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$3^a=5^b=225$, $\frac {ab}{a+b}=?$

While trying to find $\frac {ab}{a+b}$ given $3^a=5^b=225$, a student follows the method below:

$3^a=5^b=225=9\times 25=3^2\times 5^2$

So, $3^a\times 5^b=(3^2\times 5^2)\times (3^2\times 5^2)$

Or, $3^a\times 5^b=3^4\times 5^4$

Since 3 & 5 are prime numbers, a=4 & b=4

Now, $\frac {ab}{a+b}=\frac {4\times 4}{4+4}=2$

Considering above solution, select correct options. More that 1 option may be correct.

(A) a = 4 is correct
(B) b = 4 is correct
(C) $\frac {ab}{a+b}=2$ is correct
(D) $\frac {1}{a}+\frac {1}{b}=\frac {1}{2}$ is correct

Key: (C) & (D)

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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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