x ∈ - 32, 14
x ∈ - 34, 14
x ∈ - 34, 12
x < 14
If R→ R and is defined by fx = 12 - cos3xfpr each x ∈ R, then the range of f is
13, - 1
13, 1
(1, 2)
1, 2
2+ 5 - 6 - 35 + 14 - 65 = ?
1
2
3
4
If 3xx - ax - b = 2x - a + 1x - b, then a : b is equal to
1 : 2
- 2 : 1
1 : 3
3 : 1
f : [-6, 6] R is defined by f(x) = x2 - 3for x ∈ R, then(fofof) (-1) + (fofof) (0) + (fofof)(1) is equal to
f42
f32
f22
f2
If x2 + x + 1x2 + 2x + 1 = A + Bx + 1 + Cx + 12, then A - B is equal to
4C
4C + 1
3C
2C
If f : 2, 3→ R is defined by f (x) = x3 + 3x - 2, then the range f(x) iscontained in the interval
[1, 12]
[12, 34]
[35, 50]
[- 12, 12]
The longest distance of the point (a, 0) from the curve 2x2 + y2 = 2x is
1 + a
1 - a
1 - 2a +2a2
1 - 2a + 3a2
C.
Given curve is2x2 + y2 = 2x2x2 - 2x + y2 = 0⇒ 2x - 122 + y2 = 12⇒ x - 122 14 + y212 = 1which represents an ellipseHere, a = 12, b = 12, h = 12, k = 0Consider a point P(h + acosθ, k + bsinθ)= P12 + 12cosθ, 12sinθ on the ellipse from which the distanceof point (a, 0) is maximumlet Q(a, 0)Now,PQ = 12 + 12cosθ - a2 + 12sinθ - 02 = 14 + 14cos2θ + a2 + 12cosθ - acosθ - a + 12sin2θPQ = 12 + a2 - a + 12 - acosθ + 14sin2θLet y = PQ2 = 12 + a2 - a + 12 - acosθ + 14sin2θ
For maxima and minima, put dydθ = 0⇒ - 12 - asinθ + 14 . 2sinθcosθ = 0⇒ sinθ- 12 + a + 12cosθ = 0⇒ sinθ = 0 or - 12 + a + 12cosθ = 0⇒ θ = 0 or cosθ = 1 - 2a⇒ sin2θ = 1 - cos2θ = 1 - 1 - 2a2 = - 4a2 + 4a
Now, d2ydθ2 < 0 for cosθ = 1 - 2aThus, distance PQ is maximum, whencosθ = 1 - 2a and sin2θ = - 4a2 + 4aNow, required longest distance is =12 + a2 - a + 12 - a1 - 2a + 14- 4a2 + 4a= 12 + a2 - a + 12 - a - a + 2a2 - a2 + a= 2a2 - 2a + 1= 1 - 2a + 2a2
If f : R → R is defined by f(x) = x5 for x ∈ R, where [y] denotes the greatest integer not exceeding y, then fx : x < 71 is equal to
- 14, - 13, . . . 0, . . . 13, 14
- 14, - 13, . . . 0, . . . 14, 15
- 15, - 14, . . . 0, . . . 14, 15
- 15, - 14, . . . 0, . . . 13, 14
x2 + x + 1x - 1x - 2x - 3 = Ax - 1 + Bx - 2 + Cx - 3⇒ A + C =
5
6
8