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Solve for $x \in \mathbb{R}$

${(6x + 5)^2} = \frac{{35}}{{3{x^2} + 5x + 2}}$

We have, $36{x^2} + 60x + 25 = \frac{{35}}{{3{x^2} + 5x + 2}}$

$ \Rightarrow 12(3{x^2} + 5x) + 25 = \frac{{35}}{{(3{x^2} + 5x) + 2}}$

Let, $3{x^2} + 5x = t$

So, $12t + 25 = \frac{{35}}{{t + 2}}$

$ \Rightarrow 12{t^2} + 49t + 15 = 0$

$ \Rightarrow t = \frac{{ - 49 \pm \sqrt {{{49}^2} - 720} }}{{24}} = \frac{{ - 49 \pm 41}}{{24}} =  - \frac{1}{3}, - \frac{{15}}{4}$

Now, $3{x^2} + 5x =  - \frac{1}{3}, - \frac{{15}}{4}$

$9{x^2} + 15x + 1 = 0$ equation yields, $x = \frac{{ - 5 \pm \sqrt {21} }}{6}$

The other equation $12{x^2} + 20x + 15 = 0$ has discriminant < 0, hence no real solutions.

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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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