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Find the least value of

${a^2}{\sec ^2}\theta + {b^2}{\rm{cose}}{{\rm{c}}^2}\theta $

The given expression can be written as,

$({a^2}{\sec ^2}\theta  - {a^2}) + {a^2} + ({b^2}{\rm{cose}}{{\rm{c}}^2}\theta  - {b^2}) + {b^2}$

$ = {a^2}{\tan ^2}\theta  + {b^2}{\cot ^2}\theta  + {a^2} + {b^2}$

Using A.M. $\geq $ G.M.,

$\frac{{{a^2}{{\tan }^2}\theta  + {b^2}{{\cot }^2}\theta }}{2} \ge \sqrt {{a^2}{{\tan }^2}\theta .{b^2}{{\cot }^2}\theta } $

$ \Rightarrow {a^2}{\tan ^2}\theta  + {b^2}{\cot ^2}\theta  \ge 2|ab|$

$ \Rightarrow {a^2}{\tan ^2}\theta  + {b^2}{\cot ^2}\theta  + {a^2} + {b^2} \ge 2|ab| + {a^2} + {b^2}$

$\therefore{({a^2}{\sec ^2}\theta  + {b^2}{\rm{cose}}{{\rm{c}}^2}\theta )_{\min }} = {a^2} + {b^2} + 2|ab|$

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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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