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

$\frac{{(x + 2)(x + 3)(x + 4)(x + 5)}}{{(x - 2)(x - 3)(x - 4)(x - 5)}} = 1$

We have, $\frac{{({x^2} + 5x + 6)({x^2} + 9x + 20)}}{{({x^2} - 5x + 6)({x^2} - 9x + 20)}} = 1$

$ \Rightarrow \frac{{{x^2} + 9x + 20}}{{{x^2} - 5x + 6}} = \frac{{{x^2} - 9x + 20}}{{{x^2} + 5x + 6}} = \frac{{({x^2} + 9x + 20) - ({x^2} - 9x + 20)}}{{({x^2} - 5x + 6) - ({x^2} + 5x + 6)}} = \frac{{18x}}{{ - 10x}} =  - \frac{{9}}{{5}}$
(assuming $x \ne 0$)

$ \Rightarrow \frac{{{x^2} + 9x + 20}}{{{x^2} - 5x + 6}} - 1 =  - \frac{9}{5} - 1$

$ \Rightarrow \frac{{14x + 14}}{{{x^2} - 5x + 6}} =  - \frac{{14}}{5}$

$ \Rightarrow \frac{{x + 1}}{{{x^2} - 5x + 6}} =  - \frac{1}{5}$

$ \Rightarrow {x^2} - 5x + 6 + 5x + 5 = 0$ Or ${x^2} + 11 = 0$

So, no real solution if $x \neq 0$.

If x = 0, LHS of the original equation = $\frac{{2 \times 3 \times 4 \times 5}}{{( - 2) \times ( - 3) \times ( - 4) \times ( - 5)}}$ = 1 = RHS

Thus, x = 0 is the only real solution.

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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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