1. Let R= {(x,y): y= 2-x^2} if the domain of R= {-2,-1, 0, 3, 5, 7} then find the Range of R?
2. If the graph of f(x) = x^2-2x-3 is shifted so that is axis of symmetry is the line x= -2 and it orthogonal axis is the line y= 4, Then
A) Find the equation of the new curve?
B) Find the x - intercept of the new curve?
C) Draw the graph of f(x) and the new one in the co-ordinate axis?
3. Find the domain and range of (f*g) (x) if f(x)=(x+1)/(sqrtx) and g(x)=(x+2)/(sqrt(1-x^2)) .
4. Let R be a Relation defined by R= {( x,y): y<=(5/4)x+5and y<(5/4)x-5 then
A) Draw the graph of R?
B) Find the domain and Range of R?
5. Let f(x)=2+3x-2x^2. Then
A) Find the maximum value of f(x)?
B) Find the axis of symmetry and orthogonal axis?
1. The given relation R is defined by y = 2 - x^2.
To find the range of R, we need to determine the possible values of y for the given domain {-2, -1, 0, 3, 5, 7}.
Substituting each x value into the equation of R, we get:
For x = -2, y = 2 - (-2)^2 = 2 - 4 = -2
For x = -1, y = 2 - (-1)^2 = 2 - 1 = 1
For x = 0, y = 2 - 0^2 = 2 - 0 = 2
For x = 3, y = 2 - 3^2 = 2 - 9 = -7
For x = 5, y = 2 - 5^2 = 2 - 25 = -23
For x = 7, y = 2 - 7^2 = 2 - 49 = -47
Therefore, the range of R is {-47, -23, -7, -2, 1, 2}.
2. To shift the graph of f(x) = x^2 - 2x - 3 so that its axis of symmetry becomes x = -2 and its orthogonal axis becomes y = 4, we need to make the following transformations:
A) Shifting the graph horizontally:
The new equation will have the form f(x + a), where a is the distance between the original and new axis of symmetry.
In this case, the original axis of symmetry is x = 1, so we need to shift the graph 3 units to the left.
f(x) = (x - 1)^2 - 2(x - 1) - 3
B) Shifting the graph vertically:
The new equation will have the form f(x) + b, where b is the distance between the original and new orthogonal axis.
In this case, the original orthogonal axis is y = 0, so we need to shift the graph 4 units up.
f(x) = (x - 1)^2 - 2(x - 1) - 3 + 4
f(x) = (x - 1)^2 - 2(x - 1) + 1
C) The graph of f(x) and the new curve after the shifts can be drawn on a coordinate axis to visualize the transformation.
3. The domain and range of the function (f*g)(x) can be found by considering the domains and ranges of the individual functions f(x) and g(x) and their possible combinations.
For the function f(x) = (x + 1)/√x:
The domain is restricted by the square root (√x), which means that x ≥ 0.
The range is all real numbers except x = 0, since division by 0 is undefined.
For the function g(x) = (x + 2)/√(1 - x^2):
The domain is restricted by the square root (√(1 - x^2)), which means that -1 ≤ x ≤ 1.
The range is all real numbers except x = -1 and x = 1, as division by 0 is undefined.
To find the domain and range of (f*g)(x), we need to find the intersection of the domains and ranges of f(x) and g(x).
The intersection of the domains is -1 ≤ x ≤ 0.
The intersection of the ranges is all real numbers except x = 0.
Therefore, the domain of (f*g)(x) is -1 ≤ x < 0, and the range is all real numbers except x = 0.
4. A) The graph of the relation R= {(x, y): y ≤ (5/4)x + 5, and y < (5/4)x - 5} can be drawn by plotting the points satisfying these conditions and shading the region below the line y = (5/4)x + 5 and above the line y = (5/4)x - 5.
B) To find the domain and range of R, we can consider the domain as the set of all x-values that correspond to points in the relation, and the range as the set of all y-values that correspond to points in the relation.
Domain:
Since there are no restrictions on the x-values in the relation R, we can say that the domain is all real numbers, or (-∞, ∞).
Range:
For any given value of x, the y-values in the relation R are limited by the conditions y ≤ (5/4)x + 5 and y < (5/4)x - 5.
Therefore, the range of R is (-∞, ∞), since there are no restrictions on the possible y-values.
5. A) The given function f(x) = 2 + 3x - 2x^2 is a quadratic function in the form f(x) = ax^2 + bx + c, where a = -2, b = 3, and c = 2.
To find the maximum value of f(x), we need to find the vertex of the parabola. The x-coordinate of the vertex can be found using the formula x = -b/2a.
x = -3 / (2*(-2))
x = -3 / -4
x = 3/4
Substituting this value of x back into the function, we can find the corresponding y-coordinate:
f(3/4) = 2 + 3(3/4) - 2(3/4)^2
= 2 + 9/4 - 18/16
= 2 + 9/4 - 9/8
= 2 + 9/4 - 9/8
= 8/4 + 9/4 - 9/8
= 17/4 - 9/8
= (34/8) - (9/8)
= 25/8
Therefore, the maximum value of f(x) is 25/8.
B) The axis of symmetry of the parabola represented by f(x) is a vertical line passing through the vertex. The equation of the axis of symmetry is x = 3/4.
The orthogonal axis is a horizontal line passing through the vertex. Since the axis of symmetry is vertical, the orthogonal axis is the y-axis.
Therefore, the axis of symmetry is x = 3/4 and the orthogonal axis is the y-axis.