For x in R,the number of real roots of the eq | vedclass

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(A) The given equation is $3x^2 + x - 1 = 4|x^2 - 1|$.Case $1$: If $x \in [-1, 1]$,then $|x^2 - 1| = -(x^2 - 1) = 1 - x^2$.The equation becomes $3x^2

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$4$

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(A) The given equation is $3x^2 + x - 1 = 4|x^2 - 1|$.Case $1$: If $x \in [-1, 1]$,then $|x^2 - 1| = -(x^2 - 1) = 1 - x^2$.The equation becomes $3x^2 + x - 1 = 4(1 - x^2)$ $\Rightarrow 3x^2 + x - 1 = 4 - 4x^2$ $\Rightarrow 7x^2 + x - 5 = 0$.Let $f(x) = 7x^2 + x - 5$. Here $f(-1) = 7 - 1 - 5 = 1 > 0$ and $f(1) = 7 + 1 - 5 = 3 > 0$. Since $f(0) = -5 Case $2$: If $x \in (-\infty, -1] \cup [1, \infty)$,then $|x^2 - 1| = x^2 - 1$.The equation becomes $3x^2 + x - 1 = 4(x^2 - 1)$ $\Rightarrow 3x^2 + x - 1 = 4x^2 - 4$ $\Rightarrow x^2 - x - 3 = 0$.Let $g(x) = x^2 - x - 3$. The roots are $x = \frac{1 \pm \sqrt{1 + 12}}{2} = \frac{1 \pm \sqrt{13}}{2}$.Since $\sqrt{13} \approx 3.6$,the roots are $x_1 \approx \frac{4.6}{2} = 2.3$ (which is $\ge 1$) and $x_2 \approx \frac{-2.6}{2} = -1.3$ (which is $\le -1$). Both roots are valid.Thus,there are $2 + 2 = 4$ real roots in total.

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

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