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Let $f:\mathbb{R} \to \mathbb{R}$ be defined by ....

Let $f:\mathbb{R} \to \mathbb{R}$ be defined by $$f(x) = \frac{{{x^2} - 3x - 6}}{{{x^2} + 2x + 4}}$$ Then which of the following statements is (are) TRUE?

(A) f is decreasing in the interval (-2, -1)
(B) f is increasing in the interval (1, 2)
(C) f is onto
(D) Range of f is $\left[ { - \frac{3}{2},2} \right]$

Solution

We have, $f'(x) = \frac{{({x^2} + 2x + 4).(2x - 3) - ({x^2} - 3x - 6).(2x + 2)}}{{{{({x^2} + 2x + 4)}^2}}}$

$ \Rightarrow f'(x) = \frac{{5x(x + 4)}}{{{{({x^2} + 2x + 4)}^2}}}$

Sign-scheme for f'(x) will be decided by $x.(x+4)$. -4 & 0 are the points of local extremum. The figure below shows increasing/decreasing etc. details.


$f(0) = \frac {-3}{2}$, 0 being the point of local minima.

$f(-4)=\frac {11}{6}$, -4 being the point of local maxima.

Range of function is $\left[ { - \frac{3}{2},\frac{{11}}{6}} \right]$

Options (A) & (B) fall within the values shown in the above figure making them correct. Other options are incorrect.

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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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