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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The roots of $x^2 - 2x - a^2 + 1 = 0$ are given by $(x - 1)^2 = a^2$,so $x = 1 \pm a$. Let the roots be $x_1 = 1 - a$ and $x_2 = 1 + a$.Let $f(x)

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The roots of $x^2 - 2x - a^2 + 1 = 0$ are given by $(x - 1)^2 = a^2$,so $x = 1 \pm a$. Let the roots be $x_1 = 1 - a$ and $x_2 = 1 + a$.Let $f(x) = x^2 - 2(a + 1)x + a(a - 1)$. For the roots of the first equation to lie between the roots of $f(x) = 0$,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$.Setting $f(1 + a)  -1/3$.Next,$f(1 - a) = (1 - a)^2 - 2(a + 1)(1 - a) + a(a - 1) = (a - 1)^2 + 2(a + 1)(a - 1) + a(a - 1) = (a - 1)[(a - 1) + 2(a + 1) + a] = (a - 1)(a - 1 + 2a + 2 + a) = (a - 1)(4a + 1)$.Setting $f(1 - a) Taking the intersection of $a > -1/3$ and $a \in (-1/4, 1)$,we get $a \in (-1/4, 1)$.

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:

Similar Questions

Two positive distinct numbers $a$ and $b$ each differ from their reciprocal by $1$. The value of $a + b$ is

The roots of the equation $ax^2 + bx + c = 0$ will be reciprocal if

If $x+y+z=0,$ then find the value of $\frac{(x+y)(y+z)(z+x)}{x y z}$

Two real numbers $\alpha$ and $\beta$ are such that $\alpha + \beta = 3$ and $|\alpha - \beta| = 4$. Then $\alpha$ and $\beta$ are the roots of the quadratic equation:

Solve the given two equations and select the correct answer from the given options.
$I.$ $\sqrt{784} x + 1234 = 1486$
$II.$ $\sqrt{1089} y + 2081 = 2345$

Vedclass Test Series

Exam Paper Generator

Online Exam Module