What is the domain of \(f(x) = \dfrac{\sqrt{x-3}}{x-7}\,\)?
Memory√ needs ≥ 0 · denominator ≠ 0 — find both restrictions, combine with AND.
Explanation
The square root requires \(x - 3 \ge 0\), so \(x \ge 3\). The denominator requires \(x \neq 7\). Combining both: \(x \ge 3\) AND \(x \neq 7\), which gives \([3,7)\cup(7,\infty)\). Option A forgets to remove \(x=7\).
02
Composition
If \(f(x)=2x+1\) and \(g(x)=x^2-3\), what is \((f\circ g)(2)\)?
Memoryf∘g = f(g(x)) — inside function goes first, always.
Explanation
First compute \(g(2) = 4 - 3 = 1\). Then \(f(g(2)) = f(1) = 2(1)+1 = 3\). A common trap is computing \(g(f(2))\) instead — that would give \(g(5) = 22\).
03
Inverse Functions
Which of the following is the inverse of \(f(x)=\dfrac{3x-1}{2}\,\)?
MemorySwap x and y, solve for y — that's the whole algorithm.
Explanation
Set \(y=\frac{3x-1}{2}\), swap: \(x=\frac{3y-1}{2}\). Solve: \(2x=3y-1\Rightarrow 3y=2x+1\Rightarrow y=\frac{2x+1}{3}\). Option B has the wrong sign — a classic arithmetic slip.
Test \(f(-x)=-f(x)\). For D: \(f(-x)=(-x)^3-5(-x)=-x^3+5x=-(x^3-5x)=-f(x)\). ✓ Odd. Option B has both even (\(x^2\)) and odd (\(x^3\)) powers — it is neither. Option A is even.
Polynomials & Rational Functions
05
Remainder Theorem
When \(p(x)=x^3-4x^2+2x-1\) is divided by \((x-3)\), what is the remainder?
MemoryRemainder = p(divisor root) — plug in the zero of the divisor.
Explanation
By the Remainder Theorem, substitute \(x=3\): \(27 - 36 + 6 - 1 = -4\). No long division needed!
06
Rational Functions
What are the vertical and horizontal asymptotes of \(f(x)=\dfrac{2x^2}{x^2-9}\,\)?
MemoryVA: denom=0 · HA: compare degrees (n=m → ratio of leading coefficients)
Explanation
Denom \(x^2-9=(x-3)(x+3)\), so VA at \(x=\pm3\). Degrees equal (both 2), so HA = ratio of leading coefficients = \(\frac{2}{1}=2\). D is wrong because HA = 0 only when numerator degree < denominator degree.
07
Zeros of Polynomials
How many real zeros does \(f(x)=x^4+2x^2+1\) have?
MemoryFactor first, then count real solutions — complex roots come in conjugate pairs.
Explanation
\(x^4+2x^2+1=(x^2+1)^2\). Setting this to zero gives \(x^2=-1\), which has no real solutions. All 4 roots are imaginary (\(\pm i\), each with multiplicity 2). Students often guess 2 thinking of the double root.
\(\log_2 96 - \log_2 3 = \log_2\!\frac{96}{3} = \log_2 32 = \log_2 2^5 = 5\). Option C is wrong — you cannot subtract the numbers inside the logs directly (that would be \(\log_2(96-3)\)).
09
Exponential Equations
Solve for \(x\): \(\;4^{x+1}=8^{x-1}\)
MemorySame base → same exponent — rewrite both sides as powers of 2.
Using the change-of-base formula, which expression equals \(\log_5 17\)?
Memorylog_b(x) = ln(x)/ln(b) = log(x)/log(b) — new base goes on the bottom.
Explanation
Change of base: \(\log_5 17 = \frac{\ln 17}{\ln 5}\) (or \(\frac{\log 17}{\log 5}\)). Options A and C have the fraction flipped — a very common mistake. The original base (5) always goes in the denominator.
Trigonometry
11
Unit Circle
What is the exact value of \(\cos\!\left(\dfrac{5\pi}{6}\right)\)?
MemoryReference angle + quadrant sign — \(5\pi/6\) is in Q2, so cosine is negative.
Explanation
Reference angle of \(5\pi/6\) is \(\pi - 5\pi/6 = \pi/6\). \(\cos(\pi/6)=\frac{\sqrt{3}}{2}\). Since \(5\pi/6\) is in Quadrant II, cosine is negative: \(-\frac{\sqrt{3}}{2}\). Confusing with \(\sin\) gives \(\frac{1}{2}\) — option D, a classic mix-up.
12
Trig Identities
Which expression is equivalent to \(\dfrac{1-\cos^2\theta}{\sin\theta}\,\)?
\(1-\cos^2\theta = \sin^2\theta\) (Pythagorean identity). So \(\frac{\sin^2\theta}{\sin\theta}=\sin\theta\). The trap is thinking the numerator stays as is, leading to \(\tan\theta\) by accident.
13
Amplitude & Period
For \(y=3\sin(2x-\pi)+1\), what is the period?
MemoryPeriod = 2π / |B| — the coefficient of x is B, not the phase shift.
Explanation
\(B=2\), so Period \(=\frac{2\pi}{2}=\pi\). The amplitude is 3 (not the period) and the vertical shift is 1. Many students accidentally divide by 3 instead of 2.
14
Inverse Trig
Evaluate \(\arcsin\!\left(-\dfrac{1}{2}\right)\) in radians. Give the principal value.
Memoryarcsin range: [−π/2, π/2] — the answer must be in this interval, not Q3.
Explanation
\(\sin(-\pi/6) = -1/2\) and \(-\pi/6\) is in the principal range \([-\pi/2,\pi/2]\). Option C (\(-5\pi/6\)) is in Q3 — outside the range of arcsin. Students pick B thinking of reference angles in Q2.
Sequences & Series
15
Geometric Series
Find the sum of the infinite geometric series: \(12 + 4 + \dfrac{4}{3} + \cdots\)
MemoryS = a / (1 − r), valid only when |r| < 1. Check r first!
Explanation
Common ratio \(r = 4/12 = 1/3\). Since \(|r|<1\), the series converges. \(S = \frac{12}{1-\frac{1}{3}} = \frac{12}{\frac{2}{3}} = 12\cdot\frac{3}{2} = 18\). A common error is forgetting to subtract \(r\) from 1, giving 12 · 3 = 36.
16
Arithmetic Sequences
The 5th term of an arithmetic sequence is 17 and the 9th term is 33. What is the first term?
Memorya_n = a_1 + (n−1)d — two terms give two equations; solve the system.
Explanation
From terms 5 and 9: \(33-17=(9-5)d\Rightarrow d=4\). Then \(a_5=a_1+4d\Rightarrow 17=a_1+16\Rightarrow a_1=1\). Verify: 1, 5, 9, 13, 17 ✓. Students often mis-count the number of steps between terms.
Conic Sections & Vectors
17
Circle Equations
The equation \(x^2+y^2-6x+4y-3=0\) represents a circle. What is its radius?
MemoryComplete the square for both x and y — add (b/2)² to both sides each time.
Explanation
Complete the square: \((x^2-6x+9)+(y^2+4y+4)=3+9+4=16\). So \((x-3)^2+(y+2)^2=16\). Radius \(=\sqrt{16}=4\). D (16) is \(r^2\), not \(r\) — the most common mistake.
18
Parabolas
The parabola \(y=2(x-3)^2+5\) has vertex at:
Memoryy = a(x−h)² + k → vertex (h, k) — watch the sign of h: it flips!
Explanation
Vertex form \(y=a(x-h)^2+k\) gives vertex \((h,k)=(3,5)\). The most common error is using \(h=-3\) because the formula contains \((x-3)\) — students forget the sign flip. It's \(x-h=0\Rightarrow x=h=3\).
19
Vectors
Vectors \(\mathbf{u}=\langle 3,-1\rangle\) and \(\mathbf{v}=\langle 2,4\rangle\). What is \(\mathbf{u}\cdot\mathbf{v}\,\)?
MemoryDot product = multiply matching components, then ADD — result is a scalar, not a vector.
Explanation
\(\mathbf{u}\cdot\mathbf{v}=(3)(2)+(-1)(4)=6-4=2\). Option A is component multiplication without adding — that's not a dot product. Option C adds wrong: \(6+4=10\) ignores the negative sign.
20
Binomial Theorem
What is the coefficient of \(x^2\) in the expansion of \((x+3)^5\)?
Memory\(\binom{n}{k}a^{n-k}b^k\) — choose k, attach powers. Find the term where x-power = 2.
Explanation
We need \(x^2\), so \(k=3\) (since \(x^{5-k}=x^2\Rightarrow k=3\)). Term: \(\binom{5}{3}x^2\cdot3^3=10\cdot x^2\cdot 27=270x^2\). Coefficient is \(\mathbf{270}\). Option B forgets that \(3^3=27\), using 9 instead; Option C uses \(\binom{5}{2}=10\) but forgets \(3^3\).