If both the roots of the equation x^2-4ax+1-3 | vedclass

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(B) Let $f(x) = x^2 - 4ax + (4a^2 - 3a + 1)$. For both roots to be greater than $1$,the following conditions must be satisfied: $1$. Discriminant $D \

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$\left(\frac{7+\sqrt{17}}{8}, \infty\right)$

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(B) Let $f(x) = x^2 - 4ax + (4a^2 - 3a + 1)$. For both roots to be greater than $1$,the following conditions must be satisfied: $1$. Discriminant $D \geq 0$: $D = (-4a)^2 - 4(1)(4a^2 - 3a + 1) = 16a^2 - 16a^2 + 12a - 4 = 12a - 4 \geq 0 \Rightarrow a \geq \frac{1}{3}$. $2$. Vertex position: $\frac{-b}{2a} > 1$: $\frac{4a}{2} > 1$ $\Rightarrow 2a > 1$ $\Rightarrow a > \frac{1}{2}$. $3$. $f(1) > 0$: $f(1) = 1^2 - 4a(1) + 4a^2 - 3a + 1 = 4a^2 - 7a + 2 > 0$. Roots of $4a^2 - 7a + 2 = 0$ are $a = \frac{7 \pm \sqrt{49 - 32}}{8} = \frac{7 \pm \sqrt{17}}{8}$. Since the parabola opens upward,$4a^2 - 7a + 2 > 0$ for $a \in \left(-\infty, \frac{7-\sqrt{17}}{8}\right) \cup \left(\frac{7+\sqrt{17}}{8}, \infty\right)$. Taking the intersection of $a \geq \frac{1}{3}$,$a > \frac{1}{2}$,and $a \in \left(-\infty, \frac{7-\sqrt{17}}{8}\right) \cup \left(\frac{7+\sqrt{17}}{8}, \infty\right)$: Since $\frac{7+\sqrt{17}}{8} \approx \frac{7+4.12}{8} \approx 1.39 > \frac{1}{2}$,the intersection is $a \in \left(\frac{7+\sqrt{17}}{8}, \infty\right)$.

If both the roots of the equation $x^2-4ax+1-3a+4a^2=0$ exceed $1$,then $a$ lies in the interval

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