Class 10 Maths QP Worksheet Chapter 2 Polynomials | Detailed Solutions & Marking Scheme

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VEDANT WORKSHEET SERIES 

CLASS X | MATHEMATICS (041)

CHAPTER 2: POLYNOMIALS

Time Allowed: 1 Hour  |  Maximum Marks: 30

SECTION A: OBJECTIVE TYPE QUESTIONS

(1 × 7 = 7 Marks)

Multiple Choice Questions

  1. If one zero of the quadratic polynomial p(x) = x² + 3x + k is 2, then the value of k is:
    (a) 10 (b) −10 (c) 5 (d) −5

  2. The number of polynomials having zeros −2 and 5 is:
    (a) 1 (b) 2 (c) 3 (d) More than 3

  3. If α and β are the zeros of the polynomial
    f(x) = x² − p(x + 1) − c, then (α + 1)(β + 1) is equal to:
    (a) c − 1 (b) 1 − c (c) c (d) 1 + c

  4. The zeros of the quadratic polynomial x² + 99x + 127 are:
    (a) Both positive
    (b) Both negative
    (c) One positive and one negative
    (d) Both equal

  5. If the sum of the zeros of the polynomial
    f(x) = 2x³ − 3k x² + 4x − 5 is 6, then the value of k is:
    (a) 2 (b) 4 (c) −2 (d) −4

Assertion–Reasoning Questions

Directions: Choose the correct option.

(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true but R is not the correct explanation of A
(c) A is true but R is false
(d) A is false but R is true

  1. Assertion (A): The graph of y = x² − 4x + 4 touches the x-axis at exactly one point.

    Reason (R): A quadratic polynomial has two equal zeros if its discriminant D = b² − 4ac is zero.

  2. Assertion (A): If 2 − √3 is a zero of a quadratic polynomial, then the other zero is 2 + √3.

    Reason (R): Irrational zeros of a polynomial with rational coefficients always occur in conjugate pairs.

SECTION B: VERY SHORT ANSWER QUESTIONS

(2 × 4 = 8 Marks)

  1. Find a quadratic polynomial whose sum and product of zeros are √2 and 1/3 respectively.

  2. If α and β are the zeros of the polynomial p(x) = 6x² + x − 2, find the value of:

α/β + β/α

  1. If the zeros of the polynomial x² + px + q are double in value to the zeros of 2x² − 5x − 3, find the values of p and q.

  2. Find the zeros of the quadratic polynomial:

4√3 x² + 5x − 2√3

SECTION C: SHORT ANSWER QUESTIONS

(3 × 2 = 6 Marks)

  1. If α and β are the zeros of the quadratic polynomial

f(x) = 3x² − 4x + 1

find a quadratic polynomial whose zeros are:

(α² / β) and (β² / α)

  1. Verify that 3, −1, −1/3 are the zeros of the cubic polynomial

p(x) = 3x³ − 5x² − 11x − 3

and verify the relationship between the zeros and the coefficients.

SECTION D: LONG ANSWER QUESTION

(5 × 1 = 5 Marks)

  1. If α and β are the zeros of the quadratic polynomial

x² − x − 2

find a polynomial whose zeros are:

(2α + 1) and (2β + 1)

Also find the value of:

α⁴ + β⁴

SECTION E: CASE STUDY BASED QUESTION

(1 + 1 + 2 = 4 Marks)

15. The Parabolic Path

The movement of a roller coaster track can be modeled by the polynomial:

p(x) = x² − (k − 6)x + (2k + 1)

(i) If the track touches the x-axis at only one point, what is the condition for the discriminant? (1 Mark)

(ii) If one zero of the polynomial is the negative of the other, find the value of k. (1 Mark)

(iii) If α and β are the zeros of the given polynomial, find the value of k such that:

α² + β² = 10 (2 Marks)


ANSWER KEY WITH MARKING SCHEME

CHAPTER 2: POLYNOMIALS | CLASS X

SECTION A: OBJECTIVE TYPE QUESTIONS (1 × 7 = 7 Marks)

Given p(x) = x² + 3x + k and one zero = 2

p(2) = 0
⇒ 4 + 6 + k = 0
⇒ k = −10

Answer: (b) −10
Marks: 1

Infinitely many polynomials can be formed with same zeros (multiplying by constant).

Answer: (d) More than 3
Marks: 1

f(x) = x² − p(x + 1) − c
= x² − px − p − c

Sum of zeros = p
Product of zeros = −(p + c)

(α + 1)(β + 1) = αβ + α + β + 1
= −(p + c) + p + 1
= 1 − c

Answer: (b) 1 − c
Marks: 1

For x² + 99x + 127:
Sum = −99 (negative)
Product = 127 (positive)

⇒ Both zeros are negative

Answer: (b) Both negative
Marks: 1

Sum of zeros = −(coefficient of x²)/(coefficient of x³)
= −(−3k)/2 = 3k/2

Given sum = 6

3k/2 = 6
⇒ k = 4

Answer: (b) 4
Marks: 1

y = x² − 4x + 4

D = (−4)² − 4(1)(4) = 16 − 16 = 0

⇒ Equal roots → touches x-axis

Both A and R true, R explains A

Answer: (a)
Marks: 1

Given zero: 2 − √3

Conjugate root: 2 + √3

Both A and R true, and R explains A

Answer: (a)
Marks: 1

SECTION B: VERY SHORT ANSWER (2 × 4 = 8 Marks)

Sum = √2, Product = 1/3

Polynomial: x² − (sum)x + product

= x² − √2 x + 1/3

Answer: x² − √2 x + 1/3

Marking Scheme:
Formula (1) + substitution (1)

For 6x² + x − 2:

α + β = −1/6
αβ = −1/3

α/β + β/α = (α² + β²)/αβ

α² + β² = (α + β)² − 2αβ
= (−1/6)² − 2(−1/3)
= 1/36 + 2/3
= 25/36

Now:
= (25/36) ÷ (−1/3)
= −25/12

Answer: −25/12

Marking Scheme:
Formula (1) + calculation (1)

Given polynomial: 2x² − 5x − 3

Roots:
x = [5 ± √(25 + 24)] / 4
= [5 ± 7]/4

⇒ Roots: 3 and −1/2

New roots: 6 and −1

Sum = 5
Product = −6

Polynomial: x² − 5x − 6

⇒ p = −5, q = −6

Answer: p = −5, q = −6

Marking Scheme:
Find roots (1) + new polynomial (1)

Given: 4√3 x² + 5x − 2√3

Using formula:

x = [−5 ± √(25 + 96)] / (8√3)
= [−5 ± 11] / (8√3)

Roots:
6/(8√3) = √3/4
−16/(8√3) = −2/√3

Answer: √3/4 and −2/√3

Marking Scheme:
Formula (1) + simplification (1)

SECTION C: SHORT ANSWER (3 × 2 = 6 Marks)

For 3x² − 4x + 1:

α + β = 4/3
αβ = 1/3

New roots:
α²/β and β²/α

Sum = (α³ + β³)/(αβ)

α³ + β³ = (α + β)³ − 3αβ(α + β)
= (4/3)³ − 3(1/3)(4/3)
= 64/27 − 4/3
= 28/27

Sum = (28/27)/(1/3) = 28/9

Product = αβ = 1/3

Polynomial:
x² − (28/9)x + 1/3

Multiply by 9:
9x² − 28x + 3

Answer: 9x² − 28x + 3

Marking Scheme:
Find α+β, αβ (1) + new polynomial (2)

p(x) = 3x³ − 5x² − 11x − 3

Check:
p(3) = 0
p(−1) = 0
p(−1/3) = 0

Sum = 3 − 1 − 1/3 = 5/3
= −(−5)/3

Product = 1
= −(−3)/3

Verified

Marking Scheme:
Verification (2) + relation (1)

SECTION D: LONG ANSWER (5 Marks)

Given: x² − x − 2

α + β = 1
αβ = −2

New roots:
2α + 1, 2β + 1

Sum = 2(α + β) + 2 = 4
Product = 4αβ + 2(α + β) + 1
= −8 + 2 + 1 = −5

Polynomial:
x² − 4x − 5

Now:

α² + β² = 1² − 2(−2) = 5

α⁴ + β⁴ = (α² + β²)² − 2(αβ)²
= 25 − 8 = 17

Answer:
Polynomial: x² − 4x − 5
α⁴ + β⁴ = 17

Marking Scheme:
New polynomial (3) + α⁴ + β⁴ (2)

SECTION E: CASE STUDY (4 Marks)

(i) For touching x-axis:
Discriminant D = 0

Answer: D = 0
Marks: 1

(ii) Opposite roots ⇒ sum = 0

k − 6 = 0
⇒ k = 6

Marks: 1

(iii)
α + β = k − 6
αβ = 2k + 1

α² + β² = (α + β)² − 2αβ

= (k − 6)² − 2(2k + 1)

Given = 10

k² − 12k + 36 − 4k − 2 = 10

k² − 16k + 24 = 0

k = 12 or 2

Answer: k = 12 or 2

Marking Scheme:
Formula (1) + equation (1) + solution (1)

✅ TOTAL = 30 MARKS

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