Question

The roots of the equation $6{\sin }^{-1}\left(x^3-6x^2+8x+\frac{1}{2}\right)=\pi ,$ are

• A. in AP
• B. in GP
• C. in AP and GP both
• D. neither in AP nor in GP

Answer 1

Answer: (A)

Given equations is $6{\sin }^{-1}\left(x^3-6x^2+8x+\frac{1}{2}\right)=\pi$
$\Rightarrow$ ${\sin }^{-1}\left(x^3-6x^2+8x+\frac{1}{2}\right)=\frac{\pi }{6}$
$\Rightarrow$ $x^3-6x^2+8x+\frac{1}{2}=\sin \frac{\pi }{6}$
$\Rightarrow$ $x^3-6x^2+8x+\frac{1}{2}=\frac{1}{2}$
$\Rightarrow$ $x(x^2-6x+8)=0$
$\Rightarrow$ $x(x-2)(x-4)=0$
So, roots are $x=0,2,4$
Hence, these roots are in AP.