The range of a for which the roots of x^2 - 2 | vedclass

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(C) The roots of $x^2 - 2x - a^2 + 1 = 0$ are found by $(x-1)^2 - a^2 = 0$,which gives $x = 1 \pm a$.Let $f(x) = x^2 - 2(a+1)x + a(a-1)$.For the roots

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$\left( -\frac{1}{4}, 1 \right)$

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(C) The roots of $x^2 - 2x - a^2 + 1 = 0$ are found by $(x-1)^2 - a^2 = 0$,which gives $x = 1 \pm a$.Let $f(x) = x^2 - 2(a+1)x + a(a-1)$.For the roots $1+a$ and $1-a$ to lie between the roots of $f(x)$,we must have $f(1+a) First,$f(1+a) = (1+a)^2 - 2(a+1)(1+a) + a(a-1) = (1+a)^2 - 2(1+a)^2 + a^2 - a = -(1+a)^2 + a^2 - a = -1 - 2a - a^2 + a^2 - a = -3a - 1 This implies $3a > -1$,or $a > -\frac{1}{3}$.Second,$f(1-a) = (1-a)^2 - 2(a+1)(1-a) + a(a-1) = (a-1)^2 + 2(a+1)(a-1) + a(a-1) = (a-1)[(a-1) + 2(a+1) + a] = (a-1)[a-1 + 2a + 2 + a] = (a-1)(4a+1) This inequality holds for $a \in \left( -\frac{1}{4}, 1 \right)$.Taking the intersection of $a > -\frac{1}{3}$ and $a \in \left( -\frac{1}{4}, 1 \right)$,we get $a \in \left( -\frac{1}{4}, 1 \right)$.

The range of $a$ for which the roots of $x^2 - 2x - a^2 + 1 = 0$ lie between the roots (exclusive) of the equation $x^2 - 2(a + 1)x + a(a - 1) = 0$ is:

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