Question

Let $T_n$ be the area bounded by $y={\tan }^{n}x,x=0,y=0$ and $x=\frac{\pi }{4}$ where $n$ is a integer greater than $2$, then $T_{100}$ is

• A. $\frac{1}{200}<T_{100}<\frac{1}{196}$
• B. $\frac{1}{206}<T_{100}<\frac{1}{204}$
• C. $\frac{1}{204}<T_{100}<\frac{1}{202}$
• D. $\frac{1}{202}<T_{100}<\frac{1}{198}$

Answer 1

The correct option is D $\frac{1}{202}<T_{100}<\frac{1}{198}$

We know that,
$T_n=\underset{0}{\overset{\pi /4}{\int }}{\tan }^{n}xdx{T}_{n-2}=\underset{0}{\overset{\pi /4}{\int }}{\tan }^{n-2}xdx$
Therefore,
$T_n+T_{n-2}=\underset{0}{\overset{\pi /4}{\int }}{\tan }^{n-2}x({\tan }^{2}x+1)dx\Rightarrow T_n+T_{n-2}=\underset{0}{\overset{\pi /4}{\int }}{\tan }^{n-2}x\left({\sec }^{2}x\right)dx\Rightarrow T_n+T_{n-2}={\left[\frac{{\tan }^{n-1}x}{n-1}\right]}_0^{\pi /4}\Rightarrow T_n+T_{n-2}=\frac{1}{n-1}$
We know that for
$0\le x\le \pi /40\le \tan x\le 1{\tan }^{n+2}x<{\tan }^{n}x<{\tan }^{n-2}x\Rightarrow \underset{0}{\overset{\pi /4}{\int }}{\tan }^{n+2}xdx<\underset{0}{\overset{\pi /4}{\int }}{\tan }^{n}xdx<\underset{0}{\overset{\pi /4}{\int }}{\tan }^{n-2}xdx\Rightarrow T_{n+2}<T_n<T_{n-2}\Rightarrow T_n+T_{n+2}<2T_n<T_n+T_{n-2}\Rightarrow \frac{1}{2(n+1)}<T_n<\frac{1}{2(n-1)}\Rightarrow \frac{1}{202}<T_{100}<\frac{1}{198}$