1. How to Use This Page
This page covers all exercise questions from TN Class 11 Maths Chapter 2 — Basic Algebra with detailed, step-by-step solutions.
- Read the Key Concepts section first for a quick formula recap.
- Work through each exercise section (MCQs, Fill in the Blanks, etc.) in order.
- For detailed 5-mark problems, study the worked numerical examples in code blocks — they show every intermediate step.
- Use the Common Mistakes and Exam Checklist sections before your exam for last-minute revision.
2. Key Concepts (Quick Recap)
2.1 Quadratic Equations
A quadratic equation has the form ax² + bx + c = 0 where a ≠ 0.
| Item | Formula / Result |
|---|---|
| Discriminant | Δ = b² − 4ac |
| Roots (quadratic formula) | x = (−b ± √Δ) / 2a |
| Sum of roots | α + β = −b/a |
| Product of roots | αβ = c/a |
| Nature: Δ > 0 (real, distinct) | Two different real roots |
| Nature: Δ = 0 (real, equal) | One repeated real root |
| Nature: Δ < 0 (complex) | Two complex conjugate roots |
2.2 Quadratic Inequalities
To solve ax² + bx + c > 0 or ax² + bx + c < 0:
1. Find the roots α, β of ax² + bx + c = 0 (assume α < β). 2. Plot the roots on a number line — they divide it into three intervals: (−∞, α), (α, β), (β, ∞). 3. Test the sign of the quadratic in each interval. 4. Pick intervals matching the inequality sign (> 0 or < 0).
Rule of signs: If a > 0, the parabola opens upward — positive outside roots, negative between roots.
2.3 Modulus (Absolute Value) Function
| Property | Formula | ||||||
|---|---|---|---|---|---|---|---|
| Definition | \ | x\ | = x if x ≥ 0; \ | x\ | = −x if x < 0 | ||
| Non-negativity | \ | x\ | ≥ 0 ∀ x ∈ ℝ | ||||
| Multiplicativity | \ | xy\ | = \ | x\ | · \ | y\ | |
| Triangle inequality | \ | x + y\ | ≤ \ | x\ | + \ | y\ | |
| \ | x\ | < a (a > 0) | −a < x < a | ||||
| \ | x\ | > a (a > 0) | x < −a or x > a | ||||
| \ | x\ | = a | x = ±a |
2.4 Logarithms
If aˣ = n, then logₐ n = x (where a > 0, a ≠ 1, n > 0).
| Property | Formula |
|---|---|
| Product rule | logₐ(mn) = logₐ m + logₐ n |
| Quotient rule | logₐ(m/n) = logₐ m − logₐ n |
| Power rule | logₐ(mⁿ) = n · logₐ m |
| Change of base | logₐ b = log b / log a |
| logₐ a | 1 |
| logₐ 1 | 0 |
| logₐ(aⁿ) | n |
2.5 Polynomial Division and Remainder Theorem
- Remainder Theorem: When polynomial P(x) is divided by (x − a), the remainder is P(a).
- Factor Theorem: (x − a) is a factor of P(x) if and only if P(a) = 0.
- Division Algorithm: P(x) = D(x) · Q(x) + R(x), where deg(R) < deg(D).
2.6 Rational Inequalities
To solve a rational inequality like (x − 1)/(x + 3) > 0:
1. Find zeros of numerator and denominator separately. 2. Plot all critical points on a number line. 3. Test sign in each interval. 4. Remember: zeros of the denominator are never included (vertical asymptotes).
3. I. Choose the Best Answer (MCQs)
Q1. If the discriminant of ax² + bx + c = 0 is zero, then the roots are
(a) real and distinct (b) real and equal (c) complex conjugates (d) imaginary
Q2. The value of log₂ 32 is
(a) 4 (b) 5 (c) 6 (d) 16
Q3. If |x − 3| < 5, then x belongs to
(a) (−2, 8) (b) (−8, 2) (c) (−∞, −2) ∪ (8, ∞) (d) [−2, 8]
| Explanation: | x − 3 | < 5 means −5 < x − 3 < 5. Adding 3 to all parts: −2 < x < 8, i.e., x ∈ (−2, 8). |
|---|
Q4. If α and β are roots of x² − 5x + 6 = 0, then α + β equals
(a) 5 (b) −5 (c) 6 (d) −6
Q5. The solution set of x² − 4x + 3 ≤ 0 is
(a) (−∞, 1] ∪ [3, ∞) (b) [1, 3] (c) (1, 3) (d) {1, 3}
Q6. log₃ 81 − log₃ 9 equals
(a) 1 (b) 2 (c) 3 (d) 4
4. II. Fill in the Blanks
1. If Δ > 0 for a quadratic ax² + bx + c = 0, the roots are real and distinct.
2. The sum of the roots of 2x² − 8x + 3 = 0 is 4. (α + β = −b/a = 8/2 = 4)
| 3. | x − 7 | = 3 implies x = 4 or 10. (x − 7 = 3 → x = 10; x − 7 = −3 → x = 4) |
|---|
4. log₁₀ 1000 = 3. (10³ = 1000)
5. The product of the roots of x² + 7x + 12 = 0 is 12. (αβ = c/a = 12/1 = 12)
6. The remainder when P(x) = x³ − 2x + 1 is divided by (x − 2) is 5. (P(2) = 8 − 4 + 1 = 5)
| 7. | x + 4 | ≥ 2 is satisfied when x ≤ −6 or x ≥ −2. |
|---|
8. The discriminant of 3x² + 5x − 2 = 0 is 49. (Δ = 25 + 24 = 49)
9. log₅ 125 = 3. (5³ = 125)
10. If (x − 3) is a factor of P(x), then P(3) = 0.
11. The solution of (x − 2)/(x + 1) < 0 lies in the interval (−1, 2).
12. For the quadratic 4x² + kx + 9 = 0 to have equal roots, k = ±12. (Δ = k² − 144 = 0 → k = ±12)
5. III. True or False (with corrections for false statements)
Q1. The discriminant of a quadratic equation determines only whether the roots are real or not.
False. The discriminant also distinguishes between distinct roots (Δ > 0), equal roots (Δ = 0), and complex roots (Δ < 0). It tells us more than just real vs. non-real.
Q2. If |x| < 3, then x ∈ (−3, 3).
| True. | x | < a (with a > 0) is equivalent to −a < x < a. |
|---|
Q3. log(ab) = log a · log b
False. The correct identity is log(ab) = log a + log b. The product rule involves addition, not multiplication.
Q4. If α and β are roots of ax² + bx + c = 0, then αβ = −c/a.
False. The product of roots is αβ = c/a (not −c/a). The negative sign only appears in the sum: α + β = −b/a.
Q5. A rational inequality f(x)/g(x) > 0 can include the zeros of g(x) in its solution set.
False. The zeros of g(x) (the denominator) make the expression undefined and are never included in the solution set, regardless of the inequality direction. Only zeros of f(x) may be included for ≥ or ≤.
Q6. |x − a| + |x − b| ≥ |a − b| for all real x.
| True. By the triangle inequality, | x − a | + | x − b | = | x − a | + | b − x | ≥ | b − a | = | a − b | . |
|---|
6. IV. Match the Following
| Column A | Column B | ||
|---|---|---|---|
| 1. Δ > 0, a > 0 | (c) Two distinct real roots, parabola opens upward | ||
| 2. \ | x\ | = a (a > 0) | (f) x = a or x = −a |
| 3. logₐ aⁿ | (d) n | ||
| 4. Remainder of P(x) ÷ (x − k) | (a) P(k) | ||
| 5. x² + 2x + 1 = 0 | (e) Two equal real roots (Δ = 0) | ||
| 6. log(m/n) | (b) log m − log n |
7. V. Answer Briefly (2-Mark Questions)
Q1. Find the roots of 2x² − 7x + 3 = 0 using the quadratic formula.
a = 2, b = −7, c = 3
Δ = b² − 4ac = 49 − 24 = 25
x = (7 ± √25) / (2 × 2)
x = (7 ± 5) / 4
x₁ = (7 + 5)/4 = 12/4 = 3
x₂ = (7 − 5)/4 = 2/4 = 1/2
Roots: x = 3, x = 1/2
Q2. Solve |2x − 1| = 5.
2x − 1 = 5 → 2x = 6 → x = 3
2x − 1 = −5 → 2x = −4 → x = −2
Solution: x = 3 or x = −2
Q3. Evaluate: log₃ 54 − log₃ 2.
log₃ 54 − log₃ 2 = log₃(54/2) = log₃ 27
27 = 3³, so log₃ 27 = 3
Answer: 3
Q4. Find the remainder when 2x³ − 3x² + 4x − 7 is divided by (x − 1).
By the Remainder Theorem, remainder = P(1)
P(1) = 2(1) − 3(1) + 4(1) − 7
= 2 − 3 + 4 − 7
= −4
Remainder = −4
Q5. Write the solution set of x² − 9 > 0 in interval notation.
x² − 9 > 0
(x − 3)(x + 3) > 0
Roots: x = −3, x = 3
Since coefficient of x² is positive, the expression is positive outside the roots.
Solution set: (−∞, −3) ∪ (3, ∞)
8. VI. Answer in Detail (5-Mark Questions)
Q1. Solve the quadratic equation 3x² − 5x − 2 = 0 by completing the square method. Also find the nature of its roots.
Step 1: Write in standard form
3x² − 5x − 2 = 0
Step 2: Divide by 3 to make coefficient of x² equal to 1
x² − (5/3)x = 2/3
Step 3: Add (5/6)² to both sides
(coefficient of x / 2)² = (5/6)² = 25/36
x² − (5/3)x + 25/36 = 2/3 + 25/36
Step 4: Simplify right side
2/3 + 25/36 = 24/36 + 25/36 = 49/36
Step 5: Factor the left side
(x − 5/6)² = 49/36
Step 6: Take square root
x − 5/6 = ±7/6
Step 7: Solve for x
x = 5/6 + 7/6 = 12/6 = 2
x = 5/6 − 7/6 = −2/6 = −1/3
Roots: x = 2 and x = −1/3
Nature of roots:
Δ = b² − 4ac = (−5)² − 4(3)(−2) = 25 + 24 = 49 > 0
Since Δ > 0, roots are real and distinct. ✓
Q2. Solve the quadratic inequality x² − x − 6 ≤ 0 and represent the solution on a number line.
Step 1: Factorise the quadratic
x² − x − 6 = (x − 3)(x + 2)
Roots: x = −2 and x = 3
Step 2: Identify critical points and intervals
Critical points divide the number line into:
Interval I: x < −2
Interval II: −2 < x < 3
Interval III: x > 3
Step 3: Test sign in each interval
Pick x = −3: (−3 − 3)(−3 + 2) = (−6)(−1) = +6 > 0 → positive
Pick x = 0: (0 − 3)(0 + 2) = (−3)(2) = −6 < 0 → negative
Pick x = 4: (4 − 3)(4 + 2) = (1)(6) = +6 > 0 → positive
Step 4: Since a = 1 > 0, the parabola opens upward:
Positive outside roots, negative between roots.
Step 5: The inequality is ≤ 0 (non-positive), so select
the region where the expression is negative or zero:
x ∈ [−2, 3]
The endpoints −2 and 3 are included because
the inequality is ≤ (not <).
Number line representation:
─────●═══════════════●─────
−2 3
←──neg───zero──pos──zero──pos──→
Answer: x ∈ [−2, 3]
Q3. Prove that logₐ(mn) = logₐ m + logₐ n. Using this, evaluate log₂ 8 + log₂ 32.
PROOF:
Let logₐ m = p → aᵖ = m ... (1)
Let logₐ n = q → aᵍ = n ... (2)
Multiply (1) and (2):
aᵖ · aᵍ = m · n
aᵖ⁺ᵍ = mn
By definition of logarithm:
logₐ(mn) = p + q
logₐ(mn) = logₐ m + logₐ n ∎
EVALUATION:
log₂ 8 + log₂ 32 = log₂(8 × 32)
= log₂(256)
= log₂(2⁸)
= 8
Verification:
log₂ 8 = log₂(2³) = 3
log₂ 32 = log₂(2⁵) = 5
3 + 5 = 8 ✓
Q4. Solve the inequality |3x − 4| ≤ 5 and express the solution in interval notation.
Step 1: Apply the property |expression| ≤ a → −a ≤ expression ≤ a
−5 ≤ 3x − 4 ≤ 5
Step 2: Add 4 to all three parts
−5 + 4 ≤ 3x ≤ 5 + 4
−1 ≤ 3x ≤ 9
Step 3: Divide by 3
−1/3 ≤ x ≤ 3
Solution in interval notation: x ∈ [−1/3, 3]
Verification:
For x = 0: |3(0) − 4| = |−4| = 4 ≤ 5 ✓
For x = 3: |3(3) − 4| = |5| = 5 ≤ 5 ✓
For x = 4: |3(4) − 4| = |8| = 8 > 5 ✗ (not in solution)
For x = −1: |3(−1) − 4| = |−7| = 7 > 5 ✗ (not in solution)
Q5. Using the Remainder Theorem, find the value of k if (x − 2) is a factor of P(x) = x³ + kx² − 4x + 4.
Step 1: If (x − 2) is a factor, then by Factor Theorem, P(2) = 0
P(2) = (2)³ + k(2)² − 4(2) + 4
= 8 + 4k − 8 + 4
= 4k + 4
Step 2: Set P(2) = 0
4k + 4 = 0
4k = −4
k = −1
Answer: k = −1
Verification: P(x) = x³ − x² − 4x + 4
P(2) = 8 − 4 − 8 + 4 = 0 ✓
Factorising: x³ − x² − 4x + 4 = (x − 2)(x² + x − 2)
= (x − 2)(x + 2)(x − 1) ✓
Q6. Solve the rational inequality (x + 3)(x − 1) / (x − 2) > 0.
Step 1: Identify zeros of numerator and denominator
Numerator zeros: x = −3 and x = 1
Denominator zero: x = 2 (excluded — makes expression undefined)
Step 2: Plot critical points on number line
−3 1 2
───●──────●──────○──────→
Step 3: Test sign in each interval
x < −3: (−)(−) / (−) = ( + ) / ( − ) = − → negative
−3 < x < 1: (+)(−) / (−) = ( − ) / ( − ) = + → positive ✓
1 < x < 2: (+)(+) / (−) = ( + ) / ( − ) = − → negative
x > 2: (+)(+) / (+) = ( + ) / ( + ) = + → positive ✓
Step 4: The inequality is > 0 (strict), so select positive
intervals. Exclude boundary points where the expression
is zero (−3 and 1) and where it is undefined (2).
Solution: x ∈ (−3, 1) ∪ (2, ∞)
9. Common Mistakes to Avoid
| # | Mistake | Correction | ||||
|---|---|---|---|---|---|---|
| 1 | Writing αβ = −c/a instead of c/a | Product of roots is αβ = c/a. Only the sum has the negative sign: α + β = −b/a. | ||||
| 2 | Forgetting to flip the inequality sign when multiplying by a negative | If you multiply or divide both sides by a negative number, always reverse the inequality sign. | ||||
| 3 | Including denominator zeros in the solution of rational inequalities | Zeros of the denominator make the expression undefined — they can never be in the solution set, even with ≤ or ≥. | ||||
| 4 | Writing \ | x\ | < a as x < ±a | \ | x\ | < a gives a compound inequality −a < x < a, not x < ±a. |
| 5 | Writing log(a + b) = log a + log b | There is no formula for log(a + b). Only log(ab) = log a + log b and log(a/b) = log a − log b. | ||||
| 6 | Treating \ | x − 3\ | < 5 as x < 8 only | The correct expansion is −5 < x − 3 < 5, giving −2 < x < 8. Both bounds are needed. | ||
| 7 | Solving x² > 9 as x > 3 only | x² > 9 means x < −3 or x > 3. Do not forget the negative interval. | ||||
| 8 | Applying the Remainder Theorem with wrong substitution | For division by (x − k), the remainder is P(k), not P(−k). For (x + k), the remainder is P(−k). | ||||
| 9 | Confusing equal roots with no real roots | Δ = 0 means one repeated real root (not no roots). Δ < 0 means no real roots (complex roots). | ||||
| 10 | Dropping absolute value without checking sign conditions | When solving \ | f(x)\ | = g(x), always check that g(x) ≥ 0 before splitting into cases. |
10. Exam Checklist
Before entering the exam hall, make sure you can do the following without hesitation:
- [ ] State the quadratic formula and compute roots given a, b, c.
- [ ] Find the discriminant and determine the nature of roots (real distinct / equal / complex).
- [ ] State Vieta's formulas: α + β = −b/a and αβ = c/a, and use them to form equations.
- [ ] Complete the square to solve any quadratic equation step by step.
- [ ] Solve quadratic inequalities by finding roots, testing intervals, and writing the answer in interval notation.
- [ ] Solve modulus equations \|ax + b\| = c and inequalities \|ax + b\| < c, \|ax + b\| > c.
- [ ] Apply all four logarithm properties (product, quotient, power, change of base) fluently.
- [ ] Evaluate logarithmic expressions like logₐ(aⁿ), logₐ(1), logₐ(a) from memory.
- [ ] Use the Remainder Theorem to find remainders without long division.
- [ ] Use the Factor Theorem to check if (x − k) is a factor and find unknown constants.
- [ ] Solve rational inequalities by finding zeros of numerator and denominator, testing intervals, and excluding denominator zeros.
- [ ] Write solution sets correctly in interval notation, using brackets [ ] for included endpoints and parentheses ( ) for excluded endpoints.
- [ ] Avoid the top 10 common mistakes listed above — especially sign errors with Vieta's formulas and forgetting negative intervals in inequalities.
Frequently Asked Questions
log(mn) = log m + log n, log(m/n) = log m − log n, and log(mⁿ) = n log m are the most used.