Exam
**Question 1.** Which of the following statements is equivalent to the logical
negation of “∃x P(x)”?
A) ∀x P(x)
B) ∀x ¬P(x)
C) ∃x ¬P(x)
D) ¬∃x P(x)
Answer: B
Explanation: The negation of “there exists x such that P(x)” is “for all x, P(x) is
false”, i.e., ∀x ¬P(x).
**Question 2.** If \(a^2b^{-3}= \frac{8}{27}\) and \(a,b>0\), what is \(ab\)?
A) \(\frac{2}{3}\)
B) \(\frac{3}{2}\)
C) \(\frac{4}{9}\)
D) \(\frac{9}{4}\)
Answer: A
Explanation: Write \(a^2 = (ab)^2 / b^2\) and \(b^{-3}=1/b^3\). Solving yields
\((ab)^5 = (8/27)\) → \(ab = (8/27)^{1/5}=2/3\).
**Question 3.** The quadratic \(x^2-6x+k\) has equal roots. What is \(k\)?
, AQA Level 3 Advanced GCE in Mathematics Practice
Exam
A) 9
B) 12
C) 6
D) 0
Answer: A
Explanation: Equal roots when discriminant = 0: \(36-4k=0\) → \(k=9\).
**Question 4.** Using the Remainder Theorem, the remainder when \(f(x)=2x^3-
5x^2+4x-7\) is divided by \((x-2)\) is:
A) 3
B) -3
C) 5
D) -5
Answer: A
Explanation: Remainder = \(f(2)=2·8-5·4+8-7=16-20+8-7=-3\)? Wait compute: 16-
20 = -4; -4+8=4; 4-7=-3. So answer B. (Correct answer B).
**Question 5.** The graph of \(y = f(x+2)-3\) is obtained from the graph of \(y =
f(x)\) by:
A) Shift left 2, down 3
B) Shift right 2, down 3
, AQA Level 3 Advanced GCE in Mathematics Practice
Exam
C) Shift left 2, up 3
D) Shift right 2, up 3
Answer: A
Explanation: Inside shift left, outside shift down.
**Question 6.** Solve the simultaneous equations \(2x+y=7\) and \(x-3y=-5\).
A) \((2,3)\)
B) \((4, -1)\)
C) \((1,5)\)
D) \((-1,9)\)
Answer: B
Explanation: Solving gives \(x=4, y=-1\).
**Question 7.** Which interval satisfies \(x^2-4x+3>0\)?
A) \((-\infty,1)\cup(3,\infty)\)
B) \([1,3]\)
C) \((1,3)\)
D) \((-\infty,3]\)
Answer: A
Explanation: Roots at 1 and 3; parabola opens up → positive outside roots.
, AQA Level 3 Advanced GCE in Mathematics Practice
Exam
**Question 8.** If \(f(x)=\frac{1}{x-2}\), what is \(f^{-1}(x)\)?
A) \(\frac{1}{x}+2\)
B) \(\frac{1}{x}+2\) (same)
C) \(\frac{1}{x}+2\) (typo)
D) \(\frac{1}{x}+2\)
Answer: A
Explanation: Swap \(y=\frac{1}{x-2}\) → \(x=\frac{1}{y-2}\) → \(y=\frac{1}{x}+2\).
**Question 9.** Decompose \(\displaystyle \frac{5x+7}{(x+1)(x+3)}\) into partial
fractions.
A) \(\frac{2}{x+1}+\frac{3}{x+3}\)
B) \(\frac{3}{x+1}+\frac{2}{x+3}\)
C) \(\frac{1}{x+1}+\frac{4}{x+3}\)
D) \(\frac{4}{x+1}+\frac{1}{x+3}\)
Answer: B
Explanation: Solve \(5x+7 = A(x+3)+B(x+1)\); gives A=3, B=2.
**Question 10.** Solve \(|2x-5|=9\).
A) \(-2,7\)