Question

Let ${\cos }^{-1}\left(x\right)+{\cos }^{-1}\left(2x\right)+{\cos }^{-1}\left(3x\right)=\pi ,$ where $x>0.$ 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. $25$
• C. $26$
• D. $28$

Answer 1

The correct option is C $26$

We have,
${\cos }^{-1}\left(x\right)+{\cos }^{-1}\left(2x\right)+{\cos }^{-1}\left(3x\right)=\pi$
$\Rightarrow {\cos }^{-1}\left(2x\right)+{\cos }^{-1}\left(3x\right)=\pi -{\cos }^{-1}\left(x\right)$
$\Rightarrow {\cos }^{-1}\left[\right(2x\left)\right(3x)-\sqrt{1-4x^2}\sqrt{1-9x^2}]={\cos }^{-1}(-x)$
$\Rightarrow 6x^2-\sqrt{1-4x^2}\cdot \sqrt{1-9x^2}=-x$
$\Rightarrow (6x^2+x{)}^2=(1-4x^2\left)\right(1-9x^2)$
$\Rightarrow x^2+12x^3=1-13x^2$
$\Rightarrow 12x^3+14x^2-1=0\cdots \left(1\right)$
$x$ satisfies the equation $a{x}^3+b{x}^2+cx-1=0$
Comparing this equation with equation $\left(1\right),$ we get
$a=12,b=14,c=0$
$\therefore a+b+c=12+14+0=26$