Skip to main content

Visit this link for 1 : 1 LIVE Classes.

Prove that,

$\sqrt {28 + 10\sqrt 3 } - \sqrt {4 + 2\sqrt 3 } $

$> \sqrt 3 + \sqrt 2 $ or $> \pi$

LHS $ = \sqrt {25 + 3 + 2 \times 5 \times \sqrt 3 }  - \sqrt {3 + 1 + 2 \times \sqrt 3  \times 1} $

$ = \sqrt {{5^2} + {{\sqrt 3 }^2} + 2 \times 5 \times \sqrt 3 }  - \sqrt {{{\sqrt 3 }^2} + {1^2} + 2 \times \sqrt 3  \times 1} $

$ = \sqrt {{{(5 + \sqrt 3 )}^2}}  - \sqrt {{{(\sqrt 3  + 1)}^2}} $

$ = (5 + \sqrt 3 ) - (\sqrt 3  + 1) = 4$

RHS $ = \sqrt 3  + \sqrt 2  \approx 3.14  \approx \pi$

Clearly, $4 > 3.14$

Popular posts from this blog

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

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)
Read more