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$a = 1 + {2^{1/3}} + {4^{1/3}}$, $\frac{1}{{{a^3}}} + \frac{3}{a} + \frac{3}{{{a^2}}} = ?$

 We have, $\frac{1}{{{a^3}}} + \frac{3}{a} + \frac{3}{{{a^2}}} + {1^3} = {\left( {\frac{1}{a} + 1} \right)^3}$

So, $\frac{1}{{{a^3}}} + \frac{3}{a} + \frac{3}{{{a^2}}} = {\left( {\frac{1}{a} + 1} \right)^3} - 1$

Now, $a = 1 + {2^{1/3}} + {4^{1/3}} = 1 + {2^{1/3}} + {({2^{1/3}})^2}$

Also, $({2^{1/3}} - 1)\left[ {{{({2^{1/3}})}^2} + {2^{1/3}} \times 1 + {1^2}} \right] = \left[ {{{({2^{1/3}})}^3} - {1^3}} \right] = 1$

So, $a = 1 + {2^{1/3}} + {4^{1/3}} = \frac{1}{{{2^{1/3}} - 1}}$

Thus, $\frac{1}{a} = {2^{1/3}} - 1$

Or, $\frac{1}{a} + 1 = {2^{1/3}}$

Or, ${\left( {\frac{1}{a} + 1} \right)^3} = 2$

Thus, $\frac{1}{{{a^3}}} + \frac{3}{a} + \frac{3}{{{a^2}}} = {\left( {\frac{1}{a} + 1} \right)^3} - 1 = 2 - 1 = 1$

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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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