CBSE Class 10 Mathematics Polynomials Important Questions Set 01

Answer: The degree of the zero polynomial is not defined.

Teacher's Note:
1. The zero polynomial has no non-zero term, so no highest power exists.
2. A non-zero constant polynomial has degree 0.

Answer: Sum = −b/a = \( -\dfrac{7}{2} \). Product = c/a = −2.

Teacher's Note:
1. Use sum = −b/a and product = c/a.
2. Read a, b, c carefully with their signs.

Answer: Sum of zeroes = 8k/4 = 2k. Since the zeroes are α and −α, the sum is 0, so 2k = 0 and k = 0.

Teacher's Note:
1. Zeroes that are negatives of each other add up to zero.
2. Set −b/a equal to zero.

Answer: p(x) = x² − (sum)x + product = x² − 2√3 x + 2.

Teacher's Note:
1. Use x² − (α + β)x + αβ.
2. Any non-zero multiple of it also works.

Answer: Sum = 1 and product = −12, so the polynomial is x² − x − 12.

Teacher's Note:
1. Compute the sum and product of the given zeroes.
2. Substitute them in x² − (sum)x + product.

Answer: 2(4) + 2k − 10 = 0, so 2k = 2 and k = 1.

Teacher's Note:
1. A zero makes the polynomial equal to zero.
2. Substitute x = 2 and solve for k.

Answer: α + β = 4 and αβ = 1, so α + β − αβ = 4 − 1 = 3.

Teacher's Note:
1. Read the sum and product directly from the coefficients.
2. Then substitute them in the expression.

Answer: The parabola cuts the x-axis at two distinct points, so it has two zeroes.

Teacher's Note:
1. Zeroes of a polynomial are the x-coordinates where its graph meets the x-axis.
2. A parabola with its vertex below the axis and opening upwards must cross it twice.

Answer: It has one zero (two equal zeroes) and the discriminant b² − 4ac is zero.

Teacher's Note:
1. Touching the axis means the two zeroes coincide.
2. Equal zeroes correspond to a zero discriminant.

Answer: (c) \( \dfrac{x^2}{2}-\dfrac{x}{2}-6 \). Sum = 1 and product = −12, and this polynomial is the multiple (1/2)(x² − x − 12).

Teacher's Note:
1. Any constant multiple of x² − x − 12 has the same zeroes.
2. Check each option's sum and product ratios.

Answer: (b) \( -\dfrac{2}{3} \). Sum = −2/k and product = 3, so −2/k = 3 gives k = −2/3.

Teacher's Note:
1. Write the sum as −b/a and the product as c/a.
2. Equate them and solve for k.

Answer: 6x² − 7x − 3 = 6x² − 9x + 2x − 3 = 3x(2x − 3) + 1(2x − 3) = (2x − 3)(3x + 1).
Zeroes: \( \dfrac{3}{2} \) and \( -\dfrac{1}{3} \).
Sum = \( \dfrac{3}{2}-\dfrac{1}{3}=\dfrac{7}{6} \) = −b/a. Product = \( -\dfrac{1}{2} \) = c/a = \( -\dfrac{3}{6} \). Verified.

Teacher's Note:
1. Split the middle term using the product ac = −18.
2. Compare sum and product with −b/a and c/a.

Answer: x² + 7x + 10 = (x + 2)(x + 5), so the zeroes are −2 and −5.
Sum = −7 = −b/a and product = 10 = c/a. Verified.

Teacher's Note:
1. Look for two numbers with sum 7 and product 10.
2. These are 2 and 5.

Answer: x² − 2x − 8 = (x − 4)(x + 2), so the zeroes are 4 and −2.
Sum = 2 = −(−2)/1 and product = −8 = c/a. Verified.

Teacher's Note:
1. Find two numbers with product −8 and sum −2.
2. These are −4 and 2, giving zeroes 4 and −2.

Answer: Sum = \( \dfrac{1}{4}-1=-\dfrac{3}{4} \); product = \( -\dfrac{1}{4} \).
Polynomial: x² + (3/4)x − 1/4; multiplying by 4 gives 4x² + 3x − 1.

Teacher's Note:
1. Form x² − (sum)x + product first.
2. Clear fractions by multiplying by the common denominator.

Answer: α + β = 5/3 and αβ = −2/3.
1/α + 1/β = (α + β)/αβ = \( \dfrac{5/3}{-2/3}=-\dfrac{5}{2} \).

Teacher's Note:
1. Combine the fractions over the common denominator αβ.
2. Then use the sum and product from the coefficients.

Answer: α² + β² = (α + β)² − 2αβ = \( \dfrac{25}{9}+\dfrac{4}{3}=\dfrac{37}{9} \).

Teacher's Note:
1. Use the identity α² + β² = (α + β)² − 2αβ.
2. Do not try to find α and β separately.

Answer: α + β = −1/6 and αβ = k/6.
α² + β² = (α + β)² − 2αβ = \( \dfrac{1}{36}-\dfrac{k}{3}=\dfrac{25}{36} \), so \( -\dfrac{k}{3}=\dfrac{24}{36} \) and k = −2.

Teacher's Note:
1. Express the sum of squares using the sum and product.
2. Then solve the resulting linear equation for k.

Answer: Equal zeroes need discriminant = 0: (3k)² − 4(9)(k) = 9k² − 36k = 9k(k − 4) = 0.
So k = 0 or k = 4.

Teacher's Note:
1. Equal zeroes mean b² − 4ac = 0.
2. Factorise the resulting expression in k.

Answer: α + β = k − 6 and αβ = 2k + 1.
k − 6 = 2k + 1, so k = −7.

Teacher's Note:
1. Read the sum and product directly from the polynomial.
2. Equate them as the question states.

Answer: Product of zeroes = k/1 = k. For reciprocals the product is 1, so k = 1.

Teacher's Note:
1. Reciprocal pairs multiply to 1.
2. The product of zeroes is c/a.

Answer: α + β = 6, so β = 6 − α. Then 3α + 2(6 − α) = 20 gives α = 8 and β = −2.
a = αβ = −16.

Teacher's Note:
1. Use the sum of zeroes to eliminate one variable.
2. The constant term equals the product of the zeroes.

Answer: α + β = 1 and αβ = −30.
New sum = 4 − (α + β) = 3. New product = 4 − 2(α + β) + αβ = 4 − 2 − 30 = −28.
Polynomial: x² − 3x − 28.

Teacher's Note:
1. Find the new sum and new product using the old ones.
2. Do not solve for α and β.

Answer: α + β = 5/2 and αβ = −3.
New sum = 5/2 − 3 = −1/2; new product = (5/2)(−3) = −15/2.
Polynomial: x² + x/2 − 15/2, or 2x² + x − 15.

Teacher's Note:
1. Treat α + β and αβ as the two new zeroes.
2. Multiply through by 2 to clear fractions.

Answer: Let the zeroes be α and 7α. Sum: 8α = 8/3, so α = 1/3 and the zeroes are 1/3 and 7/3.
Product: 7α² = 7/9 = (2k + 1)/3, so 2k + 1 = 7/3 and k = 2/3.

Teacher's Note:
1. Use the sum to find the smaller zero first.
2. Then match the product with c/a.

Answer: Let the zeroes be α and 2α. Sum: 3α = 3/2, so α = 1/2 and the zeroes are 1/2 and 1.
Product: 2α² = 1/2 = p/2, so p = 1.

Teacher's Note:
1. Take the zeroes as α and 2α.
2. Use sum first, then product.

Answer: Let the zeroes be α and α + 1. Sum = 2α + 1 = 7, so α = 3 and the zeroes are 3 and 4.
q = product = 12.

Teacher's Note:
1. The sum of zeroes is −p.
2. Use the difference condition to write both zeroes in terms of α.