IB math papers can be a daunting task for students who are not confident with their math skills. However, with a bit of preparation and practice, students can perform well on these exams. In this article, we will provide sample solutions to some commonly asked questions on IB math papers.
Question 1: Find the domain of the function f(x) = √(3x – 2)
Solution: The domain of the function is the set of all values of x for which the function is defined. In this case, the function is defined if the expression inside the square root is non-negative. Thus, we set 3x – 2 ≥ 0 and solve for x:
3x – 2 ≥ 0
3x ≥ 2
x ≥ 2/3
Therefore, the domain of the function is x ≥ 2/3.
Question 2: Solve the equation 2x + 5 = 7x – 3
Solution: To solve the equation, we want to isolate the variable x on one side of the equation. We can do this by adding or subtracting the same quantity from both sides of the equation. In this case, we can subtract 2x from both sides to get:
5 = 5x – 3
Next, we can add 3 to both sides:
8 = 5x
Finally, we can divide both sides by 5 to get:
x = 8/5
Therefore, the solution to the equation is x = 8/5.
Question 3: Find the slope of the line passing through the points (2, 3) and (5, 9).
Solution: The slope of a line can be found using the formula:
slope = (y2 – y1) / (x2 – x1)
where (x1, y1) and (x2, y2) are the coordinates of two points on the line. Plugging in the given values, we get:
slope = (9 – 3) / (5 – 2)
slope = 2
Therefore, the slope of the line passing through the points (2, 3) and (5, 9) is 2.
Question 4: Find the equation of the line passing through the point (3, 4) and perpendicular to the line 2x + 3y = 5.
Solution: To find the equation of the line, we need to first find the slope of the line we want to find. Since the line is perpendicular to the given line, the slope of the line we want to find is the negative reciprocal of the slope of the given line. The slope of the given line can be found by solving for y in terms of x:
2x + 3y = 5
3y = -2x + 5
y = (-2/3)x + 5/3
Thus, the slope of the given line is -2/3, and the slope of the line we want to find is 3/2.
Next, we can use the point-slope form of the equation of a line:
y – y1 = m(x – x1)
where m is the slope of the line and (x1, y1) is a point on the line. Plugging in the given values, we get:
y – 4 = (3/2)(x – 3)
Simplifying, we get:
y = (3/2)x – 1/2
Therefore, the equation of the line passing through the point (3, 4) and perpendicular to the line 2x + 3y = 5 is y = (3/2)x – 1
Question 5: Find the exact value of sin(π/3).
Solution: To find the exact value of sin(π/3), we can use the unit circle. Recall that the unit circle is a circle with radius 1 centered at the origin of the coordinate plane. The angle between the positive x-axis and a point on the circle is measured in radians.
To find sin(π/3), we first locate the point on the unit circle corresponding to an angle of π/3. This point is at the intersection of the circle with the line passing through the origin and making an angle of π/3 with the positive x-axis. Using the Pythagorean theorem, we can find the y-coordinate of this point:
y² + x² = 1
y² + (1/2)² = 1
y² = 3/4
y = √3/2
Therefore, sin(π/3) = √3/2.
Question 6: Find the derivative of the function f(x) = x³ – 4x² + 2x – 1.
Solution: To find the derivative of the function, we can use the power rule, which states that if f(x) = x^n, then f'(x) = nx^(n-1). Using this rule and the sum rule, we can find the derivative of the function:
f'(x) = 3x² – 8x + 2
Therefore, the derivative of the function f(x) = x³ – 4x² + 2x – 1 is f'(x) = 3x² – 8x + 2.
Question 7: Evaluate the integral ∫(2x + 3) dx.
Solution: To evaluate the integral, we can use the power rule of integration, which states that if f(x) = x^n, then ∫f(x) dx = (1/(n+1))x^(n+1) + C, where C is the constant of integration. Using this rule and the sum rule of integration, we can evaluate the integral:
∫(2x + 3) dx = ∫2x dx + ∫3 dx
= x² + 3x + C
Therefore, the integral of (2x + 3) dx is x² + 3x + C.
Question 8: Find the equation of the parabola that passes through the points (1, 2), (3, 6), and (5, 10).
Solution: To find the equation of the parabola, we can use the general form of the equation:
y = ax² + bx + c
where a, b, and c are constants. Substituting the given points into the equation, we get a system of three equations:
a + b + c = 2
9a + 3b + c = 6
25a + 5b + c = 10
Solving for a, b, and c using elimination or substitution, we get:
a = 1
b = 0
c = 1
Therefore, the equation of the parabola that passes through the points (1, 2), (3, 6), and (5, 10) is y = x² + 1.
In conclusion, preparing for IB math papers requires practice and a solid understanding of the concepts. By studying sample solutions to commonly asked questions, students can build their confidence and perform well on these exams.
