Question

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

• A. 24
• B. 25
• C. 26
• D. 27

Answer 1

The correct option is C 26

$co{s}^{-1}\left(x\right)+co{s}^{-1}\left(2x\right)+co{s}^{-1}\left(3x\right)=\pi$
$co{s}^{-1}\left(x\right)+co{s}^{-1}\left(2x\right)=\pi -co{s}^{-1}\left(3x\right)$
$co{s}^{-1}(2x^2-\sqrt{(1-x^2)(1-4x^2)})=\pi -co{s}^{-1}\left(3x\right)$
$2x^2-\sqrt{(1-x^2)(1-4x^2)}=cos(\pi -co{s}^{-1}(3x\left)\right)$
$2x^2-\sqrt{(1-x^2)(1-4x^2)}=-3x$
$2x^2+3x=\sqrt{(1-x^2)(1-4x^2)}$
Squaring on both the sides we get
$4x^4+9x^2+12x^3=(1-x^2)(1-4x^2)$
$4x^4+9x^2+12x^3=1-5x^2+4x^4$
$14x^2+12x^3=1$
$12x^3+14x^2-1=0$
$a{x}^3+b{x}^2+cx-1=0$
Comparing coefficients, we get
$a=12$
$b=14$
$c=0$
Hence $a+b+c=12+14+0$
$=26$