CBSE Class 10 Mathematics Algebra Polynomials, Equations and Arithmetic Progressions Worksheet Set 01

Answer: Equation: \( x^2 - (\text{sum})x + \text{product} = 0 \), so x² − 5x + 3 = 0.

Teacher's Note:
1. Use x² − (sum of roots)x + (product of roots) = 0.
2. Substitute the given values directly.

Answer: (b) an odd number. For odd n = 2m + 1, n² − 1 = 4m(m + 1), and m(m + 1) is even, so 8 divides it.

Teacher's Note:
1. Test with n = 3 (8) and n = 2 (3) to see the pattern.
2. Write an odd number as 2m + 1 for a general proof.

Answer: (a) 3/2. For a cubic, product of zeroes = −d/a = 6/a. Setting 6/a = 4 gives a = 3/2.

Teacher's Note:
1. Use product of zeroes = −(constant term)/(coefficient of x³).
2. The constant term here is −6.

Answer: (d) an odd integer. Equal zeroes need k² − 4k = 0, so k = 0 or 4. Both are even integers and 4 is positive, but k cannot be an odd integer.

Teacher's Note:
1. Set the discriminant b² − 4ac equal to zero.
2. List all possible values of k, then eliminate the options.

Answer: No solution. The two lines are parallel to the x-axis and never meet.

Teacher's Note:
1. Each equation is a horizontal line.
2. Parallel lines have no common point.

Answer: a = 21, d = 21. \( a_n = 21 + (n-1)21 = 21n \). Setting 21n = 210 gives n = 10, so it is the 10th term.

Teacher's Note:
1. Use a_n = a + (n − 1)d.
2. Here the AP is simply the multiples of 21.

Answer: a = 2, d = 4, so a_n = 2 + (n − 1)4 = 4n − 2.

Teacher's Note:
1. Find the first term and the common difference.
2. Substitute them in the nth term formula.

Answer: \( \dfrac{(x-2)-x}{x(x-2)} = 3 \Rightarrow -2 = 3x^2 - 6x \Rightarrow 3x^2 - 6x + 2 = 0 \).
\( x = \dfrac{6 \pm \sqrt{36-24}}{6} = \dfrac{6 \pm 2\sqrt{3}}{6} = \dfrac{3 \pm \sqrt{3}}{3} \).
Roots: x = 1 + √3/3 and x = 1 − √3/3.

Teacher's Note:
1. Take the LCM and reduce the equation to ax² + bx + c = 0.
2. Then apply x = (−b ± √(b² − 4ac))/2a.

Answer: Split the middle term: \( x^2 + 3\sqrt{2}x - \sqrt{2}x - 6 = 0 \) (since 3√2 × (−√2) = −6).
\( x(x + 3\sqrt{2}) - \sqrt{2}(x + 3\sqrt{2}) = 0 \Rightarrow (x - \sqrt{2})(x + 3\sqrt{2}) = 0 \).
So x = √2 or x = −3√2.

Teacher's Note:
1. Find two numbers whose sum is 2√2 and product is −6.
2. Those numbers are 3√2 and −√2.

Answer: Let d be the common difference. Then 25 = 9 + 2d, so d = 8. Hence b = 9 + 8 = 17 and a = 9 − 8 = 1.
a = 1, b = 17.

Teacher's Note:
1. Use the terms 9 and 25, which are two steps apart.
2. Then work backwards and forwards with d.

Answer: The multiples are 3, 6, ..., 24: a = 3, d = 3, n = 8. \( S_8 = \dfrac{8}{2}[2(3) + 7(3)] = 4 \times 27 = \) 108.

Teacher's Note:
1. The multiples of 3 form an AP.
2. Use S_n = (n/2)[2a + (n − 1)d].

Answer: The numbers are 12, 15, ..., 99. a = 12, d = 3, a_n = 99, so 12 + (n − 1)3 = 99, which gives n = 30. 30 numbers.

Teacher's Note:
1. The smallest two-digit multiple of 3 is 12 and the largest is 99.
2. Use the nth term formula to count the terms.

Answer: 12ⁿ = (2² × 3)ⁿ = 2²ⁿ × 3ⁿ. Its prime factorisation has no factor 5. A number ending in 0 or 5 is divisible by 5, so 12ⁿ cannot end in 0 or 5.

Teacher's Note:
1. A number ending in 0 or 5 must have 5 as a prime factor.
2. Check the prime factors of 12ⁿ.

Answer: α + β = −b/a and αβ = c/a.
\( \dfrac{1}{\alpha^2}+\dfrac{1}{\beta^2} = \dfrac{\alpha^2+\beta^2}{\alpha^2\beta^2} = \dfrac{(\alpha+\beta)^2 - 2\alpha\beta}{(\alpha\beta)^2} = \dfrac{b^2/a^2 - 2c/a}{c^2/a^2} = \dfrac{b^2 - 2ac}{c^2} \).

Teacher's Note:
1. Combine the fractions and use the identity α² + β² = (α + β)² − 2αβ.
2. Then substitute the sum and the product of zeroes.

Answer: Compare ratios: \( \dfrac{a_1}{a_2} = \dfrac{2}{3}, \dfrac{b_1}{b_2} = \dfrac{4}{6} = \dfrac{2}{3}, \dfrac{c_1}{c_2} = \dfrac{10}{12} = \dfrac{5}{6} \). Since a₁/a₂ = b₁/b₂ ≠ c₁/c₂, the lines are parallel (x + 2y = 5 and x + 2y = 4), so there is no solution.

Teacher's Note:
1. Parallel lines never intersect on a graph.
2. Compare the three ratios to decide the nature of the solution.

Answer: Discriminant D = k² − 12.
(a) Two distinct real roots: D > 0, so k² > 12, i.e. k < −2√3 or k > 2√3.
(b) Equal roots: D = 0, so k = ±2√3.

Teacher's Note:
1. The nature of roots depends on the sign of b² − 4ac.
2. Here a = 1, b = −k and c = 3.

Answer: a = 6, d = 7, last term 216. 6 + (n − 1)7 = 216 gives n = 31. The middle term is the 16th term: a₁₆ = 6 + 15 × 7 = 111.

Teacher's Note:
1. Find the number of terms first.
2. With 31 terms the middle one is the 16th.

Answer: a + 18d = 3(a + 5d) gives 3d = 2a. Also a + 8d = 19. Substituting a = 3d/2: 9.5d = 19, so d = 2 and a = 3.
The AP is 3, 5, 7, 9, ...

Teacher's Note:
1. Form two equations in a and d.
2. Solve them by substitution.

Answer: The numbers are 10, 13, 16, ..., 97. a = 10, d = 3, l = 97, so 10 + (n − 1)3 = 97 gives n = 30. S = (30/2)(10 + 97) = 15 × 107 = 1605.

Teacher's Note:
1. The numbers form an AP with common difference 3.
2. Use S = (n/2)(a + l).

Answer: For the two APs, \( \dfrac{S_n}{S_n'} = \dfrac{\frac{n}{2}[2a+(n-1)d]}{\frac{n}{2}[2a'+(n-1)d']} = \dfrac{7n+1}{4n+27} \).
The mth term ratio is \( \dfrac{a+(m-1)d}{a'+(m-1)d'} \), which equals the sum ratio with \( \dfrac{n-1}{2} = m-1 \), i.e. n = 2m − 1.
Ratio = \( \dfrac{7(2m-1)+1}{4(2m-1)+27} = \) (14m − 6) : (8m + 23).

Teacher's Note:
1. Divide the numerator and denominator of the sum ratio by n/2.
2. Replace (n − 1)/2 by (m − 1) to get the mth term.

Answer: For infinitely many solutions, \( \dfrac{2}{2\alpha} = \dfrac{3}{\alpha+\beta} = \dfrac{7}{28} = \dfrac{1}{4} \).
From 1/α = 1/4, α = 4. From 3/(α + β) = 1/4, α + β = 12, so β = 8.
α = 4, β = 8.

Teacher's Note:
1. The three ratios must be equal.
2. Use 7/28 = 1/4 to fix the common ratio.

Answer: Product of zeroes = \( \dfrac{6a}{a^2+9} \). For reciprocal zeroes the product is 1: 6a = a² + 9, so a² − 6a + 9 = 0, i.e. (a − 3)² = 0. Hence a = 3.

Teacher's Note:
1. Zeroes that are reciprocals multiply to 1.
2. Equate the product to 1 and solve for a.

Answer: \( \dfrac{7}{5} \ne \dfrac{-4}{-6} = \dfrac{2}{3} \), so a unique solution exists.
Multiply the first by 6 and the second by 4: 42x − 24y = 294 and 20x − 24y = 228. Subtracting: 22x = 66, x = 3. Then 21 − 4y = 49 gives y = −7.
x = 3, y = −7.

Teacher's Note:
1. Check a₁/a₂ ≠ b₁/b₂ for a unique solution.
2. Eliminate y by making its coefficients equal.

Answer: \( \dfrac{1}{a+b+x} - \dfrac{1}{x} = \dfrac{1}{a}+\dfrac{1}{b} \Rightarrow \dfrac{-(a+b)}{x(a+b+x)} = \dfrac{a+b}{ab} \).
Since a + b ≠ 0: −ab = x(a + b + x), so x² + (a + b)x + ab = 0, i.e. (x + a)(x + b) = 0.
x = −a or x = −b.

Teacher's Note:
1. Move 1/x to the left side before combining the fractions.
2. Cancel the non-zero factor (a + b).

Answer: Let the first term be A and the common difference D. Then a = A + (p − 1)D, b = A + (q − 1)D, c = A + (r − 1)D.
a − b = (p − q)D, b − c = (q − r)D, c − a = (r − p)D.
LHS = D[(p − q)r + (q − r)p + (r − p)q] = D[pr − qr + pq − pr + qr − pq] = 0. Hence proved.

Teacher's Note:
1. Express a, b and c using the first term and common difference.
2. The differences remove A and leave only D multiples.

Answer: Take the terms a − 3d, a − d, a + d, a + 3d. Sum = 4a = 32, so a = 8.
\( \dfrac{(8-3d)(8+3d)}{(8-d)(8+d)} = \dfrac{7}{15} \Rightarrow 15(64 - 9d^2) = 7(64 - d^2) \Rightarrow 512 = 128d^2 \), so d = ±2.
The numbers are 2, 6, 10, 14 (or 14, 10, 6, 2).

Teacher's Note:
1. Using a ± d and a ± 3d makes the sum simple.
2. The product of equidistant terms is a difference of squares.

Answer: a + 2d = 15 and a + 5d = (a + 3d)/2, which gives a = −7d. So −5d = 15, d = −3 and a = 21.
S_n = (n/2)[42 − 3(n − 1)] = 66 gives n² − 15n + 44 = 0, so n = 4 or n = 11.
Both are valid: 4 terms (21, 18, 15, 12) or 11 terms (the later terms are negative and cancel).

Teacher's Note:
1. Form two equations from the given conditions to find a and d.
2. Both roots of the quadratic are acceptable here because the terms become negative.

Answer: 2160 = 847 × 2 + 466
847 = 466 × 1 + 381
466 = 381 × 1 + 85
381 = 85 × 4 + 41
85 = 41 × 2 + 3
41 = 3 × 13 + 2
3 = 2 × 1 + 1
2 = 1 × 2 + 0
HCF = 1, so the numbers are coprime.

Teacher's Note:
1. Repeat the division until the remainder is zero.
2. The last non-zero remainder is the HCF.

Answer: α + β = 5/2 and αβ = 7/2.
Sum of new zeroes = 5(α + β) = 25/2.
Product = 6α² + 13αβ + 6β² = 6(α + β)² + αβ = 6 × 25/4 + 7/2 = 41.
Polynomial: x² − (25/2)x + 41, or 2x² − 25x + 82.

Teacher's Note:
1. Write the new sum and product in terms of α + β and αβ.
2. Multiply by 2 to remove the fraction.

Answer: Let the number be 10a + b. Then 10a + b = 4(a + b), so 6a = 3b and b = 2a. Also 10a + b = 2ab gives 12a = 4a², so a = 3 and b = 6.
The number is 36. Check: 4 × 9 = 36 and 2 × 18 = 36.

Teacher's Note:
1. Form two equations from the two conditions.
2. Substitute b = 2a in the second condition.