Question

Let ${\cos }^{-1}\left(x\right)+{\cos }^{-1}\left(2x\right)+{\cos }^{-1}\left(3x\right)=\pi .$ If $x$ satisfies the cubic equation $a{x}^3+b{x}^2+cx-1=0,$ then $a+b+c$ has the value equal to

• A. $24$
• B. $28$
• C. $26$
• D. $13$

Answer 1

The correct option is C $26$

${\cos }^{-1}\left(x\right)+{\cos }^{-1}\left(2x\right)+{\cos }^{-1}\left(3x\right)=\pi .$
Let $y={\cos }^{-1}\left(x\right)+{\cos }^{-1}\left(2x\right)+{\cos }^{-1}\left(3x\right)$
Domain of $y$ is $x\in [-\frac{1}{3},\frac{1}{3}]$
If $x<0$, then $y$ should have been greater than $\frac{3\pi }{2}$ but we have $y=\pi .$
It means $x$ can't be negative.
$\therefore x>0$

And we know that for $a>0,b>0$ in their respective domain
${\cos }^{-1}a+{\cos }^{-1}b={\cos }^{-1}[ab-\sqrt{1-a^2}\sqrt{1-b^2}]$
Now, ${\cos }^{-1}\left(2x\right)+{\cos }^{-1}\left(3x\right)=\pi -{\cos }^{-1}\left(x\right)$
$\Rightarrow {\cos }^{-1}\left[\right(2x\left)\right(3x)-\sqrt{1-4x^2}\sqrt{1-9x^2}]={\cos }^{-1}(-x)$
$\Rightarrow (6x^2+x{)}^2=(1-4x^2\left)\right(1-9x^2)$
$\Rightarrow x^2+12x^3=1-13x^2$
$\Rightarrow 12x^3+14x^2-1=0$
$\therefore a=12;b=14;c=0$
$\Rightarrow a+b+c=26$