🎓 You've Got This — Let's Cover Everything!
Complete answer guide for ALL questions from all 4 past papers. Step-by-step like a teacher. Exact exam-writing style answers.
| Section | Questions | Marks | Best Strategy |
|---|---|---|---|
| I (Q1–10) | All 10 compulsory | 10×1 = 10 | ✅ Easy — do ALL. Free marks! |
| II (Q11–22) | Any 8 from 12 questions | 8×2 = 16 | ✅ Do the ones you know |
| III (Q23–31) | Any 6 from 9 questions | 6×4 = 24 | ⚡ Pick MVT, ODE, Laplace |
| IV (Q32–35) | Any 2 from 4 questions | 2×15 = 30 | 🎯 LPP (Q35) + Laplace/ODE |
| Question | Papers | Marks |
|---|---|---|
| Fourier series for f(x)=|x| in [−π,π] | L-3942, B-3195, F-2065 (ALL!) | 4–15 |
| Verify Lagrange MVT for f(x)=x²+2x+9, [1,5] | L-3942, B-3195 (twice!) | 2 |
| State Rolle's Theorem | L-3942, B-3195 (both papers) | 1 |
| Define sinh x in exponential form | L-3942, F-2065 | 1 |
| LPP graphical method | All 4 papers (Q35) | 15 |
| Laplace transforms + inverse | All papers — multiple Qs | 2–15 |
| Euler phi-function calculation | L-3942, B-3195 | 2 |
| Solve IVP y''−y'−2y=0 | B-3195 (both papers of same code) | 2 |
📐 Hyperbolic Functions
Definitions, identities, proofs and derivatives. Appears as 1-mark fill-in and 2-mark proof questions in every exam.
Define sinh x and cosh x in terms of exponential functions ALL Papers
Think of e^x as the base. sinh is the "antisymmetric part" (odd), cosh is the "symmetric part" (even). If you add e^x and e^(-x) you get 2·cosh x. Simple!
sinh x = (eˣ − e⁻ˣ)/2 cosh x = (eˣ + e⁻ˣ)/2
If sech x = 4/5, find cosh x
sech x = 1/cosh x. So just flip the fraction!
Since sech x = 1/cosh x: cosh x = 1/sech x = 1/(4/5) = 5/4
Show that cosh(−x) = cosh x (even function)
cosh(−x) = (e⁻ˣ + eˣ)/2 = cosh x ✓ (cosh x is an even function)
Show cosh(x+y) = cosh x cosh y + sinh x sinh y Repeated
Expand RHS using definitions. The cross terms (eˣ⁻ʸ etc.) cancel out, leaving exactly the LHS definition of cosh(x+y). Pure algebra!
cosh x cosh y + sinh x sinh y = [(eˣ+e⁻ˣ)/2][(eʸ+e⁻ʸ)/2] + [(eˣ−e⁻ˣ)/2][(eʸ−e⁻ʸ)/2]
= [eˣ⁺ʸ+eˣ⁻ʸ+e⁻ˣ⁺ʸ+e⁻⁽ˣ⁺ʸ⁾]/4 + [eˣ⁺ʸ−eˣ⁻ʸ−e⁻ˣ⁺ʸ+e⁻⁽ˣ⁺ʸ⁾]/4
= [2e^(x+y) + 2e^−(x+y)]/4 = cosh(x+y) ✓
Show cosh²x + sinh²x = cosh 2x
cosh²x + sinh²x = (e²ˣ+2+e⁻²ˣ)/4 + (e²ˣ−2+e⁻²ˣ)/4 = (e²ˣ+e⁻²ˣ)/2 = cosh 2x ✓
Find dy/dx where y = x sinh x − cosh x
Use product rule for x·sinh x: d/dx(uv) = u'v + uv'. Remember d/dx(sinh x) = cosh x and d/dx(cosh x) = sinh x.
dy/dx = x cosh x
Solve the ODE: dy/dx − y tanh x = 0
y = A cosh x
🔄 Differentiation
Derivatives using chain rule, product rule, log differentiation, implicit differentiation, nth derivative.
Derivative of log_a(x)
1/(x·ln a) Special case: d/dx[ln x] = 1/x
Find dy/dx where y = x sinh x − cosh x
dy/dx = x cosh x
dy/dx when y = log(sin x + cos x) Repeated
Chain rule for log: d/dx[ln f(x)] = f'(x)/f(x). Differentiate inside (sin x + cos x), divide by (sin x + cos x).
dy/dx = (cos x − sin x)/(sin x + cos x)
dy/dx when y = sinh⁻¹x
y = sinh⁻¹x means sinh y = x. Differentiate both sides implicitly!
d/dx[sinh⁻¹x] = 1/√(1+x²)
Find dy/dx if 3x² − 2y² = 1 (Implicit)
dy/dx = 3x/(2y)
dy/dx when y = cos x^(tan x) (Log Differentiation) Sec III
Variable^variable → always use Logarithmic Differentiation: take log both sides, differentiate, then multiply by y.
Let y = (cos x)^(tan x). Taking log: log y = tan x · log(cos x)
Differentiating: (1/y)(dy/dx) = sec²x·log(cos x) − tan²x
∴ dy/dx = (cos x)^(tan x)[sec²x·log(cos x) − tan²x]
dy/dx if y = ∛(x(x−2)/(x²+1)) (Log Differentiation)
Cube root with products/quotients inside → take log! Log converts products to sums and roots to fractions, making differentiation easy.
log y = ⅓[log x + log(x−2) − log(x²+1)]
dy/dx = ⅓·y·[1/x + 1/(x−2) − 2x/(x²+1)]
dy/dx when y = (cos x)^x (Log Differentiation)
dy/dx = (cos x)^x [log(cos x) − x tan x]
Find dy/dx when x³ + y³ = 3axy (Implicit)
dy/dx = (ay − x²)/(y² − ax)
Find nth differential coefficient of y = aˣ
The nth differential coefficient: yₙ = aˣ · (ln a)ⁿ
If y = cos(m sin⁻¹x), prove the recurrence relation (4 Marks)
This is a successive differentiation (Leibnitz type) problem. Find y', y'', then form a differential equation and use induction via Leibnitz theorem.
y = cos(m sin⁻¹x). y₁ = −m sin(m sin⁻¹x)/√(1−x²)
Squaring: (1−x²)y₁² = m²(1−y²). Differentiating: (1−x²)y₂ − xy₁ + m²y = 0
Differentiating n times by Leibnitz:
(1−x²)yₙ₊₂ − (2n+1)xyₙ₊₁ + (m²−n²)yₙ = 0 ✓
Open Box Problem: 16×30 cardboard, max volume (4 Marks — Optimization)
Cut squares of side x from corners. Box dimensions: length=(30−2x), width=(16−2x), height=x. Volume = lwh. Maximize V by finding dV/dx = 0!
V = x(16−2x)(30−2x). dV/dx = 12x²−184x+480 = 0 ⟹ 3x²−46x+120=0 ⟹ x = 10/3 or 12
x=12 rejected (too large). d²V/dx² < 0 at x=10/3 confirms maximum.
Square size = 10/3 inches ≈ 3.33 inches gives maximum volume.
Divide 15 into two parts such that (one)²×(other)³ is maximum
Let parts be x and 15−x. f(x) = x²(15−x)³
f'(x) = x(15−x)²(30−5x) = 0 ⟹ x = 6
Parts are 6 and 9 [gives maximum = 36×729 = 26244]
📊 Mean Value Theorems
Rolle's Theorem, Lagrange's MVT — statement, verification, proof. Appears in every paper!
State Rolle's Theorem Every Paper!
Ball thrown up and comes back to same height → at the peak, velocity = 0. That's Rolle's theorem! The peak is 'c', velocity=0 means f'(c)=0.
Rolle's Theorem: If a function f(x) satisfies:
(i) f is continuous on [a, b]
(ii) f is differentiable on (a, b)
(iii) f(a) = f(b)
Then ∃ at least one c ∈ (a, b) such that f'(c) = 0
State Lagrange's Mean Value Theorem (MVT) Every Paper!
Lagrange's MVT: If f(x) is continuous on [a,b] and differentiable on (a,b), then ∃ c ∈ (a,b) such that:
f'(c) = [f(b) − f(a)] / (b − a)
Note: Geometrically, the tangent at c is parallel to the chord from a to b.
Verify Lagrange's MVT for f(x) = x² + 2x + 9, x ∈ [1,5] MOST REPEATED!
Steps: (1) Check polynomial ✓ continuous & differentiable. (2) Calculate f(1) and f(5). (3) Compute [f(5)−f(1)]/(5−1). (4) Find f'(x)=that value. (5) Solve for c. That's it!
f(x) = x²+2x+9 is a polynomial → continuous on [1,5] and differentiable on (1,5). ✓
f(x)=x²+2x+9 is continuous on [1,5] and differentiable on (1,5). ✓
f(1)=12, f(5)=44. [f(5)−f(1)]/(5−1) = 32/4 = 8
f'(x)=2x+2, f'(c)=8 ⟹ c=3 ∈ (1,5) ✓
∴ Lagrange's MVT is verified. c = 3.
State and PROVE Lagrange's MVT (15 Marks Section IV)
The proof uses Rolle's Theorem! We construct a helper function h(x) such that h(a)=h(b), apply Rolle's to h, and arrive at the MVT result. Very elegant proof!
See M2 above for the statement.
h is continuous on [a,b] and differentiable on (a,b), and h(a)=h(b)=0.
By Rolle's Theorem, ∃ c ∈ (a,b) such that h'(c) = 0.
Define h(x) = f(x) − f(a) − [(f(b)−f(a))/(b−a)](x−a). Then h is cts on [a,b], diff on (a,b), and h(a)=h(b)=0.
By Rolle's Theorem, ∃ c ∈ (a,b): h'(c)=0.
h'(c)=0 ⟹ f'(c) = [f(b)−f(a)]/(b−a) ∎
⚙️ Ordinary Differential Equations
Separable, homogeneous, IVP, non-homogeneous, Euler-Cauchy. Major exam topic across all sections.
Give degree and order of dy/dx + (x^½ − y) = 0
Order = highest derivative (1st derivative here → order 1). Degree = power of highest derivative (dy/dx appears once, power 1). Note: Degree is only defined when the equation is polynomial in derivatives!
Order = 1 (highest derivative is dy/dx)
Degree = 1 (dy/dx has power 1)
General solution of dy/dx = 2x (1 Mark)
y = x² + C
Solve y' = −xy (Separable)
y = Ae^(−x²/2)
Obtain first order ODE associated with y = xe^(−x) (eliminate constant)
y = xe⁻ˣ. y' = e⁻ˣ(1−x). Since y/x = e⁻ˣ: y' = (y/x)(1−x)
xy' + y(x−1) = 0
Solve y'' − y' − 6y = 0 Repeated
m²−m−6=0 ⟹ (m−3)(m+2)=0 ⟹ m=3,−2
y = C₁e^(3x) + C₂e^(−2x)
Solve IVP: y'' − y' − 2y = 0, y(0) = 0, y'(0) = 1 Both B-3195 Papers!
GS: y=C₁e²ˣ+C₂e⁻ˣ. y(0)=0 ⟹ C₁+C₂=0; y'(0)=1 ⟹ 2C₁−C₂=1 ⟹ C₁=1/3, C₂=−1/3
y = (e^(2x) − e^(−x))/3
Solve general solution y'' + 2ky' + k²y = 0 (Equal roots)
m²+2km+k²=0 ⟹ (m+k)²=0 ⟹ m=−k (repeated root)
y = (C₁ + C₂x)e^(−kx)
Solve IVP: y'' + 2αy' + (α²+π²)y = 0, y(0)=3, y'(0)=−3α
Roots: m = −α ± πi. GS: y = e^(−αx)[C₁cos(πx)+C₂sin(πx)]
y(0)=3: C₁=3; y'(0)=−3α: −3α+C₂π=−3α ⟹ C₂=0
y = 3e^(−αx) cos(πx)
Solve IVP: y'' + y' − 6y = 0, y(0) = 10, y'(0) = 0
m=−3,2. GS: y=C₁e^(−3x)+C₂e^(2x). C₁=4, C₂=6
y = 4e^(−3x) + 6e^(2x)
Solve y' = (y−1) cot x (Separable) Both B-3195 papers
Separating: dy/(y−1) = cot x dx. Integrating: ln|y−1| = ln|sin x|+C
y = 1 + A sin x
Solve Non-Homogeneous: y'' + 4y = 8x² Section IV
CF: m=±2i → C₁cos2x+C₂sin2x
PI: Assume Ax²+Bx+C → A=2, B=0, C=−1 → PI=2x²−1
GS: y = C₁cos(2x) + C₂sin(2x) + 2x² − 1
Solve Non-Homogeneous IVP: y''−4y = e^(−2x)−2x, y(0)=0, y'(0)=0
CF = C₁e²ˣ+C₂e⁻²ˣ; PI₁ = −(x/4)e^(−2x); PI₂ = x/2
Applying ICs: C₁=−1/16, C₂=1/16
y = −(1/16)e^(2x) + (1/16)e^(−2x) − (x/4)e^(−2x) + x/2
Solve (D²−5D+6)y = e^(4x)
y = C₁e^(2x) + C₂e^(3x) + (1/2)e^(4x)
Solve Euler-Cauchy: x² y'' − 2.5x y' − 2y = 0
Euler-Cauchy: substitute x = e^t (or try y = x^m). The equation x²y''+bxy'+cy=0 becomes auxiliary equation m(m-1)+bm+c=0.
Auxiliary eqn: m²−3.5m−2=0 ⟹ m=4, m=−1/2
y = C₁x⁴ + C₂x^(−1/2)
🌊 Partial Differential Equations
Laplace equation verification, wave equation, separation of variables, general solutions of PDEs.
Show u = eˣ sin y is a solution of the Laplace equation
∂²u/∂x² + ∂²u/∂y² = eˣsin y − eˣsin y = 0 ✓
Show u = eˣ cos y satisfies the 2D Laplace equation Repeated
∂²u/∂x² = eˣcos y; ∂²u/∂y² = −eˣcos y
∂²u/∂x² + ∂²u/∂y² = 0 ✓ (Laplace equation satisfied)
Show u(x,t) = cos4t·sin2x is a solution of the wave equation
∂²u/∂t² = −16cos4t·sin2x ; ∂²u/∂x² = −4cos4t·sin2x
So ∂²u/∂t² = 4·∂²u/∂x² ✓ (wave equation with c=2)
Solve PDE uₓ + uᵧ = 0 by separation of variables
Separation of variables: assume u = X(x)·Y(y). Substitute, divide by XY to separate variables. Each side equals a constant.
Let u=X(x)Y(y). Then X'/X = −Y'/Y = k (constant)
X = Ae^(kx), Y = Be^(−ky)
u = Ce^(k(x−y))
If u = eˣ(x cos y − y sin y), show ∂²u/∂x² + ∂²u/∂y² = 0
∂²u/∂x² = eˣ[(x+2)cos y − y sin y]
∂²u/∂y² = eˣ[−(x+2)cos y + y sin y]
Sum = 0 ✓ (Laplace equation satisfied)
Solve Lagrange's PDE: x² ∂z/∂x + y² ∂z/∂y = (x+y)z
Lagrange's linear PDE Pp+Qq=R. Solve using auxiliary equations: dx/P = dy/Q = dz/R. Find two independent solutions u=c₁ and v=c₂. General solution is F(u,v)=0.
Auxiliary equations: dx/x² = dy/y² = dz/((x+y)z)
From dx/x² = dy/y²: −1/x = −1/y + c₁ ⟹ 1/y − 1/x = c₁
General solution: F(1/y − 1/x, ...) = 0 (or express in terms of found constant)
🎯 Laplace Transform
L{f(t)}, inverse Laplace, partial fractions, convolution, shifting theorem. High marks topic!
Define Laplace Transform + Prove L{1} = 1/s
Definition: L{f(t)} = F(s) = ∫₀^∞ e^(−st) f(t) dt (for s > 0)
Proof L{1} = 1/s:
L{1} = ∫₀^∞ e^(−st)·1 dt = [e^(−st)/(−s)]₀^∞ = 0 − 1/(−s) = 1/s ✓
Find L{e^(−at)} and L{e^(at)}
L{e^(−at)} = 1/(s+a) L{e^(at)} = 1/(s−a)
Find L{cos at} and L{sin at} Section IV Q32!
L{cos at} = s/(s²+a²) L{sin at} = a/(s²+a²)
Find L{cosh(at)}
L{cosh at} = s/(s²−a²)
Find L{(t²+1)²}
L{(t²+1)²} = 24/s⁵ + 4/s³ + 1/s
Find L⁻¹{s/(s²+36)} Repeated
L⁻¹{s/(s²+36)} = cos(6t)
Evaluate L⁻¹{(2s+1)/[s(s+1)]} — Partial Fractions Both B-3195 papers!
(2s+1)/[s(s+1)] = 1/s + 1/(s+1)
L⁻¹ = 1 + e^(−t)
Define inverse Laplace and find L⁻¹{1/(s+5)}
Definition: If L{f(t)} = F(s), then L⁻¹{F(s)} = f(t). The inverse Laplace transform recovers f(t) from F(s).
1/(s+5) = 1/(s−(−5)). Since L{e^(at)} = 1/(s−a):
L⁻¹{1/(s+5)} = e^(−5t)
Evaluate convolution: eᵗ ★ sin t
Convolution theorem: L{f★g} = F(s)·G(s). So compute in s-domain then take inverse! Alternatively, use the integral formula directly.
eᵗ★sin t = L⁻¹{1/[(s−1)(s²+1)]} = L⁻¹{(1/2)/(s−1) − (s+1)/[2(s²+1)]}
= (1/2)eᵗ − (1/2)cos t − (1/2)sin t
State and Prove Second Shifting Theorem (15 Marks Section IV)
If L{f(t)} = F(s) and g(t) = f(t−a)·u(t−a) where a > 0 and u is the unit step function, then:
Statement as above. Proof: change of variable τ=t−a in the integral gives the e^(−as) factor times F(s). ∎
🔢 Number Theory
GCD, LCM, Euler's phi-function, Fermat's Little Theorem, Wilson's Theorem, remainders, divisibility. Formula-based scoring!
When are two integers relatively prime? + State Wilson's Theorem
Relatively Prime: Two integers a and b are called relatively prime (or coprime) if gcd(a,b) = 1.
Wilson's Theorem: If p is a prime number, then
(p−1)! ≡ −1 (mod p) i.e., (p−1)! + 1 is divisible by p.
Find gcd(−8, −36) and gcd(34, 126) as linear combinations
gcd(−8,−36) = gcd(8,36) (sign doesn't matter for gcd). Use Euclidean Algorithm: keep dividing and replacing.
gcd(−8, −36) = 4
gcd(34, 126) = 2 expressed as: 2 = 26×34 − 7×126
Find gcd(12378, 3054) by Euclidean Algorithm
gcd(12378, 3054) = 6
Find lcm(272, 1479)
gcd(272, 1479) = 17
lcm(272, 1479) = 272×1479/17 = 23664
Calculate φ(5040) — Euler's Phi Function Repeated!
5040 = 2⁴·3²·5·7
φ(5040) = 5040·(½)·(⅔)·(⅘)·(6/7) = 1152
Compute φ(1575)
1575 = 3²×5²×7. φ(1575) = 1575·(⅔)·(⅘)·(6/7) = 720
Remainder of 8^103 ÷ 13 (Fermat's Little Theorem)
By FLT: 8^12 ≡ 1 (mod 13). 103=12×8+7, so 8^103 ≡ 8^7 (mod 13)
8^2≡−1, 8^4≡1, 8^7=8^4·8^2·8 ≡ −8 ≡ 5 (mod 13)
Remainder = 5
Find remainder when 2^1000 is divided by 17
2^16≡1(mod 17). 1000=16×62+8. 2^1000≡2^8=256≡1(mod 17)
Remainder = 1
Show 41 divides 2^20 − 1
2^10 = 1024 = 24×41 + 40 ≡ −1 (mod 41)
2^20 = (2^10)² ≡ (−1)² = 1 (mod 41)
∴ 2^20 − 1 ≡ 0 (mod 41) ⟹ 41 | (2^20 − 1) ✓
Find prime factorization of 1995
1995 = 3 × 5 × 7 × 19
Express 360 in standard form; find number and sum of divisors
360 = 2³×3²×5
Number of divisors = (3+1)(2+1)(1+1) = 24
Sum of divisors = 15 × 13 × 6 = 1170
Prove Fermat's Theorem: a^p ≡ a (mod p) [Section IV]
This is Fermat's Little Theorem. The proof uses mathematical induction on a, with the binomial theorem. Key idea: p | C(p,k) for 1 ≤ k ≤ p−1.
Proof by induction: Base a=1 ✓. Assume a^p≡a(mod p).
(a+1)^p = a^p + ΣC(p,k)a^k + 1 ≡ a+1 (mod p) since p|C(p,k) for 1≤k≤p−1 ✓
∴ a^p ≡ a (mod p) for all integers a ∎
🌀 Complex Numbers
Modulus, argument, polar form, roots of complex numbers using De Moivre's theorem.
Give the modulus of 3−2i Repeated
|3−2i| = √(3² + 2²) = √(9+4) = √13
Find polar form of √3 − i Repeated in B-3195, F-2065
r = 2, θ = −π/6
√3 − i = 2[cos(−π/6) + i sin(−π/6)]
Determine principal value of argument of −π − πi
Principal argument = −3π/4
Find all values of (−8i)^(1/3) L-3942!
−8i = 8[cos(−π/2)+i sin(−π/2)]. Cube roots (k=0,1,2):
z₁ = √3−i, z₂ = 2i, z₃ = −√3−i
Find all values of (−i)^(3/4)
−i = cos(−π/2)+i sin(−π/2). (−i)^(3/4) = cos((−3π/2+6kπ)/4)+i sin(...) for k=0,1,2,3
The 4 values are: cos(−3π/8)±i sin(3π/8) and cos(9π/8)±i sin(9π/8)
Find cube roots of unity
The three cube roots of unity are:
1, ω = (−1+i√3)/2, ω² = (−1−i√3)/2
where ω³=1 and 1+ω+ω²=0
Separate cos(α+iβ) into real and imaginary parts
cos(α+iβ) = cos α cosh β − i sin α sinh β
Real part: cos α cosh β Imaginary part: −sin α sinh β
〰️ Fourier Series
Euler's formulae, Fourier series for various functions. Appears in EVERY exam — sometimes worth 15 marks!
Write Euler's formulae for Fourier coefficients of 2π-periodic function ALL Papers!
For a 2π-periodic function, Fourier series: f(x) = a₀/2 + Σ[aₙcos(nx) + bₙsin(nx)]
Euler's Formulae:
a₀ = (1/π) ∫₋π^π f(x) dx
aₙ = (1/π) ∫₋π^π f(x) cos(nx) dx
bₙ = (1/π) ∫₋π^π f(x) sin(nx) dx
Find Fourier series of f(x) = |x| in [−π, π] ALL 4 PAPERS!
f(x)=|x| is EVEN (|−x|=|x|). For even functions: bₙ=0! Only compute a₀ and aₙ. Use integration by parts for aₙ — it's actually quick once you know the trick.
EVEN FUNCTION SHORTCUT: If f(x) is even → bₙ=0, and aₙ = (2/π)∫₀^π f(x)cos(nx)dx. Saves half the work!
f(x) = |x| is an even function ∴ bₙ = 0 for all n.
a₀ = (2/π)∫₀^π x dx = π
aₙ = (2/π)∫₀^π x cos(nx) dx = (2/πn²)[(−1)ⁿ−1]
For n even: aₙ=0. For n odd: aₙ = −4/(πn²)
f(x) = π/2 − (4/π)[cos x/1 + cos 3x/9 + cos 5x/25 + ...]
Find Fourier series of f(x) = x, 0 < x < π (Fourier Sine Series)
bₙ = 2(−1)^(n+1)/n
f(x) = 2[sin x − sin2x/2 + sin3x/3 − ...] = 2Σ(−1)^(n+1)sin(nx)/n
Fourier series for f(x) = {−π for −π
a₀=−π/2; aₙ=[(−1)ⁿ−1]/(πn²); and bₙ computed similarly
This is a piecewise function — compute all three using the integrals above.
Find Fourier series of f(x)=(π−x)/2 for 0
a₀=0, aₙ=0, bₙ=1/n
f(x) = Σ sin(nx)/n = sin x + sin2x/2 + sin3x/3 + ...
Fourier series for f(x) = x + x² in [−π, π]
Split: x is ODD, x² is EVEN. For the ODD part (x): only bₙ terms. For the EVEN part (x²): only a₀ and aₙ terms. Combine!
a₀ = 2π²/3, aₙ = 4(−1)ⁿ/n², bₙ = 2(−1)^(n+1)/n
f(x) = π²/3 + Σ[(4(−1)ⁿ/n²)cos nx + (2(−1)^(n+1)/n)sin nx]
📈 Linear Programming + Max/Min Problems
Graphical method for LPP — all 4 past paper problems solved with full steps and graphs. Worth 15 marks!
Maximize z = 5x₁+7x₂ subject to x₁+x₂≤4, 3x₁+8x₂≤24, 10x₁+7x₂≤35 L-3942 Q35
Constraints converted to lines. Intercepts found. Feasible region identified.
Corner points and z = 5x₁+7x₂:
| Corner Point | z = 5x₁+7x₂ |
|---|---|
| (0,0) | 0 |
| (3.5,0) | 17.5 |
| (7/3, 5/3) | 70/3 ≈ 23.3 |
| (8/5, 12/5) | 24.8 |
| (0,3) | 21 |
Maximum z = 24.8 at x₁ = 8/5, x₂ = 12/5
Minimize z = 20x₁+10x₂ subject to x₁+2x₂≤40, 3x₁+x₂≥30, 4x₁+3x₂≥60
This is a minimization with ≥ constraints (feasible region is unbounded below/right).
Key corner: A(6,12) gives z=240. This is the minimum (verify no lower corner exists).
Minimum z = 240 at x₁=6, x₂=12
Maximize z = 3x₁+4x₂ subject to 4x₁+2x₂≤80, 2x₁+5x₂≤180
| Corner | z = 3x₁+4x₂ |
|---|---|
| (0,0) | 0 |
| (20,0) | 60 |
| (2.5,35) | 147.5 |
| (0,36) | 144 |
Maximum z = 147.5 at x₁ = 2.5, x₂ = 35
Identify objective function and constraints in LPP (1 Mark)
Objective Function: The function to be maximized or minimized (e.g., z = 5x₁+7x₂)
Constraints: The restrictions/inequalities that x₁, x₂ must satisfy (e.g., x₁+x₂≤4, x₁,x₂≥0)
The constraint x₁,x₂≥0 is called the non-negativity constraint.
f(x)=x²+px+q, f(1)=3 is extreme value on [0,2] — max or min?
f'(x)=2x+p. f'(1)=0 ⟹ p=−2. f(1)=3 ⟹ q=4. So f(x)=x²−2x+4.
f''(1)=2>0 ⟹ x=1 is a MINIMUM. The minimum value is 3.
🔥 Hot Topics — Last Minute Focus
Based on frequency analysis of all 4 past papers. If you're short on time, study these first — they'll appear tomorrow!
| Question | Times Appeared | Where to Study | Expected Marks |
|---|---|---|---|
| Fourier series of f(x)=|x| | 🔥🔥🔥🔥 (ALL 4) | Fourier → F2 | 4–15 |
| Verify Lagrange MVT (x²+2x+9) | 🔥🔥🔥 (3 papers) | MVT → M3 | 2 |
| State Rolle's Theorem | 🔥🔥🔥 (3 papers) | MVT → M1 | 1 |
| LPP Graphical Method | 🔥🔥🔥🔥 (ALL 4) | LPP → LP1-LP3 | 15 |
| sinh x / cosh x definition | 🔥🔥🔥 (3 papers) | Hyp → H1 | 1 |
| Laplace transforms + table | 🔥🔥🔥🔥 (ALL 4) | Laplace → L1-L10 | 2-15 |
| L⁻¹{(2s+1)/s(s+1)} | 🔥🔥 (same code twice) | Laplace → L7 | 2 |
| IVP y''−y'−2y=0 | 🔥🔥 (same paper 2x) | ODE → O6 | 2 |
| Euler phi function calculation | 🔥🔥 (2 papers) | NT → N5, N6 | 2 |
| Polar form of √3−i | 🔥🔥 (2 papers) | Complex → C2 | 2 |
1. LPP Method (15 marks — mechanical, follow steps!)
2. Fourier f(x)=|x| (everywhere — same answer always!)
3. Laplace Table + L⁻¹ (just memorize the table)
4. Verify MVT (2 marks — very quick!)
5. Solve ODE y''−y'−6y=0 (char. equation)
6. Hyperbolic definitions (1 mark free!)
7. Euler phi function (one formula)
8. Fermat's theorem remainder (smart trick)
9. Polar form + roots (De Moivre)
10. Rolle + MVT statements (write definition)
sinh x = (eˣ-e⁻ˣ)/2
cosh x = (eˣ+e⁻ˣ)/2
cosh²-sinh² = 1
L{1}=1/s, L{eᵃᵗ}=1/(s-a)
L{sin at}=a/(s²+a²)
L{cos at}=s/(s²+a²)
L{cosh at}=s/(s²-a²)
MVT: f'(c)=[f(b)-f(a)]/(b-a)
φ(n)=n·∏(1-1/p)
Fermat: a^(p-1)≡1(mod p)
Fourier |x|: π/2-(4/π)Σcos(2m-1)x/(2m-1)²
ODE CF: char. eq. m²+pm+q=0