१२ वी (विज्ञान, कला) · गणित व संख्याशास्त्र भाग 2 · पाठ 1
Differentiation
थोडक्यात सारांश
This chapter extends derivatives to more functions. It covers the derivative of composite functions (chain rule), inverse functions, inverse trigonometric functions, logarithmic differentiation, implicit functions, parametric functions, and higher order derivatives.
महत्त्वाचे मुद्दे
Chain rule: if y = f(u) and u = g(x), then dy/dx = (dy/du)(du/dx).
Derivative of an inverse function: dx/dy = 1/(dy/dx), provided dy/dx ≠ 0.
Logarithmic differentiation is used for functions like xˣ, [f(x)]^g(x), or long products/quotients: take log on both sides first.
Implicit function: f(x, y) = 0. Differentiate both sides with respect to x, treating y as a function of x, then solve for dy/dx.
Parametric functions x = f(t), y = g(t): dy/dx = (dy/dt)/(dx/dt).
Second derivative d²y/dx² is the derivative of dy/dx. For parametric functions, d²y/dx² = d/dt(dy/dx) ÷ (dx/dt).
Substitution (e.g. x = sin θ, x = tan θ) often simplifies derivatives of inverse trigonometric expressions.
महत्त्वाच्या संज्ञा
Composite function – function of a function, f(g(x)).
Implicit function – y not expressed directly in terms of x.
Parameter – a third variable t through which x and y are related.
सूत्रे
Function
Derivative
sin⁻¹x
1/√(1 − x²)
cos⁻¹x
−1/√(1 − x²)
tan⁻¹x
1/(1 + x²)
cot⁻¹x
−1/(1 + x²)
sec⁻¹x
1/(
x
√(x² − 1))
aˣ
aˣ log a
log x
1/x
eˣ
eˣ
xⁿ
n xⁿ⁻¹
Product rule
(uv)' = u v' + v u'
Quotient rule
(u/v)' = (v u' − u v')/v²
५ महत्त्वाचे प्रश्न
Q1. Differentiate y = sin(x² + 5) with respect to x.
Let u = x² + 5. dy/dx = cos u × 2x = 2x cos(x² + 5).
Q2. Find dy/dx if y = xˣ.
log y = x log x. Differentiating: (1/y) dy/dx = x(1/x) + log x = 1 + log x. So dy/dx = xˣ (1 + log x).
Q3. Find dy/dx if x² + y² + xy = 7.
2x + 2y (dy/dx) + x (dy/dx) + y = 0. So dy/dx (2y + x) = −(2x + y), dy/dx = −(2x + y)/(x + 2y).
Q4. Find dy/dx if x = a cos³t and y = a sin³t.
dx/dt = −3a cos²t sin t; dy/dt = 3a sin²t cos t. dy/dx = (3a sin²t cos t)/(−3a cos²t sin t) = −tan t.
Q5. Differentiate y = tan⁻¹(2x/(1 − x²)) with respect to x (|x| < 1).
Put x = tan θ: 2x/(1 − x²) = tan 2θ, so y = 2θ = 2 tan⁻¹x. dy/dx = 2/(1 + x²).
परीक्षेसाठी टीप
Identify the type first (composite, implicit, parametric, log), then apply the right method. Learn the six inverse trig derivatives by heart.
१२ वी (विज्ञान, कला) · गणित व संख्याशास्त्र भाग 2 · पाठ 2
Applications of Derivatives
थोडक्यात सारांश
Derivatives measure rates of change. This chapter uses them to find tangents and normals to curves, rates of change, approximations, Rolle's theorem and the Lagrange mean value theorem, increasing and decreasing functions, and maxima and minima.
महत्त्वाचे मुद्दे
Slope of the tangent at (x₁, y₁) is m = (dy/dx) at that point. Tangent: y − y₁ = m(x − x₁).
Slope of the normal = −1/m. Normal: y − y₁ = (−1/m)(x − x₁).
dy/dt gives the rate of change of y with time; related rates use the chain rule.
Approximation: f(a + h) ≈ f(a) + h f'(a), for small h.
Rolle's theorem: if f is continuous on [a, b], differentiable on (a, b) and f(a) = f(b), there is c in (a, b) with f'(c) = 0.
Lagrange's mean value theorem (LMVT): there is c in (a, b) with f'(c) = (f(b) − f(a))/(b − a).
f is increasing where f'(x) > 0 and decreasing where f'(x) < 0.
१२ वी (विज्ञान, कला) · गणित व संख्याशास्त्र भाग 2 · पाठ 3
Indefinite Integration
थोडक्यात सारांश
Integration is the reverse process of differentiation. This chapter covers standard integrals, integration by substitution, integration by parts, integration using partial fractions, and special integrals involving a² ± x² forms.
महत्त्वाचे मुद्दे
If d/dx F(x) = f(x), then ∫ f(x) dx = F(x) + c, where c is the constant of integration.
∫ [f(x) ± g(x)] dx = ∫ f dx ± ∫ g dx; ∫ k f(x) dx = k ∫ f(x) dx.
Substitution: put a part of the integrand equal to t so that its derivative also appears.
१२ वी (विज्ञान, कला) · गणित व संख्याशास्त्र भाग 2 · पाठ 4
Definite Integration
थोडक्यात सारांश
A definite integral gives a number: the net value of an integral between two limits. This chapter covers the definite integral as the limit of a sum, the fundamental theorem of integral calculus, methods of evaluation, and important properties of definite integrals.
महत्त्वाचे मुद्दे
Fundamental theorem: if ∫ f(x) dx = F(x), then ∫ from a to b of f(x) dx = F(b) − F(a).
The constant of integration is not needed in definite integrals.
A definite integral can be seen as the limit of a sum of areas of thin rectangles.
When substituting, change the limits too.
Properties are very useful to evaluate integrals that are hard to do directly, especially the property f(x) → f(a + b − x).
For an even function, the integral from −a to a is twice the integral from 0 to a. For an odd function it is zero.
महत्त्वाच्या संज्ञा
Lower limit (a) and upper limit (b) – the end values of x.
Even function – f(−x) = f(x), e.g. x², cos x.
Odd function – f(−x) = −f(x), e.g. x³, sin x.
सूत्रे
Property
Statement (all integrals with respect to x)
P1
∫ₐᵇ f(x) dx = ∫ₐᵇ f(t) dt
P2
∫ₐᵇ f(x) dx = −∫ᵇₐ f(x) dx
P3
integral from a to b = integral from a to c + integral from c to b, where a < c < b
[−cos x] from 0 to π/2 = −cos(π/2) + cos 0 = 0 + 1 = 1.
Q3. Evaluate I = integral of √(sin x)/(√(sin x) + √(cos x)) from 0 to π/2.
Using P5, replacing x by π/2 − x gives I = integral of √(cos x)/(√(cos x) + √(sin x)) over the same limits. Adding both forms: 2I = integral of 1 dx from 0 to π/2 = π/2. So I = π/4.
Q4. Evaluate the integral of x³ cos x from −π/2 to π/2.
f(x) = x³ cos x. f(−x) = −x³ cos x = −f(x), so f is odd. By P7 the integral is 0.
Q5. Evaluate the integral of 1/(1 + x²) from 0 to 1.
[tan⁻¹x] from 0 to 1 = π/4 − 0 = π/4.
परीक्षेसाठी टीप
If a definite integral looks impossible directly, try property P4/P5 and add the two forms — this trick appears almost every year.
१२ वी (विज्ञान, कला) · गणित व संख्याशास्त्र भाग 2 · पाठ 5
Application of Definite Integration
थोडक्यात सारांश
Definite integrals can find areas of regions bounded by curves. This chapter covers the area under a curve with the x-axis or y-axis, the area of standard curves (circle, ellipse, parabola), and the area between two curves.
महत्त्वाचे मुद्दे
Area bounded by y = f(x), the x-axis and lines x = a, x = b is ∫ from a to b of y dx (when f(x) ≥ 0).
Area bounded by x = g(y), the y-axis and lines y = c, y = d is ∫ from c to d of x dy.
If the curve is below the x-axis, the integral is negative; take its absolute value for the area.
If the curve crosses the axis, split the integral at the crossing points and add absolute values.
Area between two curves y = f(x) (upper) and y = g(x) (lower) from x = a to b is ∫ from a to b of [f(x) − g(x)] dx.
Find the limits by solving the equations of the curves together (points of intersection).
Use symmetry to simplify: for a circle or ellipse, find the area of one quadrant and multiply by 4.
Always draw a rough sketch first.
महत्त्वाच्या संज्ञा
Bounded region – region enclosed completely by curves and lines.
Points of intersection – where the curves meet; they give the limits.
Symmetric region – region identical on both sides of an axis.
१२ वी (विज्ञान, कला) · गणित व संख्याशास्त्र भाग 2 · पाठ 6
Differential Equations
थोडक्यात सारांश
A differential equation (DE) contains derivatives of an unknown function. This chapter covers order and degree, formation of DEs, solving DEs by variable separation, homogeneous DEs, linear DEs, and applications such as growth and decay, Newton's law of cooling, and surface area/population problems.
महत्त्वाचे मुद्दे
Order is the order of the highest derivative in the equation. Degree is the power of the highest order derivative after making the equation free of radicals and fractions in derivatives.
A general solution contains arbitrary constants equal in number to the order. A particular solution has specific values of the constants.
Formation of DE: differentiate the given relation as many times as there are arbitrary constants, then eliminate them.
Variable separable: write f(y) dy = g(x) dx and integrate both sides.
Homogeneous DE: dy/dx = f(y/x). Put y = v x, so dy/dx = v + x dv/dx, then separate variables.
Linear DE: dy/dx + P y = Q (P, Q functions of x). Integrating factor IF = e^(∫ P dx). Solution: y × IF = ∫ Q × IF dx + c.
Growth and decay: dx/dt = k x gives x = x₀ e^(kt) (k > 0 growth, k < 0 decay).
Newton's law of cooling: dθ/dt = −k(θ − θ₀).
महत्त्वाच्या संज्ञा
Order – highest derivative present.
Degree – power of the highest order derivative.
Integrating factor (IF) – factor that makes a linear DE exactly integrable.
Arbitrary constant – constant c in the general solution.
सूत्रे
Type
Method
Variable separable
∫ f(y) dy = ∫ g(x) dx + c
Homogeneous
y = v x, dy/dx = v + x dv/dx
Linear in y
IF = e^(∫ P dx), y·IF = ∫ Q·IF dx + c
Linear in x
dx/dy + P x = Q, IF = e^(∫ P dy)
Growth/decay
x = x₀ e^(kt)
५ महत्त्वाचे प्रश्न
Q1. Find the order and degree of (d²y/dx²)³ + (dy/dx)⁴ + y = 0.
Highest derivative is d²y/dx², so order = 2. Its power is 3, so degree = 3.
Q2. Form the DE by eliminating the arbitrary constant in y = A e^(2x).
१२ वी (विज्ञान, कला) · गणित व संख्याशास्त्र भाग 2 · पाठ 7
Probability Distributions
थोडक्यात सारांश
A random variable assigns a number to each outcome of a random experiment. This chapter covers discrete and continuous random variables, probability mass function (p.m.f.), cumulative distribution function (c.d.f.), probability density function (p.d.f.), and the expected value, variance and standard deviation of a random variable.
महत्त्वाचे मुद्दे
A random variable X is a real-valued function on the sample space.
Discrete random variable: takes countable values (number of heads, number on a die).
Continuous random variable: takes any value in an interval (height, time).
p.m.f. of a discrete X: P(X = xᵢ) = pᵢ, with each pᵢ ≥ 0 and Σ pᵢ = 1.
c.d.f.: F(x) = P(X ≤ x). It is non-decreasing, from 0 to 1.
p.d.f. of a continuous X: f(x) ≥ 0 and the integral of f(x) over the whole range = 1. P(a ≤ X ≤ b) = ∫ from a to b of f(x) dx.
For a continuous variable, P(X = a) = 0 for any single value a.
१२ वी (विज्ञान, कला) · गणित व संख्याशास्त्र भाग 2 · पाठ 8
Binomial Distribution
थोडक्यात सारांश
The binomial distribution gives the probability of a number of successes in a fixed number of independent trials, each with the same probability of success. This chapter covers Bernoulli trials, the binomial distribution, and its mean, variance and standard deviation.
महत्त्वाचे मुद्दे
A Bernoulli trial has exactly two outcomes: success (probability p) and failure (probability q = 1 − p).
Conditions for binomial distribution: (1) fixed number of trials n; (2) trials are independent; (3) each trial has two outcomes; (4) p is the same in every trial.
If X ~ B(n, p), P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ, x = 0, 1, 2, …, n.
Mean = n p; variance = n p q; SD = √(n p q).
Variance is always less than the mean in a binomial distribution (since q < 1).