POST UTME ELIZADE UNIVERSITY 2023 Mathematics | Objective

Practice these randomly selected questions to test your readiness.

Question 1
Solve the equation \( x^3 - 6x^2 + 11x - 6 = 0 \).
A. 1
B. 2
C. 3
D. 4
Question 2
Find the area under the curve \( y = \frac{1}{x^2 + 1} \) from \( x = 0 \) to \( x = 1 \).
A. \frac{\pi}{4}
B. \frac{\pi}{2}
C. \frac{\pi}{3}
D. \frac{\pi}{6}
Question 3
Solve the inequality \( \frac{x^2 - 4}{x^2 - 9} > 0 \).
A. \( -3, -1 \) \cup (1, 3)
B. \( -3, -1 \) \cup (1, 3) \cup \( 4, \infty \)
C. \( -3, -1 \) \cup (1, 3) \cup \( -\infty, -4 \) \cup \( 4, \infty \)
D. \( -3, -1 \) \cup (1, 3) \cup \( -\infty, -4 \) \cup \( 4, \infty \) \cup (0, 1)
Question 4
Find the determinant of the matrix \[ \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix} \]
A. 0
B. 1
C. 2
D. 3
Question 5
Let ( X ) be a random variable with probability density function ( f(x) = egin{cases} 2x & 0 leq x leq 1 \ 0 & \text{otherwise} \end{cases} ). Find the probability that ( X ) takes a value greater than 0.5.
A. 0.25
B. 0.5
C. 0.75
D. 1
Question 6
Solve the quadratic equation \( x^2 + 4x + 4 = 0 \)
A. x = -2
B. x = -1
C. x = 0
D. x = 1
Question 7
Solve for x in the equation \( 2^x + 5^x = 7^x \)
A. \log(7) - \log(5)
B. \log(7) - \log(2)
C. \log(5) - \log(2)
D. \log(7) - \log(2) + \log(5)
Question 8
A histogram of exam scores has a mean of 60 and a s\tandard deviation of 10. If the scores are normally distributed, what is the probability that a randomly selected score will be greater than 70?
A. 0.1587
B. 0.3413
C. 0.4772
D. 0.6827
Question 9
A random variable X has a probability distribution given by P\( X = 1 \) = 0.4, P\( X = 2 \) = 0.3, P\( X = 3 \) = 0.2, and P\( X = 4 \) = 0.1. What is the expected value of X?
A. 1.2
B. 1.8
C. 2.4
D. 3.0
Question 10
Find the sum of the infinite geometric series \( \sum_{n=1}^\infty \frac{2}{3^n} \).
A. \frac{2}{3}
B. \frac{2}{3} \cdot \frac{1}{1 - \frac{1}{3}}
C. \frac{2}{3} \cdot \frac{1}{1 - \frac{1}{3}} \cdot \frac{1}{1 - \frac{1}{3}}
D. \frac{2}{3} \cdot \frac{1}{1 - \frac{1}{3}} \cdot \frac{1}{1 - \frac{1}{3}} \cdot \frac{1}{1 - \frac{1}{3}}

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