{1+x}{1-x})+1}=\frac{2x}{2}=x;x\in (1,\infty)\ldots ..(iv)$

${f}^{5}(x)=f{{f}^{4}(x)}=f(x)=\frac{x-1}{x+1};x\in (1,\infty)\ldots ..(v)$

From the nature of table which completes the 25 iterations in 6 cycles with orbit periodicity 4 :

${f}^{25}(x)=f(x)={f}^{1}(x)\Rightarrow {f}^{26}(x)={f}^{2}(x)=-\frac{1}{x}\ldots \ldots (vi)$

Equation (vi) exactly identifies the function $g(x)$ defined under SOW :

$g(x)+{f}^{26}(x)=0\Rightarrow g(x)=-{f}^{26}(x)=\frac{1}{x}\ldots \ldots (vii)$

Equation (vii) has identified the function $g(x)$ and we can mathematically plot the bounded area between the curves :

$\begin{gathered} g(x) \quad  \quad  \quad  \ \frac{1}{x} \quad  y=x-(\frac{3}{2}) \quad  X \quad  x=4 \end{gathered}$

$\begin{gathered} g(x) \quad  \quad  \quad  \ 1 \quad  \frac{1}{x} \quad  y=x-(\frac{3}{2}) \quad  \quad  A,K \end{gathered}$

$\Rightarrow \frac{1}{x}=x-(\frac{3}{2})\Rightarrow 2{x}^{2}-3x-2=0\Rightarrow x=(2,-\frac{1}{2})\Rightarrow y=(\frac{1}{2},-2)$

$\begin{gathered} A \quad  K \quad  \quad  \ (2,\frac{1}{2}) \quad  (-\frac{1}{2},-2) \quad  K \quad  \end{gathered}$

$\begin{gathered} g(x) \quad  \quad  \quad  \ 2 \quad  \frac{1}{x} \quad  x=4 \quad  \quad  B \end{gathered}$

$\Rightarrow x=4\Rightarrow y=0.25$

$\begin{gathered} B \ (4,0.25) \end{gathered}$

$\begin{gathered} \quad  \quad  \quad  \ 3 \quad  X \quad  x=4 \quad  \quad  C \end{gathered}$

$\Rightarrow x=4,y=0$

$\begin{gathered} C \ (4,0) \end{gathered}$

$\begin{gathered} \quad  \quad  \quad  \ 4 \quad  y=x-(\frac{3}{2}) \quad  X \quad  \quad  D \end{gathered}$

$\Rightarrow y=0\Rightarrow x=\frac{3}{2}$

$\begin{gathered} D \ (\frac{3}{2},0) \end{gathered}$

g(x)

X-axis

Feasible

Intersection point

Reason

1/x

y = 0

No

None

X-axis is a horizontal asymptote to g(x)

Slant line

Vertical line

Feasible

Intersection point

Reason

y = x - (3/2)

x = 4

No

(4, 5/2)

Intersects much above B and outside the scope of bounded area

Sides

Left boundary

Equation

Top boundary

Equation

Right boundary

Equation

Bottom boundary

Equation

Curves

Slant line

y = x - (3/2)

Rectangular Hyperbola

y = 1/x

Vertical line

x = 4

X-axis

y = 0

Vertices

Top left

Coordinates

Top right

Coordinates

Bottom right

Coordinates

Bottom left

Coordinates

Points

A

A (2, 1/2)

B

B (4, 1/4)

C

C (4, 0)

D

D (3/2, 0)

We can visualize that both $A$ and $D$ lie on the slant line but $A$ is ahead of $D$ along the $X$-axis. From $A$ we can drop a \perp  on the $X$-axis at $M$ to form a right angled $\Delta AMD$. From $A$ and $D$ the remaining area upto the vertical $x=4$ lies under the rectangular hyperbola $g(x)=\frac{1}{x}$ :

$\begin{gathered} =(\frac{1}{2})\cdot (DM)\cdot (AM)+{\int }_{4}^{2}g(x)dx =(\frac{1}{2})\cdot (2-1\cdot 5)\cdot (\frac{1}{2})+{\int }_{4}^{2}(\frac{1}{x})dx \ =\frac{1}{8}+[\ln |x|{]}_{4}^{2} \ =\frac{1}{8}+\ln 2 \end{gathered}$

Solutions · 1