Question

Let $A$ be the area bounded by the curve $y={\left(\tan x\right)}^n$ for $n>2$ and the lines $x=0$ and $y=0$ and $x=\frac{\pi }{4}$ then the range of $A_{10}$ 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

The correct option is D $(\frac{1}{22},\frac{1}{18})$

$y=(\tan x{)}^n$

$\Rightarrow 0<\tan x<1$

$\Rightarrow 0<x<\frac{\pi }{4}$

$\Rightarrow 0<(\tan x{)}^{n+1}<(\tan x{)}^n$

$\Rightarrow A_{n+1}<A_n$

Now, for $n=2$

$A_n+A_{n+2}={\int }_0^{\frac{\pi }{4}}\left(\right(\tan x{)}^n+\left(\tan x{)}^{n+2}\right)dx$

$A_n+A_{n+2}={\int }_0^{\frac{\pi }{4}}\left(\tan x{)}^n\right(1+{\tan }^{2}x)dx$

$A_n+A_{n+2}={\int }_0^{\frac{\pi }{4}}(\tan x{)}^n{\sec }^{2}xdx$

$A_n+A_{n+2}=[\frac{(\tan x{)}^{n+1}}{n+1}{]}_0^{\frac{\pi }{4}}$

$A_n+A_{n+2}=\frac{1}{n+1}$

Now,

$\Rightarrow A_{n+2}<A_{n+1}<A_n$

$\Rightarrow A_{n+2}+A_n<2A_n$

$\Rightarrow \frac{1}{n+1}<2A_n$

$\Rightarrow A_n>\frac{1}{2n+2}$

Also, by symmetry, we can write

$\Rightarrow A_n+A_n<A_n+A_{n-2}$

$\Rightarrow 2A_n<\frac{1}{n-1}$

$\Rightarrow A_n<\frac{1}{2n-2}$

Therefore, $\Rightarrow \frac{1}{2n+2}<A_n<\frac{1}{2n-2}$

Substitute $n=10$

$\Rightarrow \frac{1}{22}<A_n<\frac{1}{18}$