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$7f(x) + 5f\left( {\frac{1}{x}} \right) = x + 1$

$t = 24xf(x)$

${\left. {\frac{{dt}}{{dx}}} \right|_{x = - 1/7}} = ?$

We have, $7f(x) + 5f\left( {\frac{1}{x}} \right) = x + 1$ ..........(A)

Replacing x by $\frac {1}{x}$, $7f\left( {\frac{1}{x}} \right) + 5f(x) = \frac{1}{x} + 1$ ...........(B)

(A) + (B) yields, $12\left[ {f(x) + f\left( {\frac{1}{x}} \right)} \right] = x + \frac{1}{x} + 2$

$\therefore f(x) + f\left( {\frac{1}{x}} \right) = \frac{x}{{12}} + \frac{1}{{12x}} + \frac{1}{6}$ ..........(C)

(A) - (B) yields, $2\left[ {f(x) - f\left( {\frac{1}{x}} \right)} \right] = x - \frac{1}{x}$

$\therefore f(x) - f\left( {\frac{1}{x}} \right) = \frac{x}{2} - \frac{1}{{2x}}$ ..........(D)

(C) + (D) yields, $2f(x) = \frac{{7x}}{{12}} - \frac{5}{{12x}} + \frac{1}{6}$

$\therefore f(x) = \frac{{7x}}{{24}} - \frac{5}{{24x}} + \frac{1}{{12}}$

Now, $t = 24xf(x) = 7{x^2} - 5 + 2x$

$\therefore{\left. {\frac{{dt}}{{dx}}} \right|_{x =  - 1/7}} = {\left. {14x + 2} \right|_{x =  - 1/7}} = 0$

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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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