Question

If $x_1,x_2,x_3$ are positive roots of $x^3-6x^2+3px-2p=0(p\in R)$, then the value of ${\sin }^{-1}(\frac{1}{x_1}+\frac{1}{x_2})+{\cos }^{-1}(\frac{1}{x_2}+\frac{1}{x_3})-{\tan }^{-1}(\frac{1}{x_3}+\frac{1}{x_1})$ is equal to

• A. $\frac{\pi }{8}$
• B. $\frac{\pi }{6}$
• C. $\frac{\pi }{4}$
• D. $\pi$

Answer 1

The correct option is C $\frac{\pi }{4}$

$x^3-6x^2+3px-2p=0$

$x_1$

$x_2$

$x_3$

are the zeros

$x_1+x_2+x_3=6$

$x_1{x}_2+x_2{x}_3+x_3{x}_1=3p$ $[\because \frac{x_1+x_2+x_3}{3}\ge (2p{)}^{\frac{1}{3}},2\ge (2p{)}^{\frac{1}{3}}\Rightarrow p\le 4]$

$x_1{x}_2{x}_3=2p$

$\frac{3p}{3}\ge \left(\right(x_1{x}_2{x}_3{)}^2{)}^{1/3}$

$p^3\ge 4p^2$

$p\ge 4$

$\Rightarrow p=4$

$\therefore x_1{x}_2+x_2{x}_3+x_3{x}_1=12$

$x_1{x}_2{x}_3=8$

$x_1+x_2+x_3=6$

${\sin }^{-1}\left(1\right)+{\cos }^{-1}\left(1\right)-{\tan }^{-1}\left(1\right)$

$\frac{\pi }{2}+0-\frac{\pi }{4}=\frac{\pi }{4}$.