2023 WASSCE Core Mathematics: 10+ Likely Questions And Answers
Mathematics is a subject that is feared by many students, especially when it comes to the West African Senior School Certificate Examination (WASSCE).
The WASSCE is a crucial examination that determines the academic progress of students in West Africa, and mathematics is one of the core subjects that students must take.
The WASSCE mathematics paper can be challenging, but with the right preparation and practice, students can excel in this subject.
In this article, we will provide 10+ analyzed WASSCE mathematics theory questions likely to be asked in the 2023 examination, along with their answers.
1. Find the equation of the tangent to the curve y = x^3 – 2x^2 – x + 1 at the point where the curve crosses the x-axis.
Solution: The derivative of the function is y’ = 3x^2 – 4x – 1. Setting y’ equal to zero gives x = (4 ± √20) / 6. Evaluating the function at these values gives the points of intersection (1,0) and (-1/3,0). Therefore, the equation of the tangent at point (1,0) is y = -2x + 1.
2. Solve the system of equations: 3x – 2y + z = 6, x + y – 2z = -3, 2x – y + 3z = 11.
Solution: Using row operations, the augmented matrix of the system is reduced to [1 0 0 | 2], [0 1 0 | -1], [0 0 1 | 3]. Therefore, the solution is x = 2, y = -1, z = 3.
3. A circular cylinder with a radius of 5 cm and height of 10 cm is melted down and recast into a sphere. Find the radius of the sphere.
Solution: The volume of the cylinder is Vc = πr^2h = 250π cm^3. The volume of a sphere is Vs = (4/3)πr^3. Equating these volumes and solving for r gives r = 5√6 / 3 cm.
4. Find the equation of the circle passing through the points (1,2), (3,4), and (-1,0).
Solution: The general equation of a circle is (x – h)^2 + (y – k)^2 = r^2. Substituting the three points into the equation gives the system of equations: h^2 + k^2 – 2h – 4k + 5 = 0, h^2 + k^2 – 6h – 8k + 20 = 0, h^2 + k^2 + 2h = Solving this system gives h = 1, k = 2, and r = √10.
5. Find the fourth term in the expansion of (2x – 3y)^6.
Solution: The fourth term is the coefficient of (2x)^3(-3y)^1, which is given by the binomial coefficient 6C3 * (-3)^1 * (2)^3 = -960.
6. A ladder 10 meters long rests against a wall. If the base of the ladder is pulled away from the wall at a rate of 2 m/s, how fast is the top of the ladder sliding down the wall when it is 6 meters above the ground?
Solution: Using the Pythagorean theorem, the distance between the top of the ladder and the wall is √(10^2 – h^2), where h is the height above the ground. Taking the derivative of both sides with respect to time and using the chain rule, we get dh/dt = (-3h/√(100 – h^2)) m/s when h = 6.
7. Find the value of x in the equation log2(x) + log2(x – 1) = 3.
Answer: x = 5
8. Simplify the expression (3x^3 – 2x^2 + 5x) / (x^2 – 2x)
Answer: 3x + 11 / (x – 2)
9. Find the gradient of the curve y = 3x^2 – 4x + 1 at the point (2, 5).
Answer: Gradient = 10
10. Find the value of x that satisfies the equation 2sin(x) – 1 = 0 in the interval [0, 2π].
Answer: x = 5π/6, 7π/6
11. If log3(x) = 4, find the value of x.
Answer: x = 81
12. Solve for x: e^(2x) – 5e^x + 6 = 0.
Answer: x = ln(3), ln(2)
13. Find the value of x that satisfies the equation 4cos^2(x) – 3 = 0 in the interval [0, π].
Answer: x = π/6, 5π/6
14. Find the value of y in the equation 2^y = 8.
Answer: y = 3
15. Simplify the expression (2x – 3)(3x + 4) – (x – 1)(x + 2).
Answer: 5x^2 + 7x – 10
16. Find the value of x that satisfies the equation log2(x + 2) – log2(x) = 1.
Answer: x = 2
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