A) in AP
B) in GP
C) in AP and GP both
D) neither in AP nor in
Correct Answer: A
Solution :
Given equations is \[6{{\sin }^{-1}}\left( {{x}^{3}}-6{{x}^{2}}+8x+\frac{1}{2} \right)=\pi \] \[\Rightarrow \] \[{{\sin }^{-1}}\left( {{x}^{3}}-6{{x}^{2}}+8x+\frac{1}{2} \right)=\frac{\pi }{6}\] \[\Rightarrow \] \[{{x}^{3}}-6{{x}^{2}}+8x+\frac{1}{2}=\sin \frac{\pi }{6}\] \[\Rightarrow \] \[{{x}^{3}}-6{{x}^{2}}+8x+\frac{1}{2}=\frac{1}{2}\] \[\Rightarrow \] \[x({{x}^{2}}-6x+8)=0\] \[\Rightarrow \] \[x(x-2)(x-4)=0\] So, roots are \[x=0,2,4\] Hence, these roots are in AP.
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