Question

If both roots of the equation $x^2-2ax+a^2$ - 1 = 0 lie in the interval (-3,4), then sum of the integral parts of a is :

• A. 0
• B. 2
• C. 4
• D. None

Answer 1

The correct option is A

0

$x^2-2ax+a^2$ - 1 = 0. Both roots lie in (-3, 4). Roots are a - 1 and a + 1.

$\therefore$ a - 1 > -3, a + 1 < 4 i.e -2 < a < 3

[a] = - 2, -1,0,1,2, whose sum is 0.