If are the roots of the equation x2 + px + q = 0, where are real, then the roots of the equation (p2 - 4q)(p2x2 + 4px) - 16q = 0 are
The number of solutions of the equation is
0
1
2
3
C.
2
Given, = 1
Since, x = 1 and x = - 3 are satisfy the given equation. Therefore the number of solutions of the equation are two.
Let R be the set of real numbers and the functions f : R ➔ R and g : R ➔ R be defined by f(x) = x2 + 2x - 3 and g(x) = x + 1. Then, the value of x for which f(g(x)) = g(f(x)) is
- 1
0
1
2
Let f(x) = ax2 + bx + c, g(x) = px2 + qx + r such that f(1) = g(1), f(2) = g(2) and f(3) - g(3) = 2. Then, f(4) - g(4) is
4
5
6
7
The equations x2 + x + a= 0 and x2 + ax + 1 = 0 have a common real root
for no value of a
for exactly one value of a
for exactly two value of a
for exactly three value of a
The points representing the complex number z for which arg lie on
a circle
a straight line
an ellipse
a parabola
The quadratic equation 2x2 - (a3 + 8a - 1)x + a2 - 4a = 0 posses roots of opposite sign. Then,
a 0
0 < a < 4