Skip to main content

Visit this link for 1 : 1 LIVE Classes.

$\frac{1}{{a + \frac{1}{{b + \frac{1}{c}}}}} = \frac{{16}}{{115}}$

$a,b,c \in \mathbb{N}$

Select correct option(s).

(A) a, b, c are in A.P.
(B) a, b, c are in G.P.
(C) a, b, c are in H.P.
(D) a+b+c = bc

Solution

We have, $a + \frac{1}{{b + \frac{1}{c}}} = \frac{{115}}{{16}}$

$ \Rightarrow a + \frac{c}{{bc + 1}} = \frac{{115}}{{15 + 1}}$

$ \Rightarrow a + \frac{c}{{bc + 1}} = \frac{{115}}{{5 \times 3 + 1}}$

$ \Rightarrow a + \frac{c}{{bc + 1}} = \frac{{16 \times 7 + 3}}{{5 \times 3 + 1}}$

$ \Rightarrow a + \frac{c}{{bc + 1}} = 7 + \frac{3}{{5 \times 3 + 1}}$

$\therefore \{ a,b,c\}  \equiv \{ 7,5,3\} $

Hence, (A) & (D).

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