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SaitechAI Mathematics Lecture Series (Class 11–12, JEE/NEET Level)
A partial fraction expresses a rational function as a sum of simpler fractions. If \( \frac{P(x)}{Q(x)} \) is a rational function and \( \deg P(x) < \deg Q(x) \), it can be written as a sum of partial fractions.
If \( \frac{P(x)}{Q(x)} \) is proper, its decomposition depends on the factors of \( Q(x) \):
Example 1:
\( \frac{3x – 5}{(x – 1)(x + 2)} = \frac{A}{x – 1} + \frac{B}{x + 2} \)
Multiply both sides by \( (x – 1)(x + 2) \):
\( 3x – 5 = A(x + 2) + B(x – 1) \)
Let \( x = 1 \Rightarrow A = -\frac{2}{3} \);
\( x = -2 \Rightarrow B = \frac{11}{3} \)
Final form:
\( \frac{3x – 5}{(x – 1)(x + 2)} = \frac{-2/3}{x – 1} + \frac{11/3}{x + 2} \)
Example 2 (Repeated Factor):
\( \frac{2x + 3}{(x + 1)^2} = \frac{A}{x + 1} + \frac{B}{(x + 1)^2} \)
Multiply: \( 2x + 3 = A(x + 1) + B \)
Let \( x = -1 \Rightarrow B = 1 \); comparing coefficients → \( A = 2 \)
So, \( \frac{2x + 3}{(x + 1)^2} = \frac{2}{x + 1} + \frac{1}{(x + 1)^2} \)
Example 3 (Irreducible Quadratic):
\( \frac{2x^2 + 3x + 4}{(x + 1)(x^2 + 1)} = \frac{A}{x + 1} + \frac{Bx + C}{x^2 + 1} \)
After decomposition, integrate each term separately:
\( \int \frac{A}{x – a} dx = A \ln|x – a| + C \)
\( \int \frac{Bx + C}{x^2 + px + q} dx = \text{Use substitution or arctan form} \)

Video Lecture

Idea: Molecules at the surface experience a net inward cohesive pull, making the surface behave like a stretched membrane.
Surface tension (also called surface force per unit length) is defined as
$$ T \equiv \frac{F}{L} $$
Surface energy: Work required to increase the surface area by unit amount. In SI, numerical value of surface energy per unit area equals \(T\) (J m\(^{-2}\) ↔ N m\(^{-1}\)).
For a spherical drop of radius \(r\):
$$ \Delta P = \frac{2T}{r} \quad \text{(inside higher than outside)} $$
For a spherical bubble of radius \(r\):
$$ \Delta P = \frac{4T}{r} $$
These follow from mechanical equilibrium of a curved surface under tension.
Capillary rise/fall in a tube of radius \(r\):
$$ h = \frac{2T\cos\theta}{\rho g r} $$
Meniscus: Concave when adhesion \(>\) cohesion (\(\theta<90^\circ\)); convex when cohesion \(>\) adhesion (\(\theta>90^\circ\)).
To create new area \( \Delta A \) at constant \(T\):
$$ W = T\,\Delta A, \qquad \text{so} \quad \frac{dW}{dA} = T. $$
Interpretation: \(T\) is the surface free energy per unit area (isothermal, reversible addition of area).
| Liquid | Approx. \(T\) (N m\(^{-1}\)) | Remarks |
|---|---|---|
| Water | 0.072 | High; strong hydrogen bonding |
| Alcohol (ethanol) | ~0.022 | Lower than water |
| Glycerol | ~0.063 | Viscous, relatively high \(T\) |
| Mercury | ~0.485 | Very high; poor wetting on glass |
| Soap solution | ~0.025–0.040 | Reduced by surfactants |
Values are indicative for classroom use; exact values depend on temperature and purity.
For a bubble of radius \( r = 1.0\,\text{mm} \) with \( T = 0.030\,\mathrm{N\,m^{-1}} \):
$$ \Delta P = \frac{4T}{r} = \frac{4\times 0.030}{1.0\times 10^{-3}} = 120\,\text{Pa}. $$
\( r = 0.50\,\text{mm},\; T = 0.072\,\mathrm{N\,m^{-1}},\; \rho = 1000\,\mathrm{kg\,m^{-3}},\; \theta \approx 0^\circ \):
$$ h = \frac{2T\cos\theta}{\rho g r} = \frac{2 \times 0.072 \times 1}{1000 \times 9.8 \times 0.5\times 10^{-3}} \approx 0.029\,\text{m} \;=\; 2.9\,\text{cm}. $$
© SaitechAI — Prepared for Class 11 learners. You may print or save this page for study use.
Worksheet in Surface Tension, Surface Energy, Capillarity, contact angle, pressure inside the soap bubble.
Modern Periodic Table — Lecture Notes (Class 11, SaitechAI Edition)
| Feature | Periods | Groups |
|---|---|---|
| Number | 7 | 18 |
| Represents | Principal quantum number (n) | Valence shell configuration |
| Example | 2nd period → Li to Ne | Group 17 → Halogens (F, Cl, Br, I, At) |
| Block | Range | Example Elements | Characteristic |
|---|---|---|---|
| s-block | 1–2 | Na, Mg | Highly reactive metals |
| p-block | 13–18 | B, C, N, O, F | Includes nonmetals, metalloids |
| d-block | 3–12 | Fe, Cu, Zn | Transition metals |
| f-block | Lanthanoids, Actinoids | Ce, U | Inner transition metals |

where ( Z ) = atomic number, ( p ) = number of protons, ( n ) = neutrons.
Check formal charge to find the most stable structure.
Count valence electrons of all atoms.
Arrange atoms: the least electronegative atom is usually central (except H).
Form bonds by pairing electrons between atoms.
Complete octets of outer atoms first, then central atom.
Use multiple bonds (double/triple) if needed to satisfy octet.

Activities
STD12 State Board | Maths | Application of Integrals | 4050
Marks: 10 x 3 marks = 30 marks | Time: 1 hour
Level STD 12 CBSE | Maths
Antiderivative by method of inspection
Quiz in Antiderivatives – Activity
Antiderivatives – worksheet-1 | worksheet-2
Substitution Methods – Worksheet-1 | Worksheet-2 | Worksheet-3
Partial Fractions Decomposition