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f'(0) = ? for an Even Function f that is Differentiable at 0

Since f is differentiable at 0 we have,

$\mathop {\lim }\limits_{h \to 0} \frac{{f(0 + h) - f(0)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{f(0 - h) - f(0)}}{{ - h}}$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{{f(h) - f(0)}}{h} =  - \mathop {\lim }\limits_{h \to 0} \frac{{f( - h) - f(0)}}{h}$

Since f is an even function, f(-h) = f(h)

$\therefore \mathop {\lim }\limits_{h \to 0} \frac{{f(h) - f(0)}}{h} =  - \mathop {\lim }\limits_{h \to 0} \frac{{f(h) - f(0)}}{h}$

$ \Rightarrow 2\mathop {\lim }\limits_{h \to 0} \frac{{f(h) - f(0)}}{h} = 0$

$ \Rightarrow 2f'(0) = 0$

$\therefore f'(0) = 0$

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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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