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Solve for $x \in \mathbb{R}$

$7x^5 + x + 8 = 0$

-1 satisfies the given equation. Thus,

$7x^4 (x+1) - 7x^3 (x+1) + 7x^2 (x+1) -7x(x+1)+8(x+1)=0$

$\Rightarrow (x+1) (7x^4 - 7x^3 +7x^2 -7x +8) = 0$

Consider, $7x^4 - 7x^3 +7x^2 -7x +8 = 0$

Or, $7(x^4 - x^3 +x^2 -x) +8 = 0$

Let us consider III cases,

I. x < 0

Since $x^3$ and x both terms are negative, the left side is always +ve and cannot be 0. So, no solution is possible in this case.

II. x > 1

$x^4 > x^3$ and $x^2 > x$. So, left side is always +ve and cannot be 0. So, no solution is possible in this case.

III 0 < x < 1

$x^4 - x^3$ is a -ve decimal and so is $x^2 - x$. So, left side will be slightly smaller than 8 but cannot be 0. No solution in this case also.

As can be seen x cannot be 0. x cannot be 1 either.

The only solution possible is -1 for the original equation.

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${\log _{\sqrt 5 }}\left[ {3 + \cos \left( {\frac{{3\pi }}{4} + x} \right) + \cos \left( {\frac{\pi }{4} + x} \right) + \cos \left( {\frac{\pi }{4} - x} \right) - \cos \left( {\frac{{3\pi }}{4} - x} \right)} \right]$

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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