Question

Given, $f\left(x\right)=-\frac{x^3}{3}+x^2\sin 1.5a-x\sin a\cdot \sin 2a-5arc\sin (a^2-8a+17)$, then

• A. $f\left(x\right)$ is not defined at $x=sin8$
• B. $f^{\prime }\left(sin8\right)>0$
• C. $f^{\prime }\left(x\right)$ is not defined at $x=sin8$
• D. $f^{\prime }\left(sin8\right)<0$

Answer 1

Answer: (D)

$f^{\prime }\left(sin8\right)<0$

$f\left(x\right)=-\frac{x^3}{3}+x^2\sin 1.5a-x\sin a\cdot \sin 2a-5arc\sin (a^2-8a+17)$
$f\left(x\right)=\frac{-x^3}{3}+x^2\sin 6-x\sin 4\sin 8-5{\sin }^{-1}({(a-4)}^2+1)$
$f^{\prime }\left(x\right)=-x^2+2x\sin 6-\sin 4.\sin 8$ .......( $a=4$ considering the domain of inverse sine function)
$f^{\prime }\left(\sin 8\right)=-{\sin }^{2}8+2\sin 6\sin 8-\sin 4\sin 8$
$=\sin 8(-\sin 8+2\sin 6-\sin 4)$
$=-\sin 8(\sin 8+\sin 4-2\sin 6)$
$=-\sin 8\left(2\sin 6\cos 2-2\sin 6\right)$
$=2\sin 8\sin 6(1-\cos 2)$
Now,
$sin8>0$ (Since,$2\pi <8<3\pi$ )
$sin6<0$ (Since, $\pi <6<2\pi$)
$(1-\cos 2)>0$
Hence, $f^{\prime }\left(\sin 8\right)<0$