Question

If $x_1,x_2,x_3$ are the roots of $x^3-6x^2+11x-6=0$ then ${\cot }^{-1}\left(x_1\right)+{\cot }^{-1}\left(x_2\right)+{\cot }^{-1}\left(x_3\right)$ is equal to

• A. $0$
• B. $\frac{\pi }{2}$
• C. $\pi$
• D. $\frac{3\pi }{2}$

Answer 1

Answer: (B)

$\frac{\pi }{2}$

$x^3-6x^2+11x-6=0$

$(x-1)(x^2-5x+6)=0$

$(x-1)(x-2)(x-3)=0$

$\Rightarrow x=1,2,3$

$\Rightarrow x_1=1,x_2=2,x_3=3$

${\cot }^{-1}\left(x_1\right)+{\cot }^{-1}\left(x_2\right)+{\cot }^{-1}\left(x_3\right)$

$={\cot }^{-1}\left(1\right)+{\cot }^{-1}\left(2\right)+{\cot }^{-1}\left(3\right)$

$={\tan }^{-1}\left(1\right)+{\tan }^{-1}\left(1/2\right)+{\tan }^{-1}\left(1/3\right)$

$={\tan }^{-1}\left(\frac{1+\frac{1}{2}+\frac{1}{3}-1.\frac{1}{2}.\frac{1}{3}}{1-\frac{1}{2}-\frac{1}{3}-\frac{1}{6}}\right)$

$={\tan }^{-1}\left(\infty \right)=\pi /2$