Question

Find number of values of x which satisfying $\left[{\tan }^{-1}x\right]+\left[{\cot }^{-1}x\right]=2$, where [.] represents the greatest integer function.

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

The correct option is A 0

$0<{\cot }^{-1}x<\pi$ and $-\pi /2<{\tan }^{-1}x<\pi$

$\Rightarrow$ $\left[{\cot }^{-1}x\right]\in \{0,1,2,3\}$ and $\left[{\tan }^{-1}x\right]\in \{-2,-1,0,1\}$

For $\left[{\tan }^{-1}x\right]+\left[{\cot }^{-1}x\right]=2$, following cases are possible

Case (i): $\left[{\cot }^{-1}x\right]=\left[{\tan }^{-1}x\right]=1$

$\Rightarrow$ $1\le {\cot }^{-1}x<2$ and $1\le {\tan }^{-1}x<\pi /2$

$\Rightarrow$ $x\in (\cot 2,\cot 1]$ and $x\in [\tan 1,\infty )$

$\therefore$ $x\in \varphi$ as $\cot 1<\tan 1$

Case (ii): $\left[{\cot }^{-1}x\right]=2$, $\left[{\tan }^{-1}x\right]=0$

$\Rightarrow$ $2\le {\cot }^{-1}x<3$ and $0\le {\tan }^{-1}x<1$

$\Rightarrow$ $x\in (\cot 3,\cot 2]$ and $x\in [0,\tan 1)$

$\therefore$ $x\in \varphi$ as $\cot 2<0$

So no solution.

Case (iii): $\left[{\cot }^{-1}x\right]=3$, $\left[{\tan }^{-1}x\right]=-1$

$\Rightarrow$ $3\le {\cot }^{-1}x<\pi$ and $-1\le {\tan }^{-1}x<0$

$\Rightarrow$ $x\in (-\infty ,\cot 3]$ and $x\in [-\tan 1,0)$

$\therefore$ $x\in \varphi$ as $\cot 3<-\tan 1$

Therefore, no such values of $x$ exist.