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Solve the equation,

$x^3+3x+2i=0$

Where, $i=\sqrt {-1}$

$x=-i$ satisfies the given equation,

$\therefore x^2(x+i)-xi(x+i)+2(x+i)=0$

$\Rightarrow (x+i)(x^2-xi+2)=0$

Now consider, $x^2-xi+2=0$

$x=-i$ again satisfies the above equation,

$\therefore x(x+i)-2i(x+i)=0$

$\Rightarrow (x+i)(x-2i)=0$

$x=2i$ is another solution,

So, $x=-i$, $x=-i$, $x=2i$ are the three solutions.

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The range of the function $f(x) = {\log _{\sqrt 5 }}\left[ {3 + \cos \left( {\frac{{3\pi }}{4} + x} \right) + \cos \left( {\frac{\pi }{4} + x} \right) + \cos \left( {\frac{\pi }{4} - x} \right) - \cos \left( {\frac{{3\pi }}{4} - x} \right)} \right]$ is: (A) $[ - 2,2]$ (B) $\left[ {\frac{1}{{\sqrt 5 }},\sqrt 5 } \right]$ (C) $(0,\sqrt 5 )$ (D) $[ 0,2]$ Solution We have, $f(x) = {\log _{\sqrt 5 }}\left( {3 - 2\sin \frac{{3\pi }}{4}\sin x + 2\cos \frac{\pi }{4}\cos x} \right)$ $ \Rightarrow f(x) = {\log _{\sqrt 5 }}\left[ {3 + \sqrt 2 (\cos x - \sin x)} \right]$ Now, $ - \sqrt 2  \le \cos x - \sin x \le \sqrt 2 $ $\therefore - 2 \le \sqrt 2 (\cos x - \sin x) \le 2$ $\therefore 1 \le 3 + \sqrt 2 (\cos x - \sin x) \le 5$ $\therefore{\log _{\sqrt 5 }}1 \le {\log _{\sqrt 5 }}[3 + \sqrt 2 (\cos x - \sin x)] \le {\log _{\sqrt 5 }}5$ $ \Rightarrow 0 \le f(x) \le 2$ Answer: (D)
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