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
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
C.
6
Given, f(x) = ax2 + bx + c, g(x) = px2 + qx + r
Since f(1) = g(1)
...(i)
f(2) = g(2)
...(ii)
Subtracting Eq. (ii) from Eq. (i), we get
3a + b = 3p + q ...(iii)
f(3) - g(3) = 2
From Eq. (i),
(a - p) + (b - q) + (c - r) = 0
...(v)
From Eq. (ii),
4(a - p) + 2(b - q) + c - r = 0
...(vi)
Subtracting Eq.(v) from Eq. (vi), we get
(a - p) - 1 = 0
a - p = 1
From Eq. (v),
b - q = - 3
Now,
f(4) - g(4) = (16a + 4b + c) - (16p + 4q + r)
= 16(a - p) + 4(b - q) + (c - r) ...(vii)
Substituting the values of (a - p), (b - q), (c - r) from above in Eq. (vii), we get
f(4) - g(4) =
= 16 - 12 + 2 = 6
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