MCQ
Given $f(x) = -\frac{{{x^3}}}{3} + x^2 sin 1.5 a - x sin a . sin 2a - 5 \,\,sin^{-1} (a^2 - 8a + 17)$ then :
- A$f(x)$ is not defined at $x = sin 8$
- B$f ‘ (sin 8) > 0$
- C$f ‘ (x)$ is not defined at $x = sin 8$
- ✓$f ‘ (sin 8) < 0$
$f'(x) = - {x^2} + 2x\,sin\,6 - sin\,4\,sin\,8$ $ (a = 4)$
$f'(sin\,8) = - si{n^2}8 + 2\,sin\,6\,sin\,8 - sin\,4\,sin\,8$
$ = {\rm{ }}sin\,8\,\,\left[ {{\rm{ }} - {\rm{ }}sin\,8{\rm{ }} + {\rm{ }}2{\rm{ }}sin\,6{\rm{ }} - {\rm{ }}sin\,4} \right]$
$\; = {\rm{ }} - {\rm{ }}sin\,8{\rm{ }}\,\left[ {sin\,8{\rm{ }} + {\rm{ }}sin\,4{\rm{ }} - {\rm{ }}2\,sin\,6} \right]{\rm{ }} = {\rm{ }} - {\rm{ }}sin\,8{\rm{ }}\left[ {2sin\,6{\rm{ }}cos\,2{\rm{ }} - {\rm{ }}2\,sin\,6} \right]$
$ = {\rm{ }}2{\rm{ }}sin\,8{\rm{ }}sin\,6{\rm{ }}\left[ {{\rm{ }}1{\rm{ }} - {\rm{ }}cos\,2} \right]\;$
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$a+b+c $$ =x $ ; $a+b \omega+c \omega^2 $$ =y $ ; $a+b \omega^2+c \omega $$ =z .$
Then the value of $\frac{|x|^2+|y|^2+|z|^2}{|a|^2+|b|^2+|c|^2}$ is