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$3x+4y=10$

$x>0,y>0$

$(x^2y^3)_{max}=?$

For positive numbers, A.M. $\geq $ G.M.

So, $\frac{{\frac{{3x}}{2} + \frac{{3x}}{2} + \frac{{4y}}{3} + \frac{{4y}}{3} + \frac{{4y}}{3}}}{5} \ge {\left[ {{{\left( {\frac{{3x}}{2}} \right)}^2}{{\left( {\frac{{4y}}{3}} \right)}^3}} \right]^{1/5}}$

$ \Rightarrow \frac{{3x + 4y}}{5} \ge {\left( {\frac{9}{4}{x^2} \times \frac{{4 \times 16}}{{9 \times 3}}{y^3}} \right)^{1/5}}$

$ \Rightarrow \frac{{10}}{5} \ge {\left( {\frac{{16}}{3}{x^2}{y^3}} \right)^{1/5}}$

$ \Rightarrow {2^5} \ge \frac{{16}}{3}{x^2}{y^3}$

$ \Rightarrow {x^2}{y^3} \le 6$

$ \Rightarrow {\left( {{x^2}{y^3}} \right)_{\max }} = 6$

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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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