Question

Let $A_n$ be the area bounded by the curve $y={\left(\tan x\right)}^n$ and the lines $x=0,y=0$ and $x=\pi /4$ then

• A. $\frac{1}{(2n+2)}<A_n<\frac{1}{(2n-2)}$
• B. $\frac{A_n}{A_{n-2}}=\frac{1}{4n+1}$
• C. $\frac{1}{(2n-4)}<A_n<\frac{1}{(2n+4)}$
• D. $\frac{A_n}{A_{n-2}}=\frac{4n}{4n+1}$

Answer 1

The correct option is A $\frac{1}{(2n+2)}<A_n<\frac{1}{(2n-2)}$

Given curve, $y={\left(\tan x\right)}^n$

Given lines $x=0,y=0$ and $x=\frac{\pi }{4}$

At $x=\frac{\pi }{4},y={\left(\tan \left(\frac{\pi }{4}\right)\right)}^n$

$=1$

Area bounded by given curve and lines is shaded region$=A_n={\int }_0^{\pi /4}{\left(\tan x\right)}^{n}dx$

As $0<x<\frac{\pi }{4},{\left(\tan x\right)}^n>{\left(\tan x\right)}^{n+1}(\because 0<\tan x<1)$

$\therefore {\int }_0^{\pi /4}{\left(\tan x\right)}^{n}dx>{\int }_0^{\pi /4}{\left(\tan x\right)}^{n+1}dx$

$\therefore A_n>A_{n+1}$

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

$={\int }_0^{\pi /4}{\left(\tan x\right)}^n[1+{\tan }^{2}x]dx$

$={\int }_0^{\pi /4}{\left(\tan x\right)}^n\left({\sec }^{2}x\right)dx$$={\int }_0^{\pi /4}{\left(\tan x\right)}^{n}d\left(\tan x\right)\because d\left(\tan x\right)={\sec }^{2}xdx$

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

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

As $A_n>A_{n+2}$

$\therefore 2A_n>A_{n+2}+A_n$

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

$\therefore A_n>\frac{1}{2n+2}\to 1$

Similarly,

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

$={\int }_0^{\pi /4}{\left(\tan x\right)}^{n-2}d\left(\tan x\right)$

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

$=\frac{1}{n-1}$

$A_n<A_{n-2}$

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

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

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

From $1$ and $2$,

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