1
Quadratic Functions Q1–Q2
📘 Concept
A quadratic function has the form $f(x)=ax^2+bx+c$.
Quadratic Formula: $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
The discriminant $\Delta = b^2 - 4ac$ tells you the nature of roots:
• $\Delta > 0$: two real roots • $\Delta = 0$: one real root • $\Delta < 0$: no real roots
• $\Delta > 0$: two real roots • $\Delta = 0$: one real root • $\Delta < 0$: no real roots
🌟 Memorize
- Vertex: $\left(-\dfrac{b}{2a},\; f\!\left(-\dfrac{b}{2a}\right)\right)$
- Vertex form: $f(x)=a(x-h)^2+k$, vertex at $(h,k)$
- Sum of roots = $-b/a$; Product of roots = $c/a$
📝 Example
Solve $2x^2 - 3x - 2 = 0$.
$x = \dfrac{3 \pm \sqrt{9+16}}{4} = \dfrac{3 \pm 5}{4}$ → $x = 2$ or $x = -\dfrac{1}{2}$
Question 1 Medium
The quadratic equation $3x^2 - 6x + k = 0$ has exactly one real solution. What is the value of $k$?
Question 2 Hard
If the vertex of $f(x) = 2x^2 - 8x + 7$ is at $(h, k)$, which of the following is true?
2
Polynomials Q3–Q4
📘 Concept
Remainder Theorem: The remainder when $p(x)$ is divided by $(x-c)$ equals $p(c)$.
Factor Theorem: $(x-c)$ is a factor of $p(x)$ if and only if $p(c)=0$.
End Behavior: Determined by the leading term $a_n x^n$.
Factor Theorem: $(x-c)$ is a factor of $p(x)$ if and only if $p(c)=0$.
End Behavior: Determined by the leading term $a_n x^n$.
🌟 Memorize
- Sum of $n$ roots = $-a_{n-1}/a_n$ (by Vieta's formulas)
- A degree-$n$ polynomial has at most $n$ real zeros
- Multiplicity: even → touches x-axis; odd → crosses x-axis
📝 Example
If $p(x) = x^3 - 4x^2 + x + 6$, find the remainder when divided by $(x - 3)$.
$p(3) = 27 - 36 + 3 + 6 = 0$ → Remainder = $0$ (so $(x-3)$ is a factor)
Question 3 Medium
When $p(x) = 2x^3 - 3x^2 + x - 5$ is divided by $(x - 2)$, what is the remainder?
Question 4 Hard
The polynomial $p(x) = x^3 + ax^2 - 4x - 4$ has $(x + 2)$ as a factor. What is the value of $a$?
3
Rational Functions Q5–Q6
📘 Concept
$f(x) = \dfrac{p(x)}{q(x)}$ where $p, q$ are polynomials, $q(x) \ne 0$.
Vertical asymptote: where $q(x) = 0$ (after cancelling common factors).
Horizontal asymptote:
• deg($p$) $<$ deg($q$): $y = 0$ • deg($p$) $=$ deg($q$): $y = \frac{\text{leading coefficients ratio}}{}$ • deg($p$) $>$ deg($q$): oblique asymptote
Vertical asymptote: where $q(x) = 0$ (after cancelling common factors).
Horizontal asymptote:
• deg($p$) $<$ deg($q$): $y = 0$ • deg($p$) $=$ deg($q$): $y = \frac{\text{leading coefficients ratio}}{}$ • deg($p$) $>$ deg($q$): oblique asymptote
🌟 Memorize
- Hole (removable discontinuity): factor cancels in both numerator & denominator
- Domain: all real $x$ where denominator $\ne 0$
📝 Example
Find the vertical asymptote(s) of $f(x) = \dfrac{x+1}{x^2-x-6}$.
$x^2-x-6=(x-3)(x+2)$ → vertical asymptotes at $x=3$ and $x=-2$
Question 5 Medium
What is the horizontal asymptote of $f(x) = \dfrac{4x^2 - 3}{2x^2 + 5x}$?
Question 6 Hard
The function $f(x) = \dfrac{x^2 - 9}{x^2 - x - 6}$ has which of the following features?
4
Radical & Rational Exponents Q7–Q8
📘 Concept
$a^{m/n} = \left(\sqrt[n]{a}\right)^m = \sqrt[n]{a^m}$
$\sqrt[n]{a}\cdot\sqrt[n]{b} = \sqrt[n]{ab}$, $\dfrac{\sqrt[n]{a}}{\sqrt[n]{b}} = \sqrt[n]{\dfrac{a}{b}}$
To solve radical equations: isolate the radical, then raise both sides to the appropriate power. Always check for extraneous solutions!
🌟 Memorize
- $x^{1/2} = \sqrt{x}$; $x^{1/3} = \sqrt[3]{x}$; $x^{-n} = 1/x^n$
- Domain of $\sqrt[n]{f(x)}$: $f(x) \ge 0$ when $n$ is even
📝 Example
Solve: $\sqrt{2x+3} = x - 1$
Square both sides: $2x+3 = x^2-2x+1$ → $x^2-4x-2=0$... Wait — let's use simpler: $x^2-4x-2=0$, $x=2\pm\sqrt{6}$. Check: only $x=2+\sqrt{6}\approx4.45$ satisfies $x\ge1$. ✓
Question 7 Easy
Simplify: $\left(27x^6\right)^{2/3}$
Question 8 Medium
Solve for $x$: $\sqrt{3x - 2} = x - 2$. Which value(s) of $x$ are valid solutions?
5
Exponential Functions Q9–Q10
📘 Concept
$f(x)=ab^x$: $a$ = initial value, $b$ = base ($b>0, b\ne1$).
• $b>1$: exponential growth • $0Compound Interest: $A = P\!\left(1+\frac{r}{n}\right)^{nt}$, Continuous: $A = Pe^{rt}$
• $b>1$: exponential growth • $0
🌟 Memorize
- $b^x = b^y \Leftrightarrow x = y$ (same base property)
- Domain: all reals; Range: $(0,\infty)$ for $f(x)=b^x$
- $e \approx 2.718$ (Euler's number)
📝 Example
Solve: $4^{x+1} = 32$
$2^{2(x+1)} = 2^5$ → $2x+2=5$ → $x = \dfrac{3}{2}$
Question 9 Easy
Solve for $x$: $2^{3x-1} = 64$
Question 10 Medium
A population of bacteria doubles every 3 hours. Starting with 500 bacteria, which expression gives the population after $t$ hours?
6
Logarithmic Functions Q11–Q12
📘 Concept
$\log_b x = y \;\Leftrightarrow\; b^y = x$ ($b>0,\, b\ne1,\, x>0$)
Product: $\log_b(mn)=\log_b m+\log_b n$
Quotient: $\log_b\!\left(\frac{m}{n}\right)=\log_b m-\log_b n$
Power: $\log_b(m^r)=r\log_b m$
Change of Base: $\log_b x = \dfrac{\ln x}{\ln b}$
Quotient: $\log_b\!\left(\frac{m}{n}\right)=\log_b m-\log_b n$
Power: $\log_b(m^r)=r\log_b m$
Change of Base: $\log_b x = \dfrac{\ln x}{\ln b}$
🌟 Memorize
- $\log_{10}10 = 1$; $\ln e = 1$; $\log_b 1 = 0$; $\log_b b = 1$
- $b^{\log_b x} = x$; $\log_b(b^x) = x$ (inverse relationship)
📝 Example
Solve: $\log_2(x+3) + \log_2(x-1) = 5$
$\log_2[(x+3)(x-1)]=5$ → $(x+3)(x-1)=32$ → $x^2+2x-3=32$ → $x^2+2x-35=0$ → $(x+7)(x-5)=0$ → $x=5$ (reject $x=-7$ since $x-1<0$) ✓
Question 11 Medium
Solve for $x$: $\log_3(2x - 1) = 4$
Question 12 Hard
Which expression is equivalent to $\log_2 \dfrac{8\sqrt{x}}{y^3}$?
7
Complex Numbers Q13–Q14
📘 Concept
$i = \sqrt{-1}$, $i^2 = -1$, $i^3 = -i$, $i^4 = 1$ (cycle of 4)
Complex number: $a + bi$ where $a$ = real part, $b$ = imaginary part. Conjugate: $\overline{a+bi} = a-bi$ Modulus: $|a+bi|=\sqrt{a^2+b^2}$ To divide: multiply numerator & denominator by the conjugate of the denominator.
Complex number: $a + bi$ where $a$ = real part, $b$ = imaginary part. Conjugate: $\overline{a+bi} = a-bi$ Modulus: $|a+bi|=\sqrt{a^2+b^2}$ To divide: multiply numerator & denominator by the conjugate of the denominator.
🌟 Memorize
- $i^{4k}=1$, $i^{4k+1}=i$, $i^{4k+2}=-1$, $i^{4k+3}=-i$
- $(a+bi)(a-bi) = a^2+b^2$ (always real!)
📝 Example
Simplify: $\dfrac{3+2i}{1-i}$
Multiply by $\dfrac{1+i}{1+i}$: $\dfrac{(3+2i)(1+i)}{1+1} = \dfrac{3+3i+2i+2i^2}{2} = \dfrac{3+5i-2}{2} = \dfrac{1+5i}{2} = \dfrac{1}{2}+\dfrac{5}{2}i$
Question 13 Easy
What is the value of $i^{47}$?
Question 14 Medium
Which of the following is equal to $\dfrac{2-3i}{1+2i}$?
8
Sequences & Series Q15–Q16
📘 Concept
Arithmetic: $a_n = a_1 + (n-1)d$; $S_n = \dfrac{n}{2}(a_1+a_n)$
Geometric: $a_n = a_1 \cdot r^{n-1}$; $S_n = a_1\cdot\dfrac{1-r^n}{1-r}$ ($r\ne1$) Infinite Geometric Sum (only if $|r|<1$): $S = \dfrac{a_1}{1-r}$
Geometric: $a_n = a_1 \cdot r^{n-1}$; $S_n = a_1\cdot\dfrac{1-r^n}{1-r}$ ($r\ne1$) Infinite Geometric Sum (only if $|r|<1$): $S = \dfrac{a_1}{1-r}$
🌟 Memorize
- Arithmetic: common difference $d = a_{n+1} - a_n$
- Geometric: common ratio $r = a_{n+1}/a_n$
- If $|r| \ge 1$, infinite geometric series diverges (no finite sum)
📝 Example
Find the sum of the infinite geometric series: $12 + 4 + \dfrac{4}{3} + \cdots$
$r = 4/12 = 1/3$; $S = \dfrac{12}{1-1/3} = \dfrac{12}{2/3} = 18$
Question 15 Medium
The first term of a geometric sequence is $5$ and the common ratio is $-2$. What is the $6^{\text{th}}$ term?
Question 16 Hard
The sum of the first $n$ terms of an arithmetic sequence is $S_n = 3n^2 + 2n$. What is the $10^{\text{th}}$ term?
9
Conic Sections Q17–Q18
📘 Concept
Circle: $(x-h)^2+(y-k)^2=r^2$, center $(h,k)$, radius $r$
Ellipse: $\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1$
Parabola: $y-k=a(x-h)^2$ or $x-h=a(y-k)^2$
Hyperbola: $\dfrac{(x-h)^2}{a^2}-\dfrac{(y-k)^2}{b^2}=1$
Ellipse: $\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1$
Parabola: $y-k=a(x-h)^2$ or $x-h=a(y-k)^2$
Hyperbola: $\dfrac{(x-h)^2}{a^2}-\dfrac{(y-k)^2}{b^2}=1$
🌟 Memorize
- Circle: $a^2=b^2$; Ellipse: $a^2>b^2$ (horizontal major axis when $a$ under $x$)
- Complete the square to convert general form → standard form
- Hyperbola asymptotes: $y-k = \pm\dfrac{b}{a}(x-h)$
📝 Example
Write in standard form: $x^2+y^2-6x+4y-3=0$
$(x^2-6x+9)+(y^2+4y+4)=3+9+4$ → $(x-3)^2+(y+2)^2=16$. Circle, center $(3,-2)$, $r=4$.
Question 17 Medium
The equation $\dfrac{(x-2)^2}{16}+\dfrac{(y+1)^2}{9}=1$ represents which conic, and what is its center?
Question 18 Hard
The equation $x^2 - 4x - y^2 + 6y = 6$, when written in standard form, represents which conic?
10
Systems of Equations Q19–Q20
📘 Concept
Methods: Substitution, Elimination, Matrices.
For a system $Ax=b$: unique solution if $\det(A)\ne0$.
A system of linear & quadratic equations can have $0$, $1$, or $2$ intersection points.
For a system $Ax=b$: unique solution if $\det(A)\ne0$.
A system of linear & quadratic equations can have $0$, $1$, or $2$ intersection points.
🌟 Memorize
- Consistent & independent: exactly one solution
- Consistent & dependent: infinitely many solutions
- Inconsistent: no solution (parallel lines)
📝 Example
Solve: $y = x^2 - 3$ and $y = x + 1$
$x^2-3=x+1$ → $x^2-x-4=0$ → $x=\dfrac{1\pm\sqrt{17}}{2}$. Two intersection points.
Question 19 Medium
How many solutions does the system $\begin{cases}y = x^2 - 4x + 4\\y = 2x - 4\end{cases}$ have?
Question 20 Hard
For what value of $k$ does the system $\begin{cases}3x - ky = 7\\6x - 4y = 14\end{cases}$ have infinitely many solutions?
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Quiz Complete!
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📖 Answer Key & Full Solutions
Detailed step-by-step explanations for all 20 questions
Q1 — Quadratic Discriminant
✅ Correct Answer: C — $k = 3$
For exactly one real solution, the discriminant must equal zero: $\Delta = b^2 - 4ac = 0$.
Here $a=3,\; b=-6,\; c=k$: $(-6)^2 - 4(3)(k) = 0$
$36 - 12k = 0 \;\Rightarrow\; k = 3$ ✓
Q2 — Vertex of a Quadratic
✅ Correct Answer: A — $h=2,\; k=-1$
The $x$-coordinate of the vertex: $h = -\dfrac{b}{2a} = -\dfrac{-8}{2(2)} = 2$.
$k = f(2) = 2(4) - 8(2) + 7 = 8 - 16 + 7 = -1$
Vertex: $(2, -1)$ ✓
Q3 — Remainder Theorem
✅ Correct Answer: B — $1$
By the Remainder Theorem, the remainder when $p(x)$ is divided by $(x-2)$ equals $p(2)$.
$p(2) = 2(2)^3 - 3(2)^2 + (2) - 5 = 16 - 12 + 2 - 5 = 1$ ✓
Q4 — Factor Theorem
✅ Correct Answer: B — $a = 1$
If $(x+2)$ is a factor, then $p(-2) = 0$.
$p(-2) = (-2)^3 + a(-2)^2 - 4(-2) - 4 = -8 + 4a + 8 - 4 = 4a - 4$
$4a - 4 = 0 \;\Rightarrow\; a = 1$ ✓
Q5 — Horizontal Asymptote
✅ Correct Answer: B — $y = 2$
Degrees of numerator and denominator are both $2$ (equal). The horizontal asymptote is the ratio of leading coefficients.
Leading coefficient of numerator: $4$; of denominator: $2$
Horizontal asymptote: $y = \dfrac{4}{2} = 2$ ✓
Q6 — Holes vs. Asymptotes
✅ Correct Answer: B — Hole at $x=3$, vertical asymptote at $x=-2$
Factor both numerator and denominator:
Numerator: $x^2-9 = (x-3)(x+3)$
Denominator: $x^2-x-6 = (x-3)(x+2)$
$(x-3)$ cancels → hole at $x=3$; $(x+2)$ remains → vertical asymptote at $x=-2$ ✓
Q7 — Rational Exponents
✅ Correct Answer: A — $9x^4$
Apply the exponent to each factor:
$(27)^{2/3} = \left(3^3\right)^{2/3} = 3^2 = 9$
$(x^6)^{2/3} = x^{6 \cdot 2/3} = x^4$
Result: $9x^4$ ✓
Q8 — Radical Equations
✅ Correct Answer: B — $x = 6$ only
Square both sides (requires $x-2 \ge 0$, so $x \ge 2$):
$3x - 2 = (x-2)^2 = x^2 - 4x + 4$
$x^2 - 7x + 6 = 0 \;\Rightarrow\; (x-1)(x-6) = 0 \;\Rightarrow\; x=1 \text{ or } x=6$
Check $x=1$: $\sqrt{1}\ne -1$ ✗ (extraneous). Check $x=6$: $\sqrt{16}=4=6-2$ ✓
Valid solution: $x = 6$ only ✓
Q9 — Exponential Equations (Same Base)
✅ Correct Answer: B — $x = \dfrac{7}{3}$
Write $64 = 2^6$, so $2^{3x-1} = 2^6$.
Set exponents equal: $3x - 1 = 6$
$3x = 7 \;\Rightarrow\; x = \dfrac{7}{3}$ ✓
Q10 — Exponential Growth Model
✅ Correct Answer: B — $P = 500 \cdot 2^{t/3}$
The population doubles every 3 hours. After $t$ hours, the number of doubling periods is $t/3$.
$P(t) = 500 \cdot 2^{t/3}$
Check: $P(3) = 500 \cdot 2^1 = 1000$ ✓ (doubled after 3 hours)
Q11 — Logarithmic Equation
✅ Correct Answer: A — $x = 41$
Convert to exponential form: $\log_3(2x-1)=4 \;\Rightarrow\; 2x-1=3^4$.
$2x - 1 = 81$
$2x = 82 \;\Rightarrow\; x = 41$ ✓
Q12 — Log Properties (Expansion)
✅ Correct Answer: A — $3 + \dfrac{1}{2}\log_2 x - 3\log_2 y$
$\log_2\dfrac{8\sqrt{x}}{y^3} = \log_2 8 + \log_2 x^{1/2} - \log_2 y^3$
$= \log_2 2^3 + \tfrac{1}{2}\log_2 x - 3\log_2 y$
$= 3 + \tfrac{1}{2}\log_2 x - 3\log_2 y$ ✓
Q13 — Powers of $i$
✅ Correct Answer: D — $-i$
The powers of $i$ cycle with period 4: $i^1=i,\; i^2=-1,\; i^3=-i,\; i^4=1$.
$47 \div 4 = 11$ remainder $3$, so $i^{47} = i^3 = -i$ ✓
Q14 — Complex Division
✅ Correct Answer: A — $-\dfrac{4}{5} - \dfrac{7}{5}i$
Multiply numerator and denominator by the conjugate $1 - 2i$:
$\dfrac{(2-3i)(1-2i)}{(1+2i)(1-2i)} = \dfrac{2-4i-3i+6i^2}{1+4} = \dfrac{2-7i-6}{5} = \dfrac{-4-7i}{5}$
$= -\dfrac{4}{5} - \dfrac{7}{5}i$ ✓
Q15 — Geometric Sequence
✅ Correct Answer: B — $-160$
$a_n = a_1 \cdot r^{n-1}$, with $a_1=5$, $r=-2$, $n=6$:
$a_6 = 5 \cdot (-2)^5 = 5 \cdot (-32) = -160$ ✓
Q16 — Finding the $n$th Term from $S_n$
✅ Correct Answer: B — $59$
The $n$th term is $a_n = S_n - S_{n-1}$.
$S_{10} = 3(100)+2(10) = 320$
$S_9 = 3(81)+2(9) = 243+18 = 261$
$a_{10} = 320 - 261 = 59$ ✓
Alternative: Since $S_n=3n^2+2n$, the general term is $a_n=S_n-S_{n-1}=6n-1$, so $a_{10}=59$.
Q17 — Identifying Conics
✅ Correct Answer: A — Ellipse, center $(2, -1)$
The equation $\dfrac{(x-2)^2}{16}+\dfrac{(y+1)^2}{9}=1$ is in ellipse standard form (sum of two squared terms over different positive constants equals 1).
Center: $(h, k) = (2, -1)$; $a^2=16$, $b^2=9$ (not equal, so ellipse not circle) ✓
Q18 — Completing the Square for Conics
✅ Correct Answer: C — Hyperbola centered at $(2, 3)$
Complete the square for $x$ and $y$:
$(x^2-4x+4) - (y^2-6y+9) = 6 + 4 - 9 = 1$
$(x-2)^2 - (y-3)^2 = 1$
This is $\dfrac{(x-2)^2}{1}-\dfrac{(y-3)^2}{1}=1$: a hyperbola centered at $(2,3)$ ✓
Q19 — System: Linear & Quadratic
✅ Correct Answer: C — $2$ solutions
Set the expressions equal: $x^2 - 4x + 4 = 2x - 4$.
$x^2 - 4x + 4 - 2x + 4 = 0 \;\Rightarrow\; x^2 - 6x + 8 = 0$
$(x-2)(x-4) = 0 \;\Rightarrow\; x = 2$ or $x = 4$
Two distinct real solutions. Discriminant $= 36 - 32 = 4 > 0$ ✓
Q20 — Infinitely Many Solutions (Systems)
✅ Correct Answer: B — $k = 2$
For infinitely many solutions, the two equations must be identical (proportional).
Multiply equation 1 by $2$: $6x - 2ky = 14$
Compare with equation 2: $6x - 4y = 14$
For these to be identical: $2k = 4 \;\Rightarrow\; k = 2$ ✓