Concept

A set is a well-defined collection of distinct objects. Operations on sets (union ∪, intersection ∩, difference, complement) are visualised with Venn diagrams. A relation from set A to B is a subset of the Cartesian product A×B; a function is a special relation in which every element of A has exactly one image in B.

Key relation types are reflexive, symmetric, transitive; a relation that is all three is an equivalence relation.

Key Points & Formulas

• n(A∪B) = n(A) + n(B) − n(A∩B); for three sets add back n(A∩B∩C).
• De Morgan's laws: (A∪B)' = A'∩B' and (A∩B)' = A'∪B'.
• Cartesian product: if n(A)=m, n(B)=n, then n(A×B)=mn.
• A set with n elements has 2ⁿ subsets and 2ⁿ−1 proper subsets.
• Functions: one-one (injective), onto (surjective), bijective (both). Only bijective functions have an inverse.

Worked Example

Q. In a survey of 120 people, 70 read newspaper A, 55 read B and 25 read both. How many read at least one newspaper?

Use n(A∪B) = n(A)+n(B)−n(A∩B).

= 70 + 55 − 25 = 100.

So 100 people read at least one (and 120−100 = 20 read neither).

Practice MCQs

Q1. If n(A)=3 and n(B)=4, the number of elements in A×B is

(A) 7 (B) 12 (C) 9 (D) 16

Answer: B. n(A×B)=n(A)·n(B)=3·4=12.

Q2. A set has 5 elements. Its total number of subsets is

(A) 10 (B) 25 (C) 32 (D) 31

Answer: C. 2⁵ = 32 subsets (31 are proper).

Concept

Real numbers comprise rationals (terminating or recurring decimals) and irrationals (non-terminating, non-recurring). The binary system uses base 2 with digits 0 and 1; computers and several NDA questions use base conversion.

To convert decimal→binary, divide repeatedly by 2 and read remainders bottom-up; binary→decimal, multiply each digit by the place value 2ⁿ.

Key Points & Formulas

• Place values in binary: …, 16, 8, 4, 2, 1.
• Decimal→binary: successive division by 2; collect remainders in reverse.
• Binary→decimal: Σ(digit × 2^position).
• A rational number p/q is terminating only if q has prime factors 2 and/or 5 only.

Worked Example

Q. Convert the decimal number 26 to binary.

26÷2=13 r0, 13÷2=6 r1, 6÷2=3 r0, 3÷2=1 r1, 1÷2=0 r1.

Read remainders bottom to top: 11010.

Check: 16+8+0+2+0 = 26. ✓ So 26 = 11010₂.

Practice MCQs

Q1. The binary number 1101 equals which decimal value?

(A) 11 (B) 13 (C) 12 (D) 14

Answer: B. 8+4+0+1 = 13.

Q2. Which of the following is an irrational number?

(A) 0.75 (B) √9 (C) √7 (D) 22/7

Answer: C. √7 is non-terminating, non-recurring; √9=3 and 22/7 are rational.

Concept

A complex number is z = a + ib where i = √(−1), a is the real part and b the imaginary part. The modulus |z| = √(a²+b²) gives its distance from the origin in the Argand plane, and the argument is the angle it makes with the positive real axis.

The three cube roots of unity are 1, ω, ω² with 1 + ω + ω² = 0 and ω³ = 1.

Key Points & Formulas

• i² = −1, i³ = −i, i⁴ = 1 (powers of i repeat with period 4).
• |z| = √(a²+b²); arg(z)=tan⁻¹(b/a), adjusted for the quadrant.
• Conjugate of a+ib is a−ib; z·z̄ = |z|².
• Cube roots of unity: 1+ω+ω²=0, ω³=1.

Worked Example

Q. Find the modulus and the conjugate of z = 5 − 12i.

Modulus |z| = √(5² + (−12)²) = √(25+144) = √169 = 13.

Conjugate z̄ = 5 + 12i.

Practice MCQs

Q1. The value of i⁷ is

(A) i (B) −i (C) 1 (D) −1

Answer: B. i⁷ = i⁴·i³ = 1·(−i) = −i.

Q2. If ω is a cube root of unity, then 1 + ω + ω² equals

(A) 1 (B) 0 (C) 3 (D) ω

Answer: B. By definition the three cube roots of unity sum to 0.

Concept

A quadratic equation ax²+bx+c=0 (a≠0) has two roots given by the quadratic formula. The discriminant D = b²−4ac decides the nature of the roots. Inequalities are solved by finding critical points and testing sign intervals.

Sum and product of roots relate directly to the coefficients.

Key Points & Formulas

• Roots: x = [−b ± √(b²−4ac)] / 2a.
• D>0: real & distinct; D=0: real & equal; D<0: complex conjugate roots.
• Sum of roots = −b/a; product of roots = c/a.
• If roots are α,β, the equation is x² − (α+β)x + αβ = 0.

Worked Example

Q. Find the value of k for which x² − kx + 9 = 0 has equal roots.

Equal roots ⇒ discriminant D = 0.

D = k² − 4(1)(9) = k² − 36 = 0.

k² = 36 ⇒ k = ±6.

Practice MCQs

Q1. The sum of the roots of 2x² − 8x + 6 = 0 is

(A) 4 (B) −4 (C) 3 (D) 8

Answer: A. Sum = −b/a = −(−8)/2 = 4.

Q2. The roots of x² + x + 1 = 0 are

(A) Real & equal (B) Real & distinct (C) Complex (D) Zero

Answer: C. D = 1−4 = −3 < 0, so the roots are complex.

Concept

An arithmetic progression (AP) has a constant common difference d; a geometric progression (GP) has a constant common ratio r; a harmonic progression (HP) is the reciprocal of an AP. These appear in nearly every paper and reward quick formula recall.

The arithmetic, geometric and harmonic means satisfy AM ≥ GM ≥ HM for positive numbers.

Key Points & Formulas

• AP: nth term aₙ = a + (n−1)d; sum Sₙ = n/2 [2a + (n−1)d].
• GP: nth term aₙ = a r^(n−1); sum Sₙ = a(rⁿ−1)/(r−1), r≠1.
• Infinite GP (|r|<1): S∞ = a/(1−r).
• For two numbers: AM = (a+b)/2, GM = √(ab), and GM² = AM·HM.

Worked Example

Q. The 4th term of a GP is 24 and the 7th term is 192. Find the common ratio.

a r⁶ / a r³ = r³ = 192/24 = 8.

r = ∛8 = 2.

Practice MCQs

Q1. The sum of the first 10 terms of the AP 4, 7, 10, … is

(A) 175 (B) 170 (C) 180 (D) 160

Answer: A. S = 10/2[2·4 + 9·3] = 5[8+27] = 5·35 = 175.

Q2. The sum to infinity of 1 + 1/2 + 1/4 + … is

(A) 2 (B) 1 (C) 3/2 (D) ∞

Answer: A. S∞ = a/(1−r) = 1/(1−½) = 2.

Concept

Permutations count ordered arrangements; combinations count unordered selections. The fundamental counting principle (multiply the number of choices at each stage) underlies both.

Watch carefully for restrictions, repetition and whether order matters.

Key Points & Formulas

• nPr = n!/(n−r)! (arrangements of r out of n).
• nCr = n!/[r!(n−r)!] (selections of r out of n).
• nCr = nC(n−r); nC0 = nCn = 1.
• Arrangements of n objects with repeats p,q,…: n!/(p! q! …).

Worked Example

Q. In how many ways can a committee of 3 men and 2 women be chosen from 6 men and 4 women?

Choose 3 men: 6C3 = 20. Choose 2 women: 4C2 = 6.

By the counting principle, total = 20 × 6 = 120.

Practice MCQs

Q1. The number of ways to arrange the letters of the word 'LEVEL' is

(A) 120 (B) 30 (C) 60 (D) 20

Answer: B. 5!/(2!·2!) = 120/4 = 30 (two L's and two E's).

Q2. The value of 7C2 is

(A) 21 (B) 42 (C) 14 (D) 49

Answer: A. 7C2 = 7·6/2 = 21.

Concept

The binomial theorem expands (a+b)ⁿ as a sum of n+1 terms whose coefficients are the binomial coefficients nCr. It is used to find a particular term, a coefficient, or the term independent of x.

The general (r+1)th term lets you target any specific term without full expansion.

Key Points & Formulas

• (a+b)ⁿ = Σ_{r=0}^{n} nCr a^(n−r) b^r.
• General term: T_{r+1} = nCr a^(n−r) b^r.
• Number of terms = n+1; middle term is the ((n/2)+1)th if n is even.
• Sum of all binomial coefficients = 2ⁿ.

Worked Example

Q. Find the term independent of x in the expansion of (x + 1/x)⁸.

General term T_{r+1} = 8Cr x^(8−r) (1/x)^r = 8Cr x^(8−2r).

Independent of x ⇒ 8−2r = 0 ⇒ r = 4.

Term = 8C4 = 70.

Practice MCQs

Q1. The number of terms in the expansion of (2x + y)⁷ is

(A) 7 (B) 8 (C) 6 (D) 14

Answer: B. An expansion of degree n has n+1 = 8 terms.

Q2. The coefficient of x³ in (1+x)⁶ is

(A) 15 (B) 20 (C) 6 (D) 30

Answer: B. 6C3 = 20.

Concept

A logarithm answers 'to what power must the base be raised to get this number?': if a^x = N then log_a N = x. Logarithms convert multiplication/division into addition/subtraction, which simplifies many calculations.

The common base is 10; the natural base is e.

Key Points & Formulas

• log_a(mn) = log_a m + log_a n.
• log_a(m/n) = log_a m − log_a n.
• log_a(mᵏ) = k log_a m.
• Change of base: log_a m = log_b m / log_b a; and log_a a = 1, log_a 1 = 0.

Worked Example

Q. If log₁₀ 2 = 0.301, find log₁₀ 8.

log₁₀ 8 = log₁₀ 2³ = 3 log₁₀ 2.

= 3 × 0.301 = 0.903.

Practice MCQs

Q1. The value of log₃ 81 is

(A) 3 (B) 4 (C) 9 (D) 27

Answer: B. 3⁴ = 81, so log₃ 81 = 4.

Q2. log 20 + log 5 (base 10) equals

(A) 1 (B) 2 (C) 3 (D) 100

Answer: B. log(20·5)=log 100 = 2.

Concept

A matrix is a rectangular array of numbers; a determinant is a scalar associated with a square matrix. Matrices are used to represent and solve systems of linear equations (Cramer's rule, matrix-inverse method).

Matrix multiplication is not commutative (AB ≠ BA in general).

Key Points & Formulas

• Determinant of [[a,b],[c,d]] = ad − bc.
• Triangular/diagonal matrix: determinant = product of diagonal entries.
• A is singular (no inverse) ⇔ det A = 0.
• Inverse: A⁻¹ = adj(A)/det(A); product order: (m×n)(n×p) → m×p.

Worked Example

Q. For what value of k is the matrix [[k, 3], [2, 4]] singular?

Singular ⇒ determinant = 0.

det = 4k − 6 = 0 ⇒ k = 6/4 = 3/2.

Practice MCQs

Q1. The determinant of [[2,1],[5,3]] is

(A) 1 (B) 11 (C) 6 (D) −1

Answer: A. 2·3 − 1·5 = 6 − 5 = 1.

Q2. If A is 3×2 and B is 2×4, the order of AB is

(A) 3×4 (B) 2×2 (C) 4×3 (D) 3×2

Answer: A. Inner dimensions match; result is 3×4.

Concept

Trigonometric ratios relate the angles and sides of a right triangle. Identities (Pythagorean, sum/difference, multiple-angle) let you transform expressions and solve equations efficiently.

Knowing the standard-angle table and sign by quadrant is essential.

Key Points & Formulas

• sin²θ + cos²θ = 1; 1 + tan²θ = sec²θ; 1 + cot²θ = cosec²θ.
• sin(A±B) = sinA cosB ± cosA sinB; cos(A±B) = cosA cosB ∓ sinA sinB.
• sin2θ = 2 sinθ cosθ; cos2θ = 2cos²θ − 1.
• Maximum of a sinθ + b cosθ is √(a²+b²); minimum is −√(a²+b²).

Worked Example

Q. If sinθ = 3/5 and θ is acute, find cosθ and tanθ.

cos²θ = 1 − sin²θ = 1 − 9/25 = 16/25 ⇒ cosθ = 4/5.

tanθ = sinθ/cosθ = (3/5)/(4/5) = 3/4.

Practice MCQs

Q1. The value of 2 sin45° cos45° is

(A) 1 (B) ½ (C) √2 (D) 0

Answer: A. 2 sin45°cos45° = sin90° = 1.

Q2. The maximum value of 5 sinθ + 12 cosθ is

(A) 17 (B) 13 (C) 12 (D) 7

Answer: B. √(5²+12²) = √169 = 13.

Concept

Inverse trigonometric functions return the angle for a given ratio, within a fixed principal-value range. They are heavily tested for their identities and principal values.

Each inverse function has a restricted range to remain single-valued.

Key Points & Formulas

• sin⁻¹x ∈ [−π/2, π/2]; cos⁻¹x ∈ [0, π]; tan⁻¹x ∈ (−π/2, π/2).
• sin⁻¹x + cos⁻¹x = π/2; tan⁻¹x + cot⁻¹x = π/2.
• tan⁻¹x + tan⁻¹y = tan⁻¹[(x+y)/(1−xy)] when xy < 1.
• sin⁻¹(−x) = −sin⁻¹x; cos⁻¹(−x) = π − cos⁻¹x.

Worked Example

Q. Evaluate sin⁻¹(1/2) + cos⁻¹(1/2).

By the identity sin⁻¹x + cos⁻¹x = π/2 for all valid x.

So the sum = π/2 (here π/6 + π/3).

Practice MCQs

Q1. The principal value of cos⁻¹(−1) is

(A) 0 (B) π (C) π/2 (D) −π

Answer: B. cos π = −1 and π lies in [0,π], so cos⁻¹(−1)=π.

Q2. tan⁻¹(1) equals

(A) π/4 (B) π/3 (C) π/2 (D) π/6

Answer: A. tan(π/4)=1, within range, so tan⁻¹1 = π/4.

Concept

These apply trigonometry to triangles and to real-world distance problems. The sine and cosine rules relate sides and angles in any triangle; heights-and-distances problems use angles of elevation and depression.

Drawing a clear figure is half the solution.

Key Points & Formulas

• Sine rule: a/sinA = b/sinB = c/sinC = 2R.
• Cosine rule: c² = a² + b² − 2ab cosC.
• Area = ½ ab sinC = √[s(s−a)(s−b)(s−c)] (Heron), s = (a+b+c)/2.
• Angle of elevation/depression is measured from the horizontal.

Worked Example

Q. A tower subtends an angle of 30° at a point 60 m from its base on level ground. Find its height.

tan30° = height / 60.

height = 60 × tan30° = 60 × (1/√3) = 60/√3 = 20√3 ≈ 34.6 m.

Practice MCQs

Q1. In a triangle, a=8, b=6, C=90°. The side c equals

(A) 10 (B) 12 (C) 14 (D) 7

Answer: A. c² = 64+36 = 100 ⇒ c = 10.

Q2. The angle of elevation of the sun when a pole's shadow equals its height is

(A) 30° (B) 45° (C) 60° (D) 90°

Answer: B. tanθ = height/shadow = 1 ⇒ θ = 45°.

Concept

Coordinate geometry studies points, lines and curves using algebra. A straight line can be written in several forms; the slope captures its steepness and direction.

Distance, section and slope formulas are the workhorses of this topic.

Key Points & Formulas

• Distance between (x₁,y₁),(x₂,y₂) = √[(x₂−x₁)²+(y₂−y₁)²].
• Slope m = (y₂−y₁)/(x₂−x₁); slope-intercept form y = mx + c.
• Parallel lines: equal slopes; perpendicular lines: m₁m₂ = −1.
• Distance of point (x₀,y₀) from line ax+by+c=0 = |ax₀+by₀+c|/√(a²+b²).

Worked Example

Q. Find the distance of the point (3, 4) from the line 4x + 3y − 10 = 0.

Distance = |4·3 + 3·4 − 10| / √(4²+3²).

= |12 + 12 − 10| / √25 = 14/5 = 2.8 units.

Practice MCQs

Q1. The slope of the line joining (2,3) and (4,7) is

(A) 2 (B) 1 (C) ½ (D) 4

Answer: A. (7−3)/(4−2) = 4/2 = 2.

Q2. Two lines with slopes 2 and −½ are

(A) Parallel (B) Perpendicular (C) Coincident (D) Intersecting at 45°

Answer: B. Product of slopes = 2·(−½) = −1, so they are perpendicular.

Concept

A circle is the locus of points equidistant (the radius) from a fixed point (the centre). Its equation comes in standard and general forms, from which the centre and radius are read off.

Many questions just need correct extraction of centre and radius.

Key Points & Formulas

• Standard form: (x−h)² + (y−k)² = r², centre (h,k), radius r.
• General form: x²+y²+2gx+2fy+c=0; centre (−g,−f), radius √(g²+f²−c).
• A circle centred at the origin: x²+y² = r².
• Tangency / position is judged by comparing distance-from-centre with r.

Worked Example

Q. Find the centre and radius of x² + y² − 6x + 4y − 12 = 0.

Here 2g = −6 ⇒ g = −3; 2f = 4 ⇒ f = 2; c = −12.

Centre = (−g, −f) = (3, −2).

Radius = √(g²+f²−c) = √(9+4+12) = √25 = 5.

Practice MCQs

Q1. The radius of the circle x² + y² = 49 is

(A) 7 (B) 49 (C) 14 (D) √7

Answer: A. r² = 49 ⇒ r = 7.

Q2. The centre of x²+y²−4x−6y+9=0 is

(A) (2,3) (B) (−2,−3) (C) (4,6) (D) (3,2)

Answer: A. Centre = (−g,−f) = (2,3).

Concept

Conics are curves obtained by slicing a cone: the parabola, ellipse and hyperbola. Each has a standard equation, a focus–directrix definition, and an eccentricity e that classifies it.

Eccentricity is the single most-asked property.

Key Points & Formulas

• Parabola y² = 4ax: vertex (0,0), focus (a,0), e = 1.
• Ellipse x²/a² + y²/b² = 1 (a>b): e = √(1 − b²/a²), e < 1.
• Hyperbola x²/a² − y²/b² = 1: e = √(1 + b²/a²), e > 1.
• Length of latus rectum: parabola 4a; ellipse/hyperbola 2b²/a.

Worked Example

Q. Find the eccentricity of the ellipse x²/25 + y²/16 = 1.

a² = 25, b² = 16 (a>b).

e = √(1 − b²/a²) = √(1 − 16/25) = √(9/25) = 3/5.

Practice MCQs

Q1. The eccentricity of any parabola is

(A) 0 (B) 1 (C) <1 (D) >1

Answer: B. By definition a parabola has e = 1.

Q2. For the parabola y² = 12x, the focus is at

(A) (3,0) (B) (0,3) (C) (12,0) (D) (6,0)

Answer: A. 4a = 12 ⇒ a = 3; focus (a,0) = (3,0).

Concept

3D geometry extends coordinates to space (x,y,z). It deals with distances, direction cosines/ratios of a line, and equations of lines, planes and spheres. It is a frequent source of the paper's harder questions.

Direction cosines are the cosines of the angles a line makes with the axes.

Key Points & Formulas

• Distance from origin to (x,y,z) = √(x²+y²+z²).
• Direction cosines l,m,n satisfy l²+m²+n² = 1.
• Direction cosines = direction ratios divided by their magnitude.
• Plane: ax+by+cz+d = 0; (a,b,c) is the normal direction.
• Sphere centred (a,b,c) radius r: (x−a)²+(y−b)²+(z−c)² = r².

Worked Example

Q. Find the direction cosines of the line joining the origin to (2, 3, 6).

Magnitude = √(2²+3²+6²) = √(4+9+36) = √49 = 7.

Direction cosines = (2/7, 3/7, 6/7).

Check: (4+9+36)/49 = 1. ✓

Practice MCQs

Q1. The distance of the point (3, 4, 12) from the origin is

(A) 13 (B) 12 (C) 19 (D) √19

Answer: A. √(9+16+144) = √169 = 13.

Q2. If the direction ratios of a line are (1, 2, 2), its direction cosines are

(A) (1/3,2/3,2/3) (B) (1,2,2) (C) (1/5,2/5,2/5) (D) (1/2,1,1)

Answer: A. Magnitude = 3; divide each ratio by 3.

Concept

A limit describes the value a function approaches near a point. A function is continuous at a point if its limit there equals its value; it is differentiable if its graph has a well-defined tangent (no sharp corner).

Differentiability implies continuity, but not the reverse.

Key Points & Formulas

• Standard limits: lim(x→0) sinx/x = 1; lim(x→0)(eˣ−1)/x = 1.
• For 0/0 forms, factorise or use L'Hôpital's idea.
• Continuous at x=a: LHL = RHL = f(a).
• |x| is continuous everywhere but not differentiable at x = 0.

Worked Example

Q. Evaluate lim(x→3) (x² − 9)/(x − 3).

Direct substitution gives 0/0, so factorise.

(x²−9)/(x−3) = (x−3)(x+3)/(x−3) = x+3.

Limit = 3 + 3 = 6.

Practice MCQs

Q1. lim(x→0) sin5x / x equals

(A) 5 (B) 1 (C) 0 (D) 1/5

Answer: A. sin5x/x = 5·(sin5x/5x) → 5·1 = 5.

Q2. The function f(x) = |x| at x = 0 is

(A) Discontinuous (B) Continuous but not differentiable (C) Differentiable (D) Undefined

Answer: B. |x| is continuous at 0 but has a corner, so not differentiable there.

Concept

Differentiation gives the instantaneous rate of change (slope) of a function. The standard rules — power, product, quotient and chain — let you differentiate almost any combination of functions.

Mastering the rules makes calculus questions quick and mechanical.

Key Points & Formulas

• d/dx(xⁿ) = n x^(n−1); d/dx(sinx)=cosx; d/dx(eˣ)=eˣ; d/dx(lnx)=1/x.
• Product rule: (uv)' = u'v + uv'.
• Quotient rule: (u/v)' = (u'v − uv')/v².
• Chain rule: dy/dx = (dy/du)(du/dx).

Worked Example

Q. Differentiate y = x³ ln x with respect to x.

Product rule with u = x³, v = ln x.

dy/dx = (3x²)(ln x) + (x³)(1/x).

= 3x² ln x + x².

Practice MCQs

Q1. d/dx (sin x · cos x) at x = 0 is

(A) 1 (B) 0 (C) −1 (D) ½

Answer: A. Derivative = cos2x; at x=0, cos0 = 1.

Q2. The derivative of e^(3x) is

(A) 3e^(3x) (B) e^(3x) (C) 3e^x (D) e^(3x)/3

Answer: A. Chain rule: e^(3x)·3.

Concept

Derivatives are applied to find tangents and normals, to determine where a function increases or decreases, and to locate maxima and minima. Optimisation ('maximum area/volume') problems are a classic format.

At a maximum or minimum the first derivative is zero.

Key Points & Formulas

• Slope of tangent at a point = value of dy/dx there.
• f'(x) > 0 ⇒ increasing; f'(x) < 0 ⇒ decreasing.
• Critical points: solve f'(x) = 0.
• Second-derivative test: f''<0 ⇒ maximum; f''>0 ⇒ minimum.

Worked Example

Q. Find the local maximum value of f(x) = x³ − 12x.

f'(x) = 3x² − 12 = 0 ⇒ x = ±2.

f''(x) = 6x; at x = −2, f'' = −12 < 0 (maximum).

Maximum value f(−2) = −8 + 24 = 16.

Practice MCQs

Q1. The function f(x)=x²−4x has a minimum at x =

(A) 2 (B) −2 (C) 4 (D) 0

Answer: A. f'(x)=2x−4=0 ⇒ x=2; f''=2>0, minimum.

Q2. The slope of the tangent to y = x² at x = 3 is

(A) 6 (B) 9 (C) 3 (D) 12

Answer: A. dy/dx = 2x = 6 at x=3.

Concept

Integration is the reverse of differentiation and also measures area under a curve. Indefinite integrals include an arbitrary constant; definite integrals give a number using the limits.

Substitution and integration by parts (ILATE order) handle most cases.

Key Points & Formulas

• ∫xⁿ dx = x^(n+1)/(n+1) + C (n≠−1); ∫(1/x)dx = ln|x| + C.
• ∫sinx dx = −cosx + C; ∫eˣ dx = eˣ + C.
• Definite: ∫ₐᵇ f(x)dx = F(b) − F(a).
• Area under y=f(x) from a to b (above x-axis) = ∫ₐᵇ f(x) dx.

Worked Example

Q. Evaluate ∫₀² (3x² + 2) dx.

Antiderivative = x³ + 2x.

At x=2: 8 + 4 = 12; at x=0: 0.

Definite integral = 12 − 0 = 12.

Practice MCQs

Q1. ∫₀^(π/2) cos x dx equals

(A) 1 (B) 0 (C) π/2 (D) 2

Answer: A. [sin x]₀^(π/2) = 1 − 0 = 1.

Q2. The area under y = x from x=0 to x=4 is

(A) 8 (B) 16 (C) 4 (D) 2

Answer: A. ∫₀⁴ x dx = [x²/2]₀⁴ = 8.

Concept

A differential equation relates a function to its derivatives. Its order is the highest derivative present; its degree is the power of that highest derivative. First-order equations are solved by variable separation, homogeneous methods, or the integrating factor.

Growth and decay problems are common applications.

Key Points & Formulas

• Order = highest derivative; degree = power of the highest-order derivative.
• Variable separable: bring all x with dx and all y with dy, then integrate.
• Linear form dy/dx + Py = Q: integrating factor = e^(∫P dx).
• Exponential growth/decay: dy/dx = ky ⇒ y = Ce^(kx).

Worked Example

Q. Solve dy/dx = 2xy.

Separate variables: dy/y = 2x dx.

Integrate: ln|y| = x² + c.

y = C e^(x²).

Practice MCQs

Q1. The order and degree of (d²y/dx²) + 3(dy/dx)² + y = 0 are

(A) 2 and 1 (B) 2 and 2 (C) 1 and 2 (D) 2 and 3

Answer: A. Highest derivative is 2nd order, raised to power 1, so order 2, degree 1.

Q2. The general solution of dy/dx = y is

(A) y = Ce^x (B) y = Cx (C) y = x+C (D) y = Ce^(−x)

Answer: A. Separable: dy/y = dx ⇒ y = Ce^x.

Concept

A vector has both magnitude and direction. The dot product yields a scalar (and the angle between vectors); the cross product yields a vector perpendicular to both (and the area of a parallelogram).

Vectors model forces, displacements and geometric relations.

Key Points & Formulas

• Magnitude of a î + b ĵ + c k̂ = √(a²+b²+c²).
• Dot product a·b = |a||b|cosθ = a₁b₁ + a₂b₂ + a₃b₃.
• Cross product magnitude |a×b| = |a||b|sinθ.
• a·b = 0 ⇒ perpendicular; a×b = 0 ⇒ parallel.

Worked Example

Q. Find the angle between a = î + ĵ and b = ĵ + k̂.

a·b = (1)(0)+(1)(1)+(0)(1) = 1.

|a| = √2, |b| = √2, so cosθ = 1/(√2·√2) = 1/2.

θ = 60°.

Practice MCQs

Q1. The dot product (2î+3ĵ+k̂)·(î−ĵ+2k̂) is

(A) 1 (B) 2 (C) 5 (D) −1

Answer: A. 2−3+2 = 1.

Q2. î × ĵ equals

(A) k̂ (B) −k̂ (C) ĵ (D) 0

Answer: A. By right-hand rule, î × ĵ = k̂.

Concept

Statistics summarises data using measures of central tendency (mean, median, mode) and dispersion (variance, standard deviation). The NDA favours direct, formula-based questions, making this a high-scoring topic.

Understand how transformations of data affect these measures.

Key Points & Formulas

• Mean = Σx/n; median = middle value (sorted); mode = most frequent value.
• Variance = Σ(x−x̄)²/n; standard deviation = √variance.
• Adding a constant k to every value: mean shifts by k, SD unchanged.
• Multiplying every value by k: mean and SD both scale by |k|.

Worked Example

Q. Find the variance of 2, 4, 6, 8.

Mean = (2+4+6+8)/4 = 5.

Squared deviations: 9, 1, 1, 9; sum = 20.

Variance = 20/4 = 5.

Practice MCQs

Q1. The median of 12, 4, 7, 9, 2 is

(A) 7 (B) 9 (C) 4 (D) 12

Answer: A. Sorted: 2,4,7,9,12; middle value = 7.

Q2. If SD of a data set is 4 and each value is doubled, the new SD is

(A) 8 (B) 4 (C) 16 (D) 2

Answer: A. SD scales by the multiplier: 2×4 = 8.

Concept

Probability measures the chance of an event, from 0 (impossible) to 1 (certain). The classical definition is favourable outcomes over total outcomes. Conditional probability, Bayes' theorem and the binomial distribution build on this.

Identify whether events are independent or mutually exclusive before applying rules.

Key Points & Formulas

• P(E) = favourable / total; 0 ≤ P(E) ≤ 1.
• Addition rule: P(A∪B) = P(A) + P(B) − P(A∩B).
• Independent events: P(A∩B) = P(A)·P(B).
• Binomial: P(r successes) = nCr pʳ q^(n−r); mean = np.

Worked Example

Q. Two fair dice are thrown. Find the probability that the sum is 9.

Favourable pairs: (3,6),(4,5),(5,4),(6,3) = 4 outcomes.

Total outcomes = 36.

Probability = 4/36 = 1/9.

Practice MCQs

Q1. A card is drawn from a standard pack. P(it is a heart) is

(A) 1/4 (B) 1/13 (C) 1/52 (D) 1/2

Answer: A. 13 hearts out of 52 = 1/4.

Q2. If P(A)=0.4, P(B)=0.5 and A,B are independent, P(A∩B) is

(A) 0.2 (B) 0.9 (C) 0.1 (D) 0.45

Answer: A. Independent ⇒ 0.4×0.5 = 0.2.