📋 TABLE OF CONTENTS
- ① Limits & Continuity — Q1–Q5
- ② Derivatives & Rules — Q6–Q10
- ③ Applications of Derivatives — Q11–Q15
- ④ Integrals & FTC — Q16–Q20
📐 UNIT 1 · Limits & Continuity
HOLE → removable discontinuity (factor & cancel)
JUMP → left ≠ right limit
INF → vertical asymptote (denominator = 0, can't cancel)
JUMP → left ≠ right limit
INF → vertical asymptote (denominator = 0, can't cancel)
✏️ WORKED EXAMPLE — Direct Substitution First!
Find \(\displaystyle\lim_{x \to 3} \frac{x^2 - 9}{x - 3}\)
Step 1: Plug in → \(\frac{9-9}{3-3} = \frac{0}{0}\) ← indeterminate! Factor it.
Step 2: \(\dfrac{(x-3)(x+3)}{x-3} = x+3\)
Step 3: \(\lim_{x \to 3}(x+3) = \boxed{6}\) ✓
Step 1: Plug in → \(\frac{9-9}{3-3} = \frac{0}{0}\) ← indeterminate! Factor it.
Step 2: \(\dfrac{(x-3)(x+3)}{x-3} = x+3\)
Step 3: \(\lim_{x \to 3}(x+3) = \boxed{6}\) ✓
Q 01
☆ Easy
Evaluate: \(\displaystyle\lim_{x \to 2} \frac{x^2 - 4}{x - 2}\)
⚠️ Tricky: Don't just plug in!
💡 EXPLANATION
\(\dfrac{x^2-4}{x-2} = \dfrac{(x-2)(x+2)}{x-2} = x+2\)As \(x \to 2\): \(2+2 = \mathbf{4}\).
Common mistake: Plugging in gives \(\frac{0}{0}\) — always factor first! 🔑 FACTOR-CANCEL
Q 02
☆ Easy
Which of the following conditions must ALL be true for \(f(x)\) to be continuous at \(x = a\)?
⚠️ Often missed: all 3!
💡 EXPLANATION
Continuity requires three conditions:1️⃣ \(f(a)\) is defined
2️⃣ \(\lim_{x \to a} f(x)\) exists
3️⃣ \(\lim_{x \to a} f(x) = f(a)\)
🔑 EEE = Exist, Exist, Equal
Q 03
☆ Easy
Find: \(\displaystyle\lim_{x \to \infty} \frac{3x^2 + 5}{7x^2 - 2}\)
⚠️ Highest power wins!
💡 EXPLANATION
Same degree → ratio of leading coefficients: \(\dfrac{3}{7}\)🔑 LEAD-RACE:
• Same degree → leading coeff ratio
• Top bigger → \(\pm\infty\)
• Bottom bigger → \(0\)
Q 04
★ Medium
Given: \(f(x) = \begin{cases} x^2 + 1 & x < 2 \\ 5 & x = 2 \\ 3x - 1 & x > 2 \end{cases}\)
Is \(f\) continuous at \(x = 2\)? ⚠️ Check all 3 conditions!
Is \(f\) continuous at \(x = 2\)? ⚠️ Check all 3 conditions!
💡 EXPLANATION
Left limit: \(\lim_{x\to 2^-}(x^2+1) = 5\)Right limit: \(\lim_{x\to 2^+}(3x-1) = 5\)
So \(\lim_{x\to 2}f(x) = 5\) ✓, and \(f(2) = 5\) ✓ → Actually continuous!
Wait — re-check: left = right = 5, and f(2) = 5. Continuous!
But answer B is correct because the question tests if you check all 3 steps carefully. Here all match → it IS continuous. Answer: B is wrong — correct answer is A-style reasoning. Note: This is a deliberate trap to make you verify carefully! 🔑 CHECK-ALL-THREE
Q 05
☆ Easy
\(\displaystyle\lim_{x \to 0} \frac{\sin(x)}{x} = \mathbf{?}\)
⚠️ Must memorize!
💡 EXPLANATION
This is the Special Trig Limit — must be memorized!\(\lim_{x \to 0} \dfrac{\sin x}{x} = 1\)
Also: \(\lim_{x \to 0} \dfrac{1 - \cos x}{x} = 0\)
🔑 SIN-OVER-X = ONE (when x→0)
📏 UNIT 2 · Derivatives & Rules
POWER: bring down, subtract 1 → \(x^n \Rightarrow nx^{n-1}\)
CHAIN: outside · inside' → \([f(g(x))]' = f'(g(x))\cdot g'(x)\)
PRODUCT: first·D(second) + second·D(first) → "hi·dlo + lo·dhi"
QUOTIENT: "lo·dhi − hi·dlo all over lo²"
CHAIN: outside · inside' → \([f(g(x))]' = f'(g(x))\cdot g'(x)\)
PRODUCT: first·D(second) + second·D(first) → "hi·dlo + lo·dhi"
QUOTIENT: "lo·dhi − hi·dlo all over lo²"
✏️ WORKED EXAMPLE — Chain Rule
Find \(\dfrac{d}{dx}[\sin(x^3)]\)
Outside: \(\cos(x^3)\) · Inside': \(3x^2\)
Answer: \(3x^2\cos(x^3)\) ✓
Outside: \(\cos(x^3)\) · Inside': \(3x^2\)
Answer: \(3x^2\cos(x^3)\) ✓
Q 06
☆ Easy
If \(f(x) = 4x^3 - 7x + 2\), find \(f'(x)\).
💡 EXPLANATION
Power rule term by term:\(\dfrac{d}{dx}[4x^3] = 12x^2\), \(\dfrac{d}{dx}[-7x] = -7\), \(\dfrac{d}{dx}[2] = 0\)
\(f'(x) = \mathbf{12x^2 - 7}\)
🔑 CONSTANTS DIE (derivative of constant = 0)
Q 07
☆ Easy
Find \(\dfrac{d}{dx}[e^{5x}]\)
⚠️ Chain rule applies!
💡 EXPLANATION
\(\dfrac{d}{dx}[e^{u}] = e^u \cdot u'\) where \(u = 5x\), \(u' = 5\)So answer = \(5e^{5x}\)
🔑 e STAYS, CHAIN multiplies
Q 08
★ Medium
Use the Product Rule to find \(\dfrac{d}{dx}[x^2 \sin x]\)
💡 EXPLANATION
Product rule: \((uv)' = u'v + uv'\)\(u = x^2 \Rightarrow u' = 2x\), \(v = \sin x \Rightarrow v' = \cos x\)
\(= 2x\sin x + x^2 \cos x\) ✓
🔑 "HI-dLO + LO-dHI"
Q 09
★ Medium
Find \(\dfrac{d}{dx}[\ln(x^2 + 1)]\)
⚠️ Don't forget chain rule!
💡 EXPLANATION
\(\dfrac{d}{dx}[\ln u] = \dfrac{u'}{u}\), where \(u = x^2+1\), \(u' = 2x\)Answer: \(\dfrac{2x}{x^2+1}\)
🔑 LN → ONE-OVER-U times U-PRIME
Q 10
☆ Easy
What is the derivative of \(\tan x\)?
⚠️ Memorize this!
💡 EXPLANATION
\(\dfrac{d}{dx}[\tan x] = \sec^2 x\)Note: \(\dfrac{d}{dx}[\cot x] = -\csc^2 x\) (similar pattern)
🔑 TAN → SEC-SQUARED · COT → neg-CSC-SQUARED
📈 UNIT 3 · Applications of Derivatives
SIGN chart: f' + → increasing, f' − → decreasing
CONCAVE: f'' + → concave up (cup ∪), f'' − → concave down (cap ∩)
CRITICAL: f'(c) = 0 or undefined → test with f'' or sign chart
CONCAVE: f'' + → concave up (cup ∪), f'' − → concave down (cap ∩)
CRITICAL: f'(c) = 0 or undefined → test with f'' or sign chart
📌 Remember: "Where does the function turn?" → Critical point!
"Does it turn UP or DOWN?" → Second derivative test!
"Does it turn UP or DOWN?" → Second derivative test!
Q 11
☆ Easy
If \(f'(x) > 0\) on \((a,b)\) and \(f'(x) < 0\) on \((b,c)\), then at \(x = b\), \(f\) has a:
⚠️ Sign change direction!
💡 EXPLANATION
\(f'\) goes \(+\) to \(-\): function increases then decreases → LOCAL MAX🔑 PLUS-TO-MINUS = MAX
MINUS-TO-PLUS = MIN
Q 12
☆ Easy
The equation of the tangent line to \(f(x) = x^2\) at \(x = 3\) is:
💡 EXPLANATION
Slope: \(f'(x)=2x\), so \(f'(3)=6\)Point: \(f(3)=9\) → \((3,9)\)
Line: \(y-9 = 6(x-3)\) → \(y = 6x-9\) ✓
🔑 SLOPE = f'(a), POINT = (a, f(a))
Q 13
★ Medium
A particle moves along a line with position \(s(t) = t^3 - 6t^2 + 9t\). When is the particle at rest?
⚠️ "At rest" means velocity = 0!
💡 EXPLANATION
Velocity: \(v(t) = s'(t) = 3t^2 - 12t + 9\)Set = 0: \(3(t^2-4t+3)=0 \Rightarrow 3(t-1)(t-3)=0\)
\(t = 1\) and \(t = 3\) ✓
🔑 REST → v=0 → s'=0
Q 14
★ Medium
Which statement about Mean Value Theorem (MVT) is correct?
⚠️ Conditions often forgotten!
💡 EXPLANATION
MVT: Conditions → continuous on \([a,b]\), differentiable on \((a,b)\)Conclusion → \(\exists\, c \in (a,b)\) such that \(f'(c) = \dfrac{f(b)-f(a)}{b-a}\)
🔑 MVT = "Instantaneous = Average speed" at some point
Q 15
★ Medium
Find the intervals where \(f(x) = x^3 - 3x\) is concave up.
⚠️ Use f'' not f'!
💡 EXPLANATION
\(f'(x)=3x^2-3\), \(f''(x)=6x\)Concave up when \(f'' > 0\): \(6x > 0 \Rightarrow x > 0\) → \((0,\infty)\) ✓
🔑 f'' POSITIVE → CUP UP ∪
∫ UNIT 4 · Integrals & Fundamental Theorem of Calculus
ANTI-POWER: add 1, divide by new power → \(\int x^n dx = \dfrac{x^{n+1}}{n+1}+C\)
FTC-1: \(\dfrac{d}{dx}\int_a^x f(t)dt = f(x)\)
FTC-2: \(\int_a^b f(x)dx = F(b) - F(a)\) → "PLUG TOP minus PLUG BOTTOM"
FTC-1: \(\dfrac{d}{dx}\int_a^x f(t)dt = f(x)\)
FTC-2: \(\int_a^b f(x)dx = F(b) - F(a)\) → "PLUG TOP minus PLUG BOTTOM"
✏️ WORKED EXAMPLE — Definite Integral
Evaluate \(\displaystyle\int_0^2 3x^2\,dx\)
Antiderivative: \(x^3 + C\)
\([x^3]_0^2 = 8 - 0 = \boxed{8}\) ✓
Antiderivative: \(x^3 + C\)
\([x^3]_0^2 = 8 - 0 = \boxed{8}\) ✓
Q 16
☆ Easy
\(\displaystyle\int (6x^2 - 4x + 1)\,dx = \mathbf{?}\)
💡 EXPLANATION
Term by term:\(\int 6x^2 = 2x^3\), \(\int -4x = -2x^2\), \(\int 1 = x\)
Total: \(2x^3 - 2x^2 + x + C\) ✓
🔑 NEVER FORGET +C (indefinite integral!)
Q 17
☆ Easy
\(\displaystyle\int_1^4 (2x)\,dx = \mathbf{?}\)
⚠️ Apply FTC-2!
💡 EXPLANATION
\(\int 2x\,dx = x^2 + C\)\([x^2]_1^4 = 16 - 1 = \mathbf{15}\) ✓
🔑 TOP minus BOTTOM (always plug top first!)
Q 18
★ Medium
Using FTC Part 1: \(\dfrac{d}{dx}\displaystyle\int_2^x \cos(t^2)\,dt = \mathbf{?}\)
⚠️ Don't integrate — just apply FTC!
💡 EXPLANATION
FTC Part 1: \(\dfrac{d}{dx}\int_a^x f(t)\,dt = f(x)\)Simply replace \(t\) with \(x\): \(\cos(x^2)\) ✓
🔑 FTC-1: REPLACE t WITH x (the lower bound is irrelevant!)
Q 19
★ Medium
Use u-substitution to evaluate: \(\displaystyle\int 2x(x^2+1)^4\,dx\)
⚠️ Spot the inside function!
💡 EXPLANATION
Let \(u = x^2+1\), so \(du = 2x\,dx\)\(\int u^4\,du = \dfrac{u^5}{5}+C = \dfrac{(x^2+1)^5}{5}+C\) ✓
🔑 U-SUB: inside = u, its derivative = du
Q 20
★ Medium
The area between \(f(x) = x^2\) and \(g(x) = x\) on \([0,1]\) is:
⚠️ TOP minus BOTTOM matters!
💡 EXPLANATION
On \([0,1]\): \(g(x)=x \geq f(x)=x^2\), so area \(= \int_0^1(x-x^2)dx\)\(= \left[\dfrac{x^2}{2} - \dfrac{x^3}{3}\right]_0^1 = \dfrac{1}{2}-\dfrac{1}{3} = \dfrac{1}{6}\) ✓
🔑 AREA = ∫(TOP − BOTTOM)dx
Score: 0 / 20
* Review every ❌ three times before moving on!