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Self-Study Edition · All Major Units

Master
Pre-Calculus
Step by Step

20 carefully crafted problems across every core unit. Each question comes with a quick memory key — the smartest shortcut to the right answer.

20 Problems
8 Core Units
Easy–Medium Level
Unit 1
Functions & Their Properties
Q 01
Domain & Range ⚠ Common Trap
🔑 MEMORY KEY: "Square root → inside ≥ 0 · Denominator → ≠ 0"

Find the domain of the function:

\( f(x) = \dfrac{\sqrt{x-3}}{x-5} \)

⚠ Don't forget — you have TWO restrictions here!

📖 Quick Example
For \(g(x)=\sqrt{x-1}\): need \(x-1\geq 0\), so domain is \([1,\infty)\).
For \(h(x)=\frac{1}{x-2}\): need \(x\neq 2\), so domain is \((-\infty,2)\cup(2,\infty)\).
📘Explanation
We need two conditions:
① Square root: \(x - 3 \geq 0 \Rightarrow x \geq 3\)
② Denominator: \(x - 5 \neq 0 \Rightarrow x \neq 5\)
Combining both: \(x \geq 3\) AND \(x \neq 5\), which gives \([3,5)\cup(5,\infty)\).
Common mistake: Forgetting to exclude \(x=5\) since \(\sqrt{5-3}=\sqrt{2}\neq 0\) but the denominator becomes 0.
Q 02
Composition 🔥 High-Yield
🔑 MEMORY KEY: "(f∘g)(x) = f( g(x) ) — plug g INTO f, not the other way!"

Given \(f(x) = 2x + 1\) and \(g(x) = x^2 - 3\), find \((f \circ g)(2)\).

📖 Quick Example
If \(f(x)=x+1\) and \(g(x)=3x\), then \((f\circ g)(2) = f(g(2)) = f(6) = 7\).
📘Explanation
Step 1: Compute \(g(2) = 2^2 - 3 = 4 - 3 = 1\).
Step 2: Compute \(f(g(2)) = f(1) = 2(1)+1 = 3\).
Answer: \(\mathbf{3}\).
Common mistake: Reversing order — computing \(g(f(2))\) gives \(g(5) = 22\), which is wrong.
Unit 2
Polynomials & Factoring
Q 03
Zeros ⚠ Common Trap
🔑 MEMORY KEY: "Set f(x)=0, factor, solve each piece → zeros are your x-intercepts"

Find all real zeros of \(f(x) = x^3 - 4x\).

📖 Quick Example
\(g(x) = x^2 - 9 = (x-3)(x+3)\) → zeros at \(x=3\) and \(x=-3\).
📘Explanation
Factor out \(x\): \(f(x) = x(x^2 - 4) = x(x-2)(x+2)\).
Set each factor to zero: \(x=0\), \(x=2\), \(x=-2\).
Common mistake: Dividing both sides by \(x\) and losing \(x=0\) as a solution!
Q 04
Remainder Theorem 🔥 High-Yield
🔑 MEMORY KEY: "Remainder Theorem: divide p(x) by (x−a) → remainder = p(a)"

When \(p(x) = x^3 - 2x^2 + 3x - 1\) is divided by \((x-2)\), what is the remainder?

📖 Quick Example
Dividing \(q(x)=x^2+3x\) by \((x-1)\): remainder \(= q(1) = 1+3 = 4\). Done — no long division needed!
📘Explanation
By the Remainder Theorem, the remainder when dividing by \((x-2)\) is \(p(2)\).
\(p(2) = 2^3 - 2(2^2) + 3(2) - 1 = 8 - 8 + 6 - 1 = 5\).
Answer: \(\mathbf{5}\).
Unit 3
Rational Functions & Asymptotes
Q 05
Horizontal Asymptote ⚠ Common Trap
🔑 MEMORY KEY: "Compare degrees: n<m → y=0, n=m → y=leading ratio, n>m → NO HA"

Find the horizontal asymptote of \( f(x) = \dfrac{3x^2 + 1}{x^2 - 4} \).

📖 Quick Example
\(\dfrac{5x^2}{2x^2+1}\): degrees equal → HA is \(y = \dfrac{5}{2}\).
\(\dfrac{2x}{x^2+1}\): degree top < bottom → HA is \(y = 0\).
📘Explanation
Numerator degree = 2, denominator degree = 2 → equal degrees.
Horizontal asymptote = ratio of leading coefficients = \(\dfrac{3}{1} = 3\).
So \(y = 3\).
Common mistake: Thinking \(y=0\) is always the HA — that only happens when the numerator degree is strictly less than the denominator degree.
Q 06
Vertical Asymptote vs. Hole ⚠ Common Trap
🔑 MEMORY KEY: "Cancel factor = HOLE · Remaining zero in denom = VERTICAL ASYMPTOTE"

For \( f(x) = \dfrac{x^2 - 1}{x^2 - x - 2} \), identify any holes and vertical asymptotes.

Factor both numerator and denominator first!

📘Explanation
Factor: numerator \(= (x-1)(x+1)\), denominator \(= (x-2)(x+1)\).
The \((x+1)\) cancels → Hole at \(x = -1\).
Remaining denominator factor \((x-2)=0\) → VA at \(x=2\).
Answer: Hole at \(x=-1\), VA at \(x=2\).
Unit 4
Exponential & Logarithmic Functions
Q 07
Log Properties 🔥 High-Yield
🔑 MEMORY KEY: "log(AB)=logA+logB · log(A/B)=logA−logB · log(Aⁿ)=n·logA"

Simplify: \( \log_2 8 + \log_2 4 \)

📖 Quick Example
\(\log_3 9 + \log_3 3 = \log_3(9 \cdot 3) = \log_3 27 = 3\), since \(3^3=27\).
📘Explanation
Method 1 (Direct): \(\log_2 8 = 3\) (since \(2^3=8\)), \(\log_2 4 = 2\) (since \(2^2=4\)). Sum \(= 5\).
Method 2 (Log rule): \(\log_2 8 + \log_2 4 = \log_2(8\cdot 4) = \log_2 32 = 5\) (since \(2^5=32\)). ✓
Q 08
Exponential Equation ⚠ Common Trap
🔑 MEMORY KEY: "Same base → set exponents equal · Different base → take log of both sides"

Solve for \(x\): \( 4^x = 8 \)

Hint: Write both sides with base 2.

📖 Quick Example
\(9^x = 27\): Write as \((3^2)^x = 3^3\), so \(3^{2x} = 3^3\), giving \(2x=3\), \(x=\frac{3}{2}\).
📘Explanation
Rewrite: \(4^x = (2^2)^x = 2^{2x}\) and \(8 = 2^3\).
So \(2^{2x} = 2^3 \Rightarrow 2x = 3 \Rightarrow x = \dfrac{3}{2}\).
Common mistake: Writing \(4^x = 8\) as \(x = 8/4 = 2\) — this ignores the exponent structure entirely!
Q 09
Log Equation 🔥 High-Yield
🔑 MEMORY KEY: "log_b(x) = y ↔ b^y = x — always CHECK for extraneous solutions!"

Solve: \( \log_3(x+6) + \log_3 x = 3 \)

⚠ Check your answer — negative values inside a log are undefined!

📘Explanation
Combine logs: \(\log_3[x(x+6)] = 3 \Rightarrow x(x+6) = 3^3 = 27\).
\(x^2 + 6x - 27 = 0 \Rightarrow (x+9)(x-3) = 0 \Rightarrow x=-9\) or \(x=3\).
Check: \(x=-9\) makes \(\log_3(-9)\) undefined → rejected.
Only \(x = 3\) is valid.
Unit 5
Trigonometry
Q 10
Unit Circle ⚠ Common Trap
🔑 MEMORY KEY: "All Students Take Calculus → Q1:All+, Q2:Sin+, Q3:Tan+, Q4:Cos+"

If \(\sin\theta = -\dfrac{\sqrt{3}}{2}\) and \(\theta\) is in Quadrant III, find \(\cos\theta\).

📖 Quick Example
\(\sin\theta = \frac{1}{2}\) in Q2 → \(\cos\theta = -\frac{\sqrt{3}}{2}\) (cosine is negative in Q2).
📘Explanation
Use identity: \(\sin^2\theta + \cos^2\theta = 1\).
\(\left(\frac{\sqrt{3}}{2}\right)^2 + \cos^2\theta = 1 \Rightarrow \frac{3}{4} + \cos^2\theta = 1 \Rightarrow \cos^2\theta = \frac{1}{4}\).
So \(\cos\theta = \pm\frac{1}{2}\). In Quadrant III, cosine is negative.
Therefore \(\cos\theta = -\dfrac{1}{2}\).
Q 11
Trig Identity 🔥 High-Yield
🔑 MEMORY KEY: "Pythagorean: sin²+cos²=1 · 1+tan²=sec² · 1+cot²=csc²"

Simplify: \( \dfrac{\sin^2\theta + \cos^2\theta}{\cos^2\theta} \)

📘Explanation
\(\dfrac{\sin^2\theta + \cos^2\theta}{\cos^2\theta} = \dfrac{1}{\cos^2\theta} = \sec^2\theta\).
(Used Pythagorean identity: \(\sin^2\theta + \cos^2\theta = 1\).)
Alternatively, split the fraction: \(\dfrac{\sin^2\theta}{\cos^2\theta} + \dfrac{\cos^2\theta}{\cos^2\theta} = \tan^2\theta + 1 = \sec^2\theta\). ✓
Q 12
Amplitude & Period ⚠ Common Trap
🔑 MEMORY KEY: "y=A·sin(Bx+C)+D → |A|=amplitude, Period=2π/|B|"

For \(y = 3\sin(2x - \pi) + 1\), state the amplitude and period.

📘Explanation
In \(y = A\sin(Bx + C) + D\): here \(A = 3\), \(B = 2\).
Amplitude \(= |A| = |3| = 3\).
Period \(= \dfrac{2\pi}{|B|} = \dfrac{2\pi}{2} = \pi\).
The \(+1\) is a vertical shift (affects midline, not amplitude). The \(-\pi\) is a phase shift (affects horizontal position, not period).
Unit 6
Sequences & Series
Q 13
Arithmetic Sequence nth Term
🔑 MEMORY KEY: "Arithmetic nth term: aₙ = a₁ + (n−1)d — find d FIRST"

In an arithmetic sequence, \(a_1 = 5\) and the common difference \(d = -3\). Find \(a_{10}\).

📖 Quick Example
If \(a_1 = 2\), \(d = 4\): \(a_5 = 2 + (5-1)(4) = 2 + 16 = 18\).
📘Explanation
\(a_{10} = a_1 + (10-1)d = 5 + 9(-3) = 5 - 27 = -22\).
Common mistake: Using \(n\) instead of \((n-1)\), giving \(5 + 10(-3) = -25\) — off by one!
Q 14
Geometric Series 🔥 High-Yield
🔑 MEMORY KEY: "Infinite Geometric Sum = a₁/(1−r), valid only when |r| < 1"

Find the sum of the infinite geometric series: \( 12 + 4 + \dfrac{4}{3} + \cdots \)

First find \(r = \dfrac{\text{2nd term}}{\text{1st term}}\).

📘Explanation
Common ratio: \(r = \dfrac{4}{12} = \dfrac{1}{3}\). Since \(|r| < 1\), the series converges.
\(S = \dfrac{a_1}{1-r} = \dfrac{12}{1 - \frac{1}{3}} = \dfrac{12}{\frac{2}{3}} = 12 \times \dfrac{3}{2} = 18\).
Unit 7
Conic Sections
Q 15
Parabola ⚠ Common Trap
🔑 MEMORY KEY: "Vertex form: y=a(x−h)²+k → vertex is (h, k) · sign of h flips!"

Find the vertex of \( f(x) = 2(x-3)^2 + 5 \).

⚠ Watch the sign — it's \((x-3)\), not \((x+3)\)!

📘Explanation
In vertex form \(y = a(x-h)^2 + k\), the vertex is \((h, k)\).
Here: \(a=2\), \(h=3\), \(k=5\) → vertex is \((3, 5)\).
Common mistake: Reading \((x-3)\) as \(h=-3\). Remember: the sign flips! \((x-h)\) with \(h=3\) gives \((x-3)\).
Q 16
Ellipse 🔥 High-Yield
🔑 MEMORY KEY: "Ellipse: x²/a² + y²/b² = 1 → larger denom = major axis direction"

For the ellipse \( \dfrac{x^2}{25} + \dfrac{y^2}{9} = 1 \), what are the lengths of the major and minor axes?

📘Explanation
\(a^2 = 25 \Rightarrow a = 5\); \(b^2 = 9 \Rightarrow b = 3\).
Length of major axis \(= 2a = 10\) (along x-axis since 25 > 9).
Length of minor axis \(= 2b = 6\).
Common mistake: Confusing the semi-axis (\(a=5\)) with the full axis length (\(2a=10\)).
Unit 8
Introduction to Limits
Q 17
Evaluating Limits ⚠ Common Trap
🔑 MEMORY KEY: "0/0 indeterminate → FACTOR and cancel · then plug in"

Evaluate: \( \displaystyle\lim_{x \to 2} \dfrac{x^2 - 4}{x - 2} \)

⚠ Direct substitution gives \(\frac{0}{0}\) — that's your signal to factor!

📖 Quick Example
\(\displaystyle\lim_{x\to 3}\frac{x^2-9}{x-3} = \lim_{x\to 3}\frac{(x-3)(x+3)}{x-3} = \lim_{x\to 3}(x+3) = 6\).
📘Explanation
Factor: \(\dfrac{x^2-4}{x-2} = \dfrac{(x-2)(x+2)}{x-2} = x+2\) (for \(x \neq 2\)).
Now substitute: \(\displaystyle\lim_{x\to 2}(x+2) = 4\).
Common mistake: Saying the limit is "undefined" because \(f(2)\) is undefined. Limits describe behavior approaching the point, not AT it.
Q 18
Limits at Infinity 🔥 High-Yield
🔑 MEMORY KEY: "Divide top & bottom by highest power of x → small terms → 0"

Evaluate: \( \displaystyle\lim_{x \to \infty} \dfrac{5x^2 + 3}{2x^2 - 7} \)

📘Explanation
Divide numerator and denominator by \(x^2\):
\(\dfrac{5 + 3/x^2}{2 - 7/x^2}\). As \(x\to\infty\), \(\frac{3}{x^2}\to 0\) and \(\frac{7}{x^2}\to 0\).
Result: \(\dfrac{5+0}{2-0} = \dfrac{5}{2}\).
This matches the "leading coefficients ratio" shortcut for equal-degree rational functions.
Q 19
Inverse Functions ⚠ Common Trap
🔑 MEMORY KEY: "To find inverse: swap x and y, then solve for y — f⁻¹ is NOT 1/f(x)!"

Find the inverse of \( f(x) = 3x - 6 \).

📖 Quick Example
\(f(x)=2x+4\): swap → \(x=2y+4\) → \(y=\frac{x-4}{2}\) → \(f^{-1}(x)=\frac{x-4}{2}\).
📘Explanation
Let \(y = 3x - 6\). Swap \(x\) and \(y\): \(x = 3y - 6\).
Solve for \(y\): \(3y = x + 6 \Rightarrow y = \dfrac{x+6}{3}\).
So \(f^{-1}(x) = \dfrac{x+6}{3}\).
Common mistake (A): Writing \(\frac{1}{f(x)}\) as the inverse — that's the reciprocal, not the inverse function!
Q 20
Binomial Theorem 🔥 Finale
🔑 MEMORY KEY: "rth term of (a+b)ⁿ = C(n, r−1)·aⁿ⁻ʳ⁺¹·bʳ⁻¹ — use Pascal's or nCr"

Find the coefficient of \(x^2\) in the expansion of \((x + 3)^4\).

Use the Binomial Theorem: \((a+b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k\)

📖 Quick Example
In \((x+2)^3\): the \(x^1\) term has \(k=2\), so \(\binom{3}{2}x^1 \cdot 2^2 = 3\cdot x\cdot 4 = 12x\). Coefficient is 12.
📘Explanation
We want the term with \(x^2\): need \(x^{4-k} = x^2\), so \(k=2\).
Term \(= \binom{4}{2} x^2 \cdot 3^2 = 6 \cdot x^2 \cdot 9 = 54x^2\).
Coefficient is 54.
\(\binom{4}{2} = \dfrac{4!}{2!\,2!} = 6\). Then multiply by \(3^2=9\). So \(6\times 9 = 54\).
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