Question

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

• A. $f\left(x\right)$ is not defined at $x=\sin 8$
• B. $f^{\prime }\left(\sin 8\right)>0$
• C. $f^{\prime }\left(x\right)$ is not defined at $x=\sin 8$
• D. $f^{\prime }\left(\sin 8\right)<0$

Answer 1

Answer: (D)

$f^{\prime }\left(\sin 8\right)<0$

Given:

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

$f\left(x\right)=\frac{-x^3}{3}+x^2\sin 1.5a-x\sin a\times \sin 2a-5{\sin }^{-1}(a^2-8a+17)$ $(\sin c+\sin d=2\sin \frac{c+d}{2}\cos \frac{c-d}{2})$

For the function to be defined

$-1\le a^2-8a+17\le 1$

$-1\le (a-4{)}^2+1\le 1$

$-1-1\le (a-4{)}^2\le 1-1$

$-2\le (a-4{)}^2\le 0$

$a-4=0\Rightarrow a=4$

Putting value of '$a$' in the +ve $f\left(x\right)$

$f\left(x\right)=\frac{-x^3}{3}+x^2\sin 6-x\sin 4\times \sin 8-5\frac{\pi }{2}$ $[\because {\sin }^{-1}1=\frac{\pi }{2}]$

domain $x\in R\because$ polynomial function

$f^{\prime }\left(x\right)=-\frac{3x^2}{3}+2x\sin 6-\sin 4\times \sin 8$

$=-x^2+2\sin 6x-\sin 4\times \sin 8$

Putting $x=\sin 8$

$f^{\prime }\left(\sin 8\right)=-{\sin }^{2}8+2\sin 6\sin 8-\sin 4\sin 8$

$=\sin 8[2\sin 6-\sin 4-\sin 8]$

$=\sin 8[2\sin 6-(\sin 4+\sin 8\left)\right]$

$=\sin 8[2\sin 6-2\sin 6\cos 2]$

$=2\sin 6\sin 8[1-\cos 2]$

Value of $\cos 2=0.999$

$\therefore 1-\cos 2=+ve$

$6rad\approx {344}^o$

${344}^o$ is in $I{V}^{th}$ quadrant

$\therefore \sin 6$ is (-ve)

$8rad\approx {548.4}^o$

${548.4}^o$ is in $I{I}^{th}$ quadrant

$\therefore \sin 8$ is (+ve)

$\therefore f^{\prime }\left(\sin 8\right)=2\underset{-ve}{\underset{}{\sin 6}}\underset{+ve}{\underset{}{\sin 8}}\underset{+ve}{\underset{}{(1-\cos 2)}}$

$\therefore f^{\prime }\left(\sin 8\right)$ is +ve

$\therefore f^{\prime }\left(\sin 8\right)<0$ $\to$ option D