EJU EJU Mathematics Course 1 Practice 1 — Questions and Answers
Question 1: A quadratic equation x² - 5x + 6 = 0 has roots α and β. What is the value of α² + β²?
- 13 (Correct answer)
- 25
- 11
- 7
Correct answer: 13
By Vieta's formulas, α + β = 5 and αβ = 6. So α² + β² = (α + β)² - 2αβ = 25 - 12 = 13.
For a quadratic x² - 5x + 6 = 0, Vieta's formulas give us α + β = 5 and αβ = 6. We use the algebraic identity α² + β² = (α + β)² - 2αβ = 5² - 2(6) = 25 - 12 = 13. This identity avoids solving for the roots individually and is a standard technique on the EJU.
Question 2: If log₂(x) = 3 and log₂(y) = 5, what is log₂(x²y)?
- 11 (Correct answer)
- 16
- 13
- 8
Correct answer: 11
log₂(x²y) = 2·log₂(x) + log₂(y) = 2(3) + 5 = 6 + 5 = 11.
Applying logarithm properties: log₂(x²y) = log₂(x²) + log₂(y) = 2·log₂(x) + log₂(y). Substituting the given values: 2(3) + 5 = 6 + 5 = 11. Remember that log(aⁿ) = n·log(a) and log(ab) = log(a) + log(b).
Question 3: A function f(x) = x² - 4x + 7. What is the minimum value of f(x)?
- 3 (Correct answer)
- 7
- 4
- 1
Correct answer: 3
Completing the square: f(x) = (x-2)² + 3. The minimum value is 3, occurring at x = 2.
f(x) = x² - 4x + 7 = (x² - 4x + 4) - 4 + 7 = (x - 2)² + 3. Since (x - 2)² ≥ 0 for all real x, the minimum value of f(x) is 3, achieved when x = 2. This vertex form (x - h)² + k directly shows the vertex of the parabola.
Question 4: In triangle ABC, if angle A = 60°, b = 4, and c = 6, what is the value of side a using the law of cosines?
- 2√7 (Correct answer)
- √52
- 4√3
- √28
Correct answer: 2√7
a² = b² + c² - 2bc·cos(A) = 16 + 36 - 2(4)(6)cos(60°) = 52 - 48(0.5) = 52 - 24 = 28. So a = √28 = 2√7.
Using the law of cosines: a² = b² + c² - 2bc·cos(A) = 4² + 6² - 2(4)(6)·cos(60°) = 16 + 36 - 48 × (1/2) = 52 - 24 = 28. Therefore a = √28 = 2√7. The law of cosines generalizes the Pythagorean theorem to non-right triangles.
Question 5: The sum of the first n terms of an arithmetic sequence is Sₙ = 3n² + 2n. What is the 5th term of the sequence?
- 29 (Correct answer)
- 35
- 27
- 32
Correct answer: 29
The nth term is aₙ = Sₙ - Sₙ₋₁ = (3n² + 2n) - (3(n-1)² + 2(n-1)) = 6n - 1. For n=5: a₅ = 6(5) - 1 = 29.
Given Sₙ = 3n² + 2n, the nth term is aₙ = Sₙ - Sₙ₋₁ for n ≥ 2. Sₙ₋₁ = 3(n-1)² + 2(n-1) = 3n² - 6n + 3 + 2n - 2 = 3n² - 4n + 1. So aₙ = (3n² + 2n) - (3n² - 4n + 1) = 6n - 1. For n = 5: a₅ = 6(5) - 1 = 29.
Question 6: A bag contains 3 red and 5 blue balls. Two balls are drawn at random without replacement. What is the probability that both are red?
- 3/28 (Correct answer)
- 1/8
- 3/56
- 6/64
Correct answer: 3/28
P(both red) = C(3,2)/C(8,2) = 3/28.
Total ways to choose 2 balls from 8: C(8,2) = 8!/(2!·6!) = 28. Ways to choose 2 red balls from 3: C(3,2) = 3!/(2!·1!) = 3. Probability = 3/28. Without replacement means we cannot use the simple multiplication of independent probabilities; combinations are the correct approach.
Question 7: The point P(3, -1) is reflected across the line y = x. What are the coordinates of the image P'?
- (-1, 3) (Correct answer)
- (3, 1)
- (-3, 1)
- (1, -3)
Correct answer: (-1, 3)
Reflection across y = x swaps the x and y coordinates. So P(3, -1) maps to P'(-1, 3).
When a point (a, b) is reflected across the line y = x, the x and y coordinates are swapped, giving the image (b, a). For P(3, -1), swapping coordinates gives P'(-1, 3). This is because the line y = x is the perpendicular bisector of the segment from (a, b) to (b, a), which can be verified by checking the midpoint ((a+b)/2, (a+b)/2) lies on y = x.
EJU EJU Mathematics Course 1 Practice 1
A quadratic equation x² - 5x + 6 = 0 has roots α and β.
What is the value of α² + β²?