10 Questions × 2 Marks = 20 Marks | Answer ALL questions
6 Questions × 5 Marks | Answer ANY FOUR | 200–250 Words
(i) f(x) = x² (ii) g(x) = √x
4 Questions × 10 Marks | Answer ANY THREE | 400–500 Words
(i) lim(x→0) sin(x)/x (ii) lim(x→∞) (1 + 1/x)ˣ
(i) f(x) = eˣ · sin(x) (ii) f(x) = ln(x² + 1)
Every question explained from absolute scratch — no prior knowledge needed!
Q1. One-to-One Function
Imagine a vending machine. You press a button (input), and you get exactly ONE item (output). A function works the same way — each input gives exactly one output.
Example: f(x) = x + 3 is a function. Put in 2, get 5. Put in 7, get 10. Each input has one output.
💡 Now, what is a ONE-TO-ONE (Injective) function?
A stricter rule: no two different inputs can give the same output. Every output is unique to its input.
Real-life analogy: Think of Aadhaar numbers. Each person has a unique number. Two people can't have the same Aadhaar → that's one-to-one!
Let's test each option carefully:
- a) y = x² → Try x = 2: y = 4. Try x = -2: y = 4. Two different inputs (2 and -2) gave the SAME output (4). NOT one-to-one. ❌
- b) y = x + 3 → Try x = 1: y = 4. Try x = 2: y = 5. Try x = 3: y = 6. Every input gives a different output. No repeats possible! If x₁ + 3 = x₂ + 3, then x₁ = x₂. ONE-TO-ONE ✅
- c) y = x³ → This is actually also one-to-one (different cubes for different inputs), but the question asks for the clearest example, and a linear function like (b) is the textbook answer.
- d) y = |x| → Try x = 5: y = 5. Try x = -5: y = 5. Two inputs, same output. NOT one-to-one. ❌
✅ Answer: b) y = x + 3
Method 1 (Algebraic): Assume f(a) = f(b). If this forces a = b, the function is one-to-one.
Method 2 (Graphical — Horizontal Line Test): Draw any horizontal line through the graph. If it hits the curve at more than ONE point, it's NOT one-to-one.
🚨 Exam tip: Linear functions (y = mx + c where m ≠ 0) are ALWAYS one-to-one. Quadratic functions (y = x²) are NEVER one-to-one on all reals.
Q2. Composite Functions (f∘g)
Think of it as a two-step machine. You feed input into the FIRST machine (g), take its output, and feed that into the SECOND machine (f).
(f∘g)(x) = f(g(x))
Read it as: "f of g of x" — g goes first, then f.
Memory trick: The function closest to x acts first. In f(g(x)), g is closest to x, so g acts first.
Given: f(x) = √x and g(x) = x + 4
We need: (f∘g)(x) = f(g(x))
Step 1: First, find what g does to x: g(x) = x + 4. So if x = 5, g(5) = 9.
Step 2: Now feed g(x) into f. Wherever f has "x", replace it with g(x):
f(x) = √x → f(g(x)) = f(x + 4) = √(x + 4)
Step 3 (Verify): Let x = 5. g(5) = 9. f(9) = √9 = 3. And √(5+4) = √9 = 3. ✅ Matches!
✅ Answer: b) √(x + 4)
√x + 4 means: "first take square root of x, THEN add 4".
√(x + 4) means: "first add 4 to x, THEN take square root".
These are completely different! With x = 5:
• √x + 4 = √5 + 4 = 2.24 + 4 = 6.24
• √(x+4) = √9 = 3
Parentheses change everything in math!
Q3. Derivative of 1/x
Imagine you're driving a car. Your speedometer shows how fast you're going at each moment. The derivative is like a speedometer for any function — it tells you how fast the function's value is changing at any point.
If the function is your position on the road, the derivative is your speed.
If the function is your bank balance, the derivative is how fast you're earning (or losing) money.
💡 Power Rule — the most important derivative rule:
d/dx [xⁿ] = n × xⁿ⁻¹
Translation: "Bring the exponent down as a multiplier, then reduce the exponent by 1."
Simple examples:
• d/dx [x³] = 3x² (bring 3 down, power becomes 3-1=2)
• d/dx [x⁵] = 5x⁴ (bring 5 down, power becomes 5-1=4)
Now solve: d/dx [1/x]
Step 1: Rewrite 1/x using negative exponents. The rule is: 1/xⁿ = x⁻ⁿ
So: 1/x = 1/x¹ = x⁻¹
Step 2: Now apply the power rule with n = -1:
d/dx [x⁻¹] = (-1) × x⁻¹⁻¹ = -1 × x⁻²
Step 3: Convert back to fraction form:
-1 × x⁻² = -1 × (1/x²) = -1/x²
✅ Answer: a) -1/x²
• d/dx [1/x] = -1/x²
• d/dx [1/x²] = -2/x³
• d/dx [√x] = 1/(2√x)
All come from the power rule!
Q4. Derivative of ln(sin(x))
When you have a function INSIDE another function (like a Russian nesting doll), you need the chain rule.
Think of it like peeling layers:
ln(sin(x)) has TWO layers:
• Outer layer: ln(▢) — the natural log
• Inner layer: sin(x) — what's inside the log
Chain Rule: d/dx [f(g(x))] = f'(g(x)) × g'(x)
Translation: "Take derivative of outside (keep inside untouched) × derivative of inside"
Solve: d/dx [ln(sin(x))]
Step 1: Identify the layers:
Outer function f(u) = ln(u), where u = sin(x)
Inner function g(x) = sin(x)
Step 2: Derivative of outer: d/du [ln(u)] = 1/u = 1/sin(x)
Step 3: Derivative of inner: d/dx [sin(x)] = cos(x)
Step 4: Multiply them together (chain rule):
= (1/sin(x)) × cos(x) = cos(x)/sin(x) = cot(x)
✅ Answer: a) cot(x)
• cos(x)/sin(x) = cot(x) (cotangent)
• sin(x)/cos(x) = tan(x) (tangent)
• 1/sin(x) = csc(x) (cosecant)
• 1/cos(x) = sec(x) (secant)
Note: Options (a) cot(x) and (b) cos(x)/sin(x) are the SAME thing written differently. The simplified answer is cot(x).
Q5. Value of sin(90°)
Imagine a point moving around a circle of radius 1 (unit circle). The sine tells you the height of that point above/below the center.
• At 0° (starting position, 3 o'clock): height = 0 → sin(0°) = 0
• At 30°: height = halfway up → sin(30°) = 1/2
• At 90° (12 o'clock, straight up): height is at its MAXIMUM → sin(90°) = 1
• At 180° (9 o'clock): back to center height → sin(180°) = 0
• At 270° (6 o'clock, straight down): lowest point → sin(270°) = -1
📚 Complete trig value table (MUST MEMORIZE for exam):
| Angle | 0° | 30° | 45° | 60° | 90° |
|---|---|---|---|---|---|
| sin | 0 | 1/2 | 1/√2 | √3/2 | 1 |
| cos | 1 | √3/2 | 1/√2 | 1/2 | 0 |
| tan | 0 | 1/√3 | 1 | √3 | ∞ |
That gives: 0, 1/2, 1/√2, √3/2, 1 — which matches the table above!
For cos, it's the reverse order (read sin values backwards).
✅ Answer: c) 1
Q6. ln(e³) = ?
Think of logarithm as asking a question: "What power do I need?"
ln(8) asks: "e raised to what power gives 8?"
ln(e³) asks: "e raised to what power gives e³?" Obviously the answer is 3!
The golden rule: ln(eˣ) = x (always!)
Because ln and e are inverse operations — they undo each other, like + and -, or × and ÷.
More examples:
• ln(e⁵) = 5
• ln(e¹) = 1
• ln(e⁰) = 0 (because e⁰ = 1, so ln(1) = 0)
• ln(e) = 1 (because e = e¹)
Solve: ln(e³)
Step 1: Recognize the pattern: this is ln(esomething)
Step 2: Apply the rule ln(eˣ) = x. Here the exponent is 3.
Step 3: Therefore ln(e³) = 3
✅ Answer: a) 3
• ln(eˣ) = x (log cancels exponential)
• eln(x) = x (exponential cancels log)
• ln(1) = 0
• ln(ab) = ln(a) + ln(b)
• ln(a/b) = ln(a) − ln(b)
• ln(aⁿ) = n × ln(a)
Q7. Product Rule
When you need to find the derivative of two functions MULTIPLIED together.
Why can't you just differentiate each separately?
Because d/dx [f(x) × g(x)] ≠ f'(x) × g'(x). That's WRONG!
The correct Product Rule:
(u × v)' = u' × v + u × v'
Memory trick: "First's derivative times Second + First times Second's derivative"
Or even simpler: "d-1 times 2 + 1 times d-2" (d = derivative)
Full worked example: Find d/dx [x² × sin(x)]
Step 1: Identify: u = x², v = sin(x)
Step 2: Find derivatives: u' = 2x, v' = cos(x)
Step 3: Apply formula: u'v + uv'
= (2x)(sin(x)) + (x²)(cos(x))
= 2x·sin(x) + x²·cos(x)
✅ Answer: a) u'v + uv'
Actually, options (a) u'v + uv' and (c) uv' + vu' are mathematically the SAME (addition is commutative: a+b = b+a). But the standard, universally accepted notation is (a) u'v + uv'. Always write it this way in exams.
Q8. What does f'(x) > 0 mean?
Imagine you're hiking on a mountain trail:
• Going uphill → your altitude is INCREASING → slope is positive → f'(x) > 0 📈
• Going downhill → your altitude is DECREASING → slope is negative → f'(x) < 0 📉
• At the peak or valley → momentarily flat → slope is zero → f'(x) = 0 (critical point)
The derivative IS the slope. Positive slope = function climbing up = increasing.
Concrete example: f(x) = x²
• f'(x) = 2x
• When x = 3: f'(3) = 6 > 0 → function is increasing at x = 3 ✅
• When x = -2: f'(-2) = -4 < 0 → function is decreasing at x = -2 ✅
• When x = 0: f'(0) = 0 → function is flat (this is the minimum of x²) ✅
✅ Answer: a) f(x) is increasing
• f'(x) > 0 everywhere in interval → f is increasing
• f'(x) < 0 everywhere in interval → f is decreasing
• f'(x) = 0 at a point → possible maximum, minimum, or inflection point
• f'(x) changes from + to - → local maximum
• f'(x) changes from - to + → local minimum
Q9. Inflection Point of f(x) = x³
It's where a curve changes its "bending direction".
Think of driving on a road:
• Concave up (∪ shape): Like the bottom of a bowl. You're turning LEFT. f''(x) > 0
• Concave down (∩ shape): Like the top of a hill. You're turning RIGHT. f''(x) < 0
• Inflection point: Where you switch from turning left to turning right (or vice versa). f''(x) = 0 AND concavity actually changes.
How to find inflection points:
1. Find the second derivative f''(x)
2. Set f''(x) = 0 and solve
3. Check that concavity actually changes at that point
Solve for f(x) = x³:
Step 1: Find the first derivative: f'(x) = 3x²
Step 2: Find the second derivative: f''(x) = 6x
Step 3: Set f''(x) = 0: 6x = 0 → x = 0
Step 4: Verify concavity changes:
• Pick x = -1 (left of 0): f''(-1) = 6(-1) = -6 < 0 → concave DOWN ∩
• Pick x = 1 (right of 0): f''(1) = 6(1) = 6 > 0 → concave UP ∪
• Concavity changed from DOWN to UP → YES, x = 0 is an inflection point! ✅
✅ Answer: b) x = 0
Q10. Integration by Parts — Choosing u
It's a technique for integrating products of two functions that can't be done normally.
Formula: ∫ u dv = uv − ∫ v du
The big question: Which function should be "u"?
u should be the function that gets SIMPLER when you differentiate it. You're going to differentiate u, so pick something that simplifies.
💡 LIATE Rule — choose u in this priority:
L → Logarithmic: ln(x), log(x)
I → Inverse trig: sin⁻¹(x), tan⁻¹(x)
A → Algebraic: x, x², x³
T → Trigonometric: sin(x), cos(x)
E → Exponential: eˣ, 2ˣ
Full worked example: ∫ x × eˣ dx
Step 1: Identify: we have x (Algebraic) and eˣ (Exponential). By LIATE, A comes before E, so u = x and dv = eˣ dx
Step 2: Find du and v:
u = x → du = dx (x becomes simpler: just 1!)
dv = eˣ dx → v = eˣ
Step 3: Apply the formula ∫ u dv = uv − ∫ v du:
= x·eˣ − ∫ eˣ dx = x·eˣ − eˣ + C = eˣ(x − 1) + C
✅ Answer: a) The one that simplifies on differentiation
Complete solutions with every single step explained — write exactly this in your exam!
Q11. Domain and Range of Functions
Think of a function as a machine:
• Domain = all the inputs you're ALLOWED to feed into the machine (what values of x work?)
• Range = all the outputs the machine can possibly produce (what values of y come out?)
How to find Domain: Ask "What values of x would BREAK this function?" Things that break functions:
• Division by zero (denominator = 0) ❌
• Square root of negative number ❌
• Logarithm of zero or negative ❌
How to find Range: Ask "What values can y possibly take?"
(i) f(x) = x²
Step 1 (Domain): Can x be any number? Let's check:
• Can x = 5? Yes, 5² = 25 ✅
• Can x = -3? Yes, (-3)² = 9 ✅
• Can x = 0? Yes, 0² = 0 ✅
• Can x = 1000000? Yes! No restrictions at all.
Domain = all real numbers = (−∞, ∞) or ℝ
Step 2 (Range): What outputs can x² produce?
• Can y = 4? Yes, x = 2 gives 2² = 4 ✅
• Can y = 0? Yes, x = 0 gives 0² = 0 ✅
• Can y = -1? NO! Squaring any number always gives ≥ 0. No x gives x² = -1. ❌
Range = all non-negative reals = [0, ∞)
(ii) g(x) = √x
Step 1 (Domain): What breaks √x?
• √4 = 2 ✅ (positive works)
• √0 = 0 ✅ (zero works)
• √(-4) = ??? ❌ (square root of negative is NOT a real number!)
So x must be ≥ 0. Domain = [0, ∞)
Step 2 (Range): √x always gives ≥ 0 (square root is never negative)
Range = [0, ∞)
• ( ) = open bracket = does NOT include that endpoint
• [ ] = closed bracket = DOES include that endpoint
• [0, ∞) means: includes 0, goes to infinity (infinity always gets open bracket because you can never reach it)
Q12. Finding Composite Functions (f∘g and g∘f)
• (f∘g)(x) = f(g(x)) → First do g, then feed result into f
• (g∘f)(x) = g(f(x)) → First do f, then feed result into g
Given: f(x) = 2x + 1, g(x) = x²
Part 1: Find (f∘g)(x) = f(g(x))
Step 1: Start from the inside: g(x) = x²
Step 2: Now plug x² into f wherever f has "x":
f(x) = 2x + 1
f(x²) = 2(x²) + 1 = 2x² + 1
Verify: If x = 3, g(3) = 9, then f(9) = 2(9) + 1 = 19. And 2(3)² + 1 = 19 ✅
Part 2: Find (g∘f)(x) = g(f(x))
Step 1: Start from the inside: f(x) = 2x + 1
Step 2: Now plug (2x + 1) into g wherever g has "x":
g(x) = x²
g(2x+1) = (2x+1)²
Step 3: Expand (2x+1)² using (a+b)² = a² + 2ab + b²:
= (2x)² + 2(2x)(1) + (1)² = 4x² + 4x + 1
= 4x² + 4x + 1
Verify: If x = 3, f(3) = 7, then g(7) = 49. And 4(9) + 4(3) + 1 = 36+12+1 = 49 ✅
Composition is NOT commutative. The order in which you compose matters!
Q13. Derivative of f(x) = x² by First Principle
Instead of using shortcut formulas, we derive the derivative from its basic definition. This is how Newton and Leibniz invented calculus!
The idea: Pick two points on the curve very close together. Find the slope between them. Then make the gap infinitely small.
f'(x) = lim(h→0) [f(x+h) − f(x)] / h
Where:
• f(x) = value at point x
• f(x+h) = value at a nearby point (x+h), where h is a tiny distance
• [f(x+h) − f(x)] / h = slope between these two points
• lim(h→0) = make h approach zero (points get infinitely close)
Given: f(x) = x². Find f'(x).
Step 1: Find f(x+h). Replace every x with (x+h):
f(x+h) = (x+h)²
Expand using (a+b)² = a² + 2ab + b²:
= x² + 2xh + h²
Step 2: Calculate f(x+h) − f(x):
= (x² + 2xh + h²) − (x²)
= x² + 2xh + h² − x²
= 2xh + h² (x² cancelled out!)
Step 3: Divide by h:
= (2xh + h²) / h
= h(2x + h) / h (factor out h)
= 2x + h (cancel h)
Step 4: Take the limit as h → 0:
lim(h→0) (2x + h) = 2x + 0 = 2x
✅ Therefore, f'(x) = 2x
The first principle always gives the same result as shortcut rules — it just takes more steps.
Q14. Partial Fraction Decomposition
It's the REVERSE of adding fractions! In school you learned:
1/2 + 1/3 = 3/6 + 2/6 = 5/6 (combining simple fractions into one)
Partial fractions does the OPPOSITE:
5/6 → break it back into 1/2 + 1/3
Why is this useful? Complex fractions are hard to integrate. But simple fractions like 1/(x+1) are easy: ∫ 1/(x+1) dx = ln|x+1| + C
Decompose: (2x + 3) / (x² + 3x + 2)
Step 1: Factor the denominator. We need two numbers that multiply to 2 and add to 3:
x² + 3x + 2 = (x + 1)(x + 2) [because 1×2 = 2 and 1+2 = 3]
Step 2: Set up the partial fractions:
(2x + 3) / [(x+1)(x+2)] = A/(x+1) + B/(x+2)
(A and B are unknown constants we need to find)
Step 3: Multiply BOTH sides by (x+1)(x+2) to clear fractions:
2x + 3 = A(x+2) + B(x+1)
Step 4: Use the "cover-up" method — choose clever values of x:
Put x = -1 (this makes (x+1) = 0, killing the B term):
2(-1) + 3 = A(-1+2) + B(0)
-2 + 3 = A(1)
1 = A → A = 1
Put x = -2 (this makes (x+2) = 0, killing the A term):
2(-2) + 3 = A(0) + B(-2+1)
-4 + 3 = B(-1)
-1 = -B → B = 1
Step 5: Write the answer:
(2x + 3) / (x² + 3x + 2) = 1/(x+1) + 1/(x+2)
1/(x+1) + 1/(x+2) = [(x+2) + (x+1)] / [(x+1)(x+2)] = (2x+3) / (x²+3x+2) ✅
Q15. Definite Integral of f(x) = 2x from x = 1 to x = 3
Think of it as calculating the area under a curve between two points.
• Indefinite integral (no limits): ∫ f(x) dx = F(x) + C — gives you a formula
• Definite integral (with limits a to b): ∫ᴁᵦ f(x) dx = F(b) − F(a) — gives you a NUMBER
∫ᴁᵦ f(x) dx = F(b) − F(a)
where F(x) is the antiderivative of f(x) (reverse of differentiation).
Find: ∫₁³ 2x dx
Step 1: Find the antiderivative of 2x.
Question: "What function, when differentiated, gives 2x?"
Answer: x² (because d/dx [x²] = 2x)
So F(x) = x²
Step 2: Evaluate F(b) − F(a) where a = 1, b = 3:
F(3) = 3² = 9
F(1) = 1² = 1
Step 3: Subtract:
F(3) − F(1) = 9 − 1 = 8
✅ Answer: 8
The area under the line y = 2x from x = 1 to x = 3 is exactly 8 square units. You can verify this: the shape is a trapezoid with bases y=2 (at x=1) and y=6 (at x=3), height = 2, area = (2+6)/2 × 2 = 8 ✅
Q16. Evaluate ∫ sin²(x) dx
You can't integrate sin²(x) directly — there's no basic formula for it! The trick is to convert it into something integrable using a trigonometric identity.
The power-reducing identity:
sin²(x) = (1 − cos(2x)) / 2
Where does this come from? From the double angle formula:
cos(2x) = 1 − 2sin²(x) → Rearranging: sin²(x) = (1 − cos(2x)) / 2
Solve: ∫ sin²(x) dx
Step 1: Replace sin²(x) with the identity:
= ∫ (1 − cos(2x)) / 2 dx
Step 2: Split into two separate integrals:
= 1/2 ∫ 1 dx − 1/2 ∫ cos(2x) dx
Step 3: Integrate each part:
• 1/2 ∫ 1 dx = x/2
• 1/2 ∫ cos(2x) dx = ?
We know ∫ cos(u) du = sin(u). Here u = 2x, so du = 2dx, meaning dx = du/2
∫ cos(2x) dx = sin(2x)/2
So: 1/2 × sin(2x)/2 = sin(2x)/4
Step 4: Combine:
∫ sin²(x) dx = x/2 − sin(2x)/4 + C
cos²(x) = (1 + cos(2x)) / 2 (note the + sign)
∫ cos²(x) dx = x/2 + sin(2x)/4 + C
Complete solutions with every working step — this is exactly what you write in your exam for full marks!
Q17. Evaluating Two Important Limits (10 marks)
(i) lim(x→0) sin(x)/x
If you directly plug x = 0, you get sin(0)/0 = 0/0 — which is undefined (you can't divide by zero!).
This 0/0 form is called an "indeterminate form". It doesn't mean the answer is 0 or undefined — it means we need to work harder to find the actual value.
This is one of the most fundamental limits in calculus. The answer is 1.
Method 1: Using L'Hôpital's Rule
When you get 0/0 or ∞/∞, you can differentiate the top and bottom separately and try again:
lim [f(x)/g(x)] = lim [f'(x)/g'(x)] (when direct substitution gives 0/0 or ∞/∞)
Step 1: Check: sin(0)/0 = 0/0 ✅ (indeterminate, so L'Hôpital applies)
Step 2: Differentiate numerator: d/dx [sin(x)] = cos(x)
Step 3: Differentiate denominator: d/dx [x] = 1
Step 4: New limit: lim(x→0) cos(x)/1 = cos(0)/1 = 1/1 = 1
✅ Answer: lim(x→0) sin(x)/x = 1
Method 2: Numerical verification
• x = 0.5: sin(0.5)/0.5 = 0.4794/0.5 = 0.9589
• x = 0.1: sin(0.1)/0.1 = 0.0998/0.1 = 0.9983
• x = 0.01: sin(0.01)/0.01 = 0.00999/0.01 = 0.99998
As x → 0, the ratio → 1 ✅
(ii) lim(x→∞) (1 + 1/x)ˣ
This limit defines Euler's number e ≈ 2.71828..., the most important constant in mathematics after π.
Where does e appear?
• Compound interest: If a bank compounds interest continuously, your money grows by a factor of e
• Population growth, radioactive decay, electricity in circuits
• The derivative of eˣ is itself: d/dx [eˣ] = eˣ (only function with this property!)
Numerical verification:
• x = 1: (1 + 1/1)¹ = 2¹ = 2
• x = 10: (1 + 1/10)¹⁰ = (1.1)¹⁰ = 2.5937...
• x = 100: (1 + 0.01)¹⁰⁰ = 2.7048...
• x = 1000: (1 + 0.001)¹⁰⁰⁰ = 2.7169...
• x → ∞: approaches e ≈ 2.71828...
✅ Answer: lim(x→∞) (1 + 1/x)ˣ = e
• lim(x→0) sin(x)/x = 1
• lim(x→0) (eˣ − 1)/x = 1
• lim(x→0) (1 − cos x)/x = 0
• lim(x→0) log(1+x)/x = 1
• lim(x→∞) (1 + 1/x)ˣ = e
Q18. Derivatives using Product Rule and Chain Rule (10 marks)
(i) f(x) = eˣ · sin(x)
We have two functions MULTIPLIED together: eˣ and sin(x). That means Product Rule!
(u × v)' = u'v + uv'
Step 1: Identify the two functions:
u = eˣ v = sin(x)
Step 2: Find their derivatives:
u' = d/dx [eˣ] = eˣ (eˣ is special — its derivative is itself!)
v' = d/dx [sin(x)] = cos(x)
Step 3: Apply the product rule formula:
f'(x) = u'v + uv'
= eˣ · sin(x) + eˣ · cos(x)
Step 4: Factor out eˣ:
= eˣ(sin(x) + cos(x))
✅ f'(x) = eˣ(sin(x) + cos(x))
(ii) f(x) = ln(x² + 1)
We have a function INSIDE another function: ln(▢) contains (x² + 1). That means Chain Rule!
d/dx [f(g(x))] = f'(g(x)) × g'(x)
"Derivative of outside × derivative of inside"
Step 1: Identify the layers:
Outer function: f(u) = ln(u) → f'(u) = 1/u
Inner function: g(x) = x² + 1 → g'(x) = 2x
Step 2: Apply chain rule: f'(g(x)) × g'(x)
= (1/(x² + 1)) × (2x)
Step 3: Simplify:
= 2x / (x² + 1)
✅ f'(x) = 2x / (x² + 1)
Q19. Local Maxima and Minima of f(x) = x³ − 6x² + 9x + 1 (10 marks)
• Local maximum: A hilltop — the function goes UP then DOWN. Like the peak of a mountain.
• Local minimum: A valley — the function goes DOWN then UP. Like the bottom of a bowl.
How to find them (3-step method):
1. Find f'(x) and set it = 0 to get critical points (where slope is zero)
2. Find f''(x) and plug in the critical points:
• f''(c) < 0 → curve bends DOWN → local MAXIMUM (hilltop)
• f''(c) > 0 → curve bends UP → local MINIMUM (valley)
3. Find the actual y-values at the critical points
Step 1: Find the first derivative and set it to zero
f(x) = x³ − 6x² + 9x + 1
f'(x) = 3x² − 12x + 9 (using power rule on each term)
• d/dx [x³] = 3x²
• d/dx [−6x²] = −12x
• d/dx [9x] = 9
• d/dx [1] = 0 (constant disappears)
Set f'(x) = 0:
3x² − 12x + 9 = 0
Divide everything by 3: x² − 4x + 3 = 0
Factor: (x − 1)(x − 3) = 0 [because (−1)×(−3) = 3 and (−1)+(−3) = −4]
Critical points: x = 1 and x = 3
Step 2: Second derivative test
f''(x) = d/dx [3x² − 12x + 9] = 6x − 12
At x = 1:
f''(1) = 6(1) − 12 = 6 − 12 = −6
f''(1) < 0 → Curve bends DOWN → LOCAL MAXIMUM at x = 1
At x = 3:
f''(3) = 6(3) − 12 = 18 − 12 = 6
f''(3) > 0 → Curve bends UP → LOCAL MINIMUM at x = 3
Step 3: Find the y-values (function values at critical points)
At x = 1:
f(1) = (1)³ − 6(1)² + 9(1) + 1 = 1 − 6 + 9 + 1 = 5
At x = 3:
f(3) = (3)³ − 6(3)² + 9(3) + 1 = 27 − 54 + 27 + 1 = 1
• Local Maximum at the point (1, 5) — the hilltop
• Local Minimum at the point (3, 1) — the valley
💡 In words: The function rises to a peak value of 5 at x = 1, then falls to a valley value of 1 at x = 3, then rises again.
Q20. Area Under the Upper Semicircle of x² + y² = 4 (10 marks)
x² + y² = 4 is the equation of a circle centered at the origin with radius r = 2 (because r² = 4).
The "upper semicircle" is the top half of the circle. Solving for y:
y² = 4 − x² → y = √(4 − x²) (taking positive root for upper half)
We need to find the area under this curve from x = −2 to x = 2.
Method 1: Using Geometry (Quick and Smart!)
Area of full circle = πr²
Area of upper semicircle = half the circle = πr² / 2
Here r = 2, so:
Area = π(2)² / 2 = 4π / 2 = 2π ≈ 6.283
Method 2: Using Integration (Required to show in exam!)
Step 1: Set up the integral:
Area = ∫⁻²² √(4 − x²) dx
Step 2: Use trigonometric substitution. Let x = 2sin(θ)
Then: dx = 2cos(θ) dθ
When x = −2: sin(θ) = −1 → θ = −π/2
When x = 2: sin(θ) = 1 → θ = π/2
Step 3: Simplify √(4 − x²):
√(4 − 4sin²θ) = √(4(1 − sin²θ)) = √(4cos²θ) = 2cosθ
(using the identity: 1 − sin²θ = cos²θ)
Step 4: Substitute everything into the integral:
= ∫ (2cosθ)(2cosθ) dθ = ∫ 4cos²θ dθ
Limits: from θ = −π/2 to θ = π/2
Step 5: Use the identity cos²θ = (1 + cos(2θ))/2:
= 4 ∫ (1 + cos(2θ))/2 dθ = 2 ∫ (1 + cos(2θ)) dθ
Step 6: Integrate:
= 2[θ + sin(2θ)/2] evaluated from −π/2 to π/2
Step 7: Evaluate at the limits:
At θ = π/2: 2[π/2 + sin(π)/2] = 2[π/2 + 0] = π
At θ = −π/2: 2[−π/2 + sin(−π)/2] = 2[−π/2 + 0] = −π
Step 8: Subtract:
π − (−π) = π + π = 2π
🔎 Cross-check: Both methods give 2π ✅
• Geometry: πr²/2 = π(4)/2 = 2π
• Integration: ∫⁻²² √(4−x²) dx = 2π
🚨 Exam tip: In exams, always mention BOTH methods. Show the integration for marks, and mention the geometry as a verification. This shows deep understanding.