Question

Let $x$ be a real number satisfying both the equations ${\tan }^{-1}(x-1)+{\tan }^{-1}x+{\tan }^{-1}(x+1)={\tan }^{-1}3x$ and $x^3+b{x}^2+cx+d=0$. Then which of the following is/are correct:

• A. $1+4c=b+d$
• B. $b+c+d<0$
• C. $1>b=d>c$
• D. $b^2+c^2+d^2=4$

Answer 1

Answer: (A), (B), (C)

$1>b=d>c$

We have,
${\tan }^{-1}(x-1)+{\tan }^{-1}(x+1)={\tan }^{-1}3x-{\tan }^{-1}x$
Applying tangent on both sides,
$\Rightarrow \frac{x-1+x+1}{1-(x^2-1)}=\frac{3x-x}{1+3x^2}\Rightarrow \frac{2x}{2-x^2}=\frac{2x}{1+3x^2}\Rightarrow 2x(4x^2-1)=0\therefore x=0,\pm \frac{1}{2}$
Checking all the values of $x$ in the given equation.
All values satisfies it.
Comparing $x^3+b{x}^2+cx+d=0$ with $8x^3-2x=0$, we get
$b=d=0c=-\frac{1}{4}$
Checking the given options, we get
$1+4c=0=b+db+c+d=\frac{-1}{4}1>b=d>c{b}^2+c^2+d^2\ne 4$