Question

Let $A_n$ be the area bounded by the curve $y=(\tan x{)}^n$ and the lines $x=0,y=0$ and $x=\frac{\pi }{4}$ then the range of $A_{1}0$ is

• A. $(\frac{1}{20},\frac{1}{16})$
• B. $(\frac{1}{18},\frac{1}{16})$
• C. $(\frac{1}{14},\frac{1}{10})$
• D. $(\frac{1}{22},\frac{1}{18})$

Answer 1

Answer: (A), (B)

The correct options are
A $(\frac{1}{20},\frac{1}{16})$
B $(\frac{1}{18},\frac{1}{16})$

The region bounded by the curve $y=(\tan x{)}^n$ and the lines $x=0,y=0$ and $x=\frac{\pi }{4}$ is given by
$Area=\stackrel{\pi /4}{\underset{0}{\int }}(\tan x{)}^{n}dx=A_n$
$\therefore A_n+A_{n-2}=\stackrel{\pi /4}{\underset{0}{\int }}(\tan x{)}^{n}dx+\stackrel{\pi /4}{\underset{0}{\int }}(\tan x{)}^{n-2}dx$
$=\stackrel{\pi /4}{\underset{0}{\int }}\left(\tan x{)}^{n-2}\right[1+\left({\tan }^{2}x\right)]dx$

$=\stackrel{\pi /4}{\underset{0}{\int }}\left(\tan x{)}^{n-2}\right[{\sec }^{2}x]dx$
$=\stackrel{\pi /4}{\underset{0}{\int }}(\tan x{)}^{n-2}d\tan x$

$={\left[\frac{(\tan x{)}^{n-1}}{n-1}\right]}_0^{\pi /4}$
$=\frac{1}{n-1}$