TMUA Algebraic Techniques & Manipulation 5 — Questions and Answers
Question 1: Rationalise the denominator of: (2 + √3)/(2 - √3)
- 7 + 4√3 (Correct answer)
- 1 + 4√3
- 7 - 4√3
- 4 + 7√3
Correct answer: 7 + 4√3
Multiply by (2+√3)/(2+√3): numerator = (2+√3)² = 7+4√3; denominator = 4-3 = 1.
Question 2: Solve: 2^(2x) - 5·2^x + 4 = 0
- x = 0 or x = 2 (Correct answer)
- x = 1 or x = 2
- x = 0 or x = 1
- x = -1 or x = 2
Correct answer: x = 0 or x = 2
Let u = 2^x: u²-5u+4=0 gives (u-1)(u-4)=0, so 2^x=1→x=0 or 2^x=4→x=2.
Question 3: If x + y = 5 and xy = 4, find x³ + y³.
- 65 (Correct answer)
- 80
- 53
- 125
Correct answer: 65
x³+y³ = (x+y)(x²-xy+y²) = (x+y)[(x+y)²-3xy] = 5[25-12] = 65.
Question 4: Simplify: log₂(32) - log₂(4) + log₂(2)
- 4 (Correct answer)
- 3
- 5
- 6
Correct answer: 4
log₂(32)=5, log₂(4)=2, log₂(2)=1; so 5-2+1=4.
Question 5: Find the value of the expression: (n+1)! / [(n-1)! · n]
- n+1 (Correct answer)
- n
- (n+1)/n
- n(n+1)
Correct answer: n+1
(n+1)! = (n+1)·n·(n-1)!, so the expression becomes (n+1)·n·(n-1)! / [(n-1)!·n] = n+1.
Question 6: Solve the inequality: (x-1)(x+3) < 0
- -3 < x < 1 (Correct answer)
- x < -3 or x > 1
- x > 1
- -1 < x < 3
Correct answer: -3 < x < 1
The product is negative when the factors have opposite signs, which occurs for -3 < x < 1.
Question 7: Given f(x) = x² - 4 and g(x) = x + 2, find f(x)/g(x) simplified, stating any restrictions.
- x - 2, x ≠ -2 (Correct answer)
- x + 2, x ≠ 2
- x - 2, x ≠ 2
- (x-2)(x+2), x ≠ -2
Correct answer: x - 2, x ≠ -2
f(x)/g(x) = (x-2)(x+2)/(x+2) = x-2, provided x ≠ -2 (since g(-2)=0).
TMUA Algebraic Techniques & Manipulation 5
Rationalise the denominator of: (2 + √3)/(2 - √3)