CLASS XI | FOM - II

Fundamental of Mathematics – II

1. Remainder Theorem of Polynomial

1.1 Zeroes of a Polynomial

A number a is called a zero (or root) of the polynomial p(x) if p(a) = 0. Zeroes give insights into the behavior and factorization of polynomials.

Practice Questions

• Find the zeroes of p(x) = x² – 5x + 6. Show Answer
  Solve x² – 5x + 6 = 0 ⇒ (x – 2)(x – 3) = 0 ⇒ x = 2, 3.

1.2 Roots of an Equation

The roots of the equation p(x) = 0 coincide with the zeroes of the polynomial. Solving the polynomial equation yields its roots, which may be real or complex.

Practice Questions

• Determine the roots of x³ – 2x² – x + 2 = 0. Show Answer
  By grouping: x²(x–2) –1(x–2) = (x–2)(x²–1) ⇒ (x–2)(x–1)(x+1)=0 ⇒ x=2,1,−1.

1.3 Factors of a Polynomial

If a is a zero of p(x), then (x – a) is a factor of p(x). Repeated roots lead to repeated factors.

Practice Questions

• Verify that (x – 3) is a factor of x³ – 6x² + 11x – 6. Show Answer
  p(3)=27 –54+33 –6=0, so (x–3) is a factor.

1.4 Remainder when Factor is Known

When dividing p(x) by (x – a), the remainder is p(a). This follows directly from the Remainder Theorem.

Practice Questions

• Find the remainder on dividing 2x³ + 3x² – x + 5 by (x + 2). Show Answer
  Here a = –2 ⇒ p(–2) = 2(–8)+3(4) –(–2)+5 = –16+12+2+5 = 3.

2. Remainder of Higher Power of a Number

2.1 Positive Remainder Concept

When an integer a is divided by m, the positive remainder r satisfies 0 ≤ r < m and a = qm + r for some integer q.

Practice Questions

• Divide 47 by 8 and state the positive remainder. Show Answer
  47 ÷ 8 = 5 with remainder 7.

2.2 Negative Remainder Concept

Sometimes we allow a negative remainder r' satisfying –(m–1) ≤ r' < 0. Then a = q'm + r'.

Practice Questions

• Express 47 divided by 8 with a negative remainder. Show Answer
  47 = 6×8 –1, so negative remainder = –1.

2.3 Addition of Remainder Rule

If a ≡ r₁ (mod m) and b ≡ r₂ (mod m), then (a + b) ≡ (r₁ + r₂) mod m, adjust back into [0, m).

Practice Questions

• Find the remainder of (27 + 35) ÷ 7 by first finding individual remainders. Show Answer
  27≡6, 35≡0 ⇒ sum ≡6 ⇒ remainder 6.

2.4 Subtraction of Remainder Rule

Similarly, (a – b) ≡ (r₁ – r₂) mod m, then adjust to the standard range.

Practice Questions

• Compute the remainder of (42 – 19) ÷ 5 using remainders. Show Answer
  42≡2, 19≡4 ⇒ 2–4=–2 ≡ 3 (mod 5).

2.5 Multiplication of Remainder Rule

If a ≡ r₁ and b ≡ r₂ (mod m), then ab ≡ r₁·r₂ (mod m), adjust as needed.

Practice Questions

• Find remainder of 14×18 divided by 5 using remainders. Show Answer
  14≡4, 18≡3 ⇒ product ≡12 ≡2 (mod 5).

2.6 Remainder 1 or –1 Concept

When a ≡ ±1 (mod m), powers of a cycle quickly: aⁿ ≡ (±1)ⁿ.

Practice Questions

• What is remainder of 7⁵ ÷ 6? Show Answer
  7≡1 ⇒1⁵=1 ⇒ remainder 1.

2.7 When Remainder is Not ±1

For a ≡ r (mod m) with |r|>1, compute powers and reduce at each step to keep numbers small.

Practice Questions

• Find remainder of 4⁶ ÷ 7 by reducing intermediate powers. Show Answer
  4²=16≡2; 4⁴≡2²=4; 4⁶≡4·2=8≡1 (mod 7).

2.8 Remainder with Common Factor

If gcd(a,m)=d>1, dividing by m may share factors. Always reduce fraction a/m by d first.

Practice Questions

• Compute remainder of 18 ÷ 12 considering common factor. Show Answer
  gcd(18,12)=6 ⇒ 18=1·12+6 ⇒ remainder 6.

3. Brackets

3.1 Types of Brackets

We use three bracket styles: discrete curly { }, continuous round ( ), and continuous square [ ].

3.1.1 Discrete Bracket { }

Used to denote sets, e.g. {1,2,3}.

Practice Questions

• Write the set of prime numbers less than 10 using discrete brackets. Show Answer
  {2, 3, 5, 7}

3.1.2 Continuous Brackets ( ) & [ ]

Denote grouping in algebra (x+2)(x–3) or nested expressions [a(b+c)].

Practice Questions

• Simplify (x+1)[2(x–1)+3]. Show Answer
  (x+1)[2x–2+3]=(x+1)(2x+1)=2x²+x+2x+1=2x²+3x+1.

3.1.3 Use of Open Bracket

An open bracket marks the beginning of a group; it must be matched by a corresponding closing bracket.

3.1.4 Use of Closed Bracket

The closing bracket ends a grouping. Always ensure brackets nest correctly.

Practice Questions

• Identify the matching pairs in {[3(x+2)] – (x–1)}. Show Answer
  { … } contains [ … ], which contains ( … ).

4. Interval

4.1 Open Interval

Denoted (a, b), meaning all x with a < x < b.

Practice Questions

• Express the set of real x between 2 and 5, excluding endpoints. Show Answer
  (2, 5)

4.2 Closed Interval

Denoted [a, b], meaning a ≤ x ≤ b.

Practice Questions

• Write x such that 0 ≤ x ≤ 1 in interval notation. Show Answer
  [0, 1]

4.3 Representation of Interval in Inequality

(a, b) ↔ a < x < b, [a, b] ↔ a ≤ x ≤ b.

4.4 Representation on Number Line

Open circles for endpoints not included; filled dots for included endpoints.

4.5 Conversion between Interval & Inequality

Replace parentheses & bracket symbols with the appropriate <, ≤ signs.

5. Linear Equations

5.1 In One Variable

A linear equation in one variable has form ax + b = 0. Solve by isolating x: x = –b/a (a≠0).

Practice Questions

• Solve 3x – 7 = 5. Show Answer
  3x = 12 ⇒ x = 4.

5.2 Graphical & Interval Representation

On number line, mark the solution point. In interval form, a single solution x₀ is [x₀, x₀].

5.3 In Two Variables

Form ax + by + c = 0. Solutions form a straight line in the plane.

Practice Questions

• Graph 2x + 3y – 6 = 0 by finding intercepts. Show Answer
  x–int: y=0⇒x=3; y–int: x=0⇒y=2 ⇒ draw line through (3,0),(0,2).

5.4 Parallel, Intersecting & Coincidental Lines

• Parallel: same slope, different intercept.
• Intersecting: slopes differ, meet at one point.
• Coincidental: identical equations (infinite solutions).

Practice Questions

• Determine relationship of 2x+4y=6 and x+2y=3. Show Answer
  Second is half of first ⇒ coincidental.

6. Linear Inequality

6.1 In One Variable

Form ax+b < 0 (or ≤, ≥, >). Solve by isolating x; flip inequality when multiplying/dividing by negative.

Practice Questions

• Solve and graph –2x + 5 > 1. Show Answer
  –2x > –4 ⇒ x < 2. Graph: open circle at 2, arrow left.

6.2 In Two Variables

Form ax + by + c < 0. Graph boundary line (dashed if strict), shade the half-plane satisfying inequality.

Practice Questions

• Sketch region for y – x ≥ 2. Show Answer
  Boundary y=x+2 (solid). Shade above.

7. Inequalities

7.1 Rules of Inequalities

• Adding/subtracting same term keeps direction.
• Multiplying/dividing by positive keeps direction; by negative reverses it.

7.2 Wavy Curve Method

A graphical technique for solving polynomial inequalities by marking zeros on number line and testing sign in each interval.

Practice Questions

• Solve (x–1)(x+2)>0 by wavy curve. Show Answer
  Roots at x=–2,1. Sign pattern: + for x<–2, – between, + for x>1 ⇒ x<–2 or x>1.

7.3 Extensions of Wavy Curve

• Division by linear factors: treat sign flips carefully.
• Higher powers: even multiplicity doesn’t change sign across root; odd does.

Practice Questions

• Solve (x–3)²(x+1)<0. Show Answer
  Root x=3 (even) doesn’t change sign; x=–1 (odd) flips. Negative region is x<–1.