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$y = \frac{{{x^2} - \alpha \beta }}{{2x - \alpha - \beta }}$

$x \in \mathbb{R} - \left\{ {\frac{{\alpha + \beta }}{2}} \right\}$

Assuming $\beta  > \alpha $ which of the following option(s) is/are correct?

(A) $y \le \alpha $
(B) $y \ge \beta $
(C) $\alpha  \le y \le \beta $
(D) $y \in \mathbb{R}$

Solution

${x^2} - 2xy + [(\alpha  + \beta )y - \alpha \beta ] = 0$

$\because x \in \mathbb{R},\Delta  \ge 0$

$\therefore 4{y^2} - 4[(\alpha  + \beta )y - \alpha \beta ] \ge 0$

$ \Rightarrow {y^2} - \alpha y - \beta y + \alpha \beta  \ge 0$

$ \Rightarrow (y - \alpha )(y - \beta ) \ge 0$

The wave-curve is shown below for the expression $ (y - \alpha )(y - \beta ) $ assuming $\beta > \alpha $.


Hence, (A) & (B).

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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)