Question

If $A_n$ be the area bounded by the curve $y={\left(\tan x\right)}^n$ and the lines $x=0,y=0$ and $x=\frac{\pi }{4}$, then for $n>2$

• A. $A_n+A_{n-2}=\frac{1}{n-1}$
• B. $A_n+A_{n22}=\frac{1}{n+1}$
• C. $\frac{1}{2(n+1)}<A_n<\frac{1}{2n-2}$
• D. $\frac{1}{2n-1}<A_n<\frac{1}{2n}$

Answer 1

Answer: (C)

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

We have,

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

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

Let $\tan x=t,$ that ${\sec }^{2}xdx=dt$

$\therefore A_n+A_{n-2}={\int }_0^1{t}^{n-2}dt={\left(\frac{t^{n-1}}{n-1}\right)}_0^1=\frac{1}{n-1}$

Now,since $0\le x\le \frac{\pi }{4}$

$\therefore 0\le \tan x\le 1.$

$\Rightarrow {\tan }^{n+2}x<{\tan }^{n}x<{\tan }^{n-2}x$

$\Rightarrow {\int }_0^{\frac{\pi }{4}}{\tan }^{n+2}xdx<{\int }_0^{\frac{\pi }{4}}{\tan }^{n}xdx<{\int }_0^{\frac{\pi }{4}}{\tan }^{n-2}xdx$

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

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

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

or, $\frac{1}{2(n+1)}<A_n<\frac{1}{2(n-1)}$