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Roots of

$3{x^2} - 4|{x^2} - 1| + x - 1 = 0$

For $x \in \mathbb{R}$, the number of real roots of the equation $$3{x^2} - 4|{x^2} - 1| + x - 1 = 0$$ is ___ .

Solution

Case I

$x \le  - 1 \cup x \ge 1$

The equation becomes, $3x^2-4(x^2-1)+x-1=0$

$\Rightarrow -x^2+x+3=0$

$\Rightarrow x = \frac{{1 \mp \sqrt {13} }}{2}$

Both above values satisfy the condition $x \le  - 1 \cup x \ge 1$ and hence are accepted.

Case II

$-1 < x < 1$

The equation becomes, $3x^2-4(1-x^2)+x-1=0$

$\Rightarrow 7x^2+x-5=0$

$x = \frac{{ - 1 \pm \sqrt {141} }}{{14}}$

Both above values satisfy the condition $-1 < x < 1$ and hence are accepted.

So, 4 solutions in total.

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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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