Class 10 Maths Chapter 2 Polynomials Practice Paper 1

Class: X Subject: Mathematics Session: 2025-26 Chapter: 02 - Polynomials Time: 1½ Hours Max. Marks: 40

General Instructions:

1. All questions are compulsory.

2. This question paper contains 20 questions divided into five sections A, B, C, D and E.

3. Section A contains 10 MCQs of 1 mark each.

4. Section B contains 4 questions of 2 marks each.

5. Section C contains 3 questions of 3 marks each.

6. Section D contains 1 question of 5 marks.

7. Section E contains 2 Case Study Based questions of 4 marks each.

SECTION A - Multiple Choice Questions (1 mark each)

1. If α and β are the zeroes of the polynomial f(x) = x² + x - 2, then the value of (1/α + 1/β) is:
(a) -1/2
(b) 1/2
(c) -2
(d) 2

2. The zeroes of the polynomial p(x) = 3x² - 2 are:
(a) ±√(2/3)
(b) ±√(3/2)
(c) ±√6
(d) ±2/3

3. If one zero of the polynomial 2x² + 3x + k is -1, then the value of k is:
(a) 1
(b) -1
(c) 2
(d) -2

4. A quadratic polynomial whose zeroes are 3 and -4 is:
(a) x² - x - 12
(b) x² + x - 12
(c) x² + x + 12
(d) x² - x + 12

5. If the sum of zeroes of the polynomial 3x² - kx + 6 is 3, then the value of k is:
(a) 3
(b) 6
(c) 9
(d) 12

6. The graph of y = p(x) is given. The number of zeroes of p(x) from the graph is: [A parabola opening upward with vertex below the x-axis, intersecting the x-axis at two points]
(a) 0
(b) 1
(c) 2
(d) 3

7. If α, β are the zeroes of polynomial f(x) = x² - 5x + 6, then the value of α² + β² is:
(a) 13
(b) 25
(c) 5
(d) 12

8. The zeroes of the polynomial x² - 9 are:
(a) 3, 3
(b) -3, -3
(c) 3, -3
(d) No real zeroes

9. Assertion
(a) : If the sum of zeroes of the quadratic polynomial x² - 2x + k is equal to half their product, then k = 4. Reason (R): For quadratic polynomial ax² + bx + c, sum of zeroes = -b/a and product of zeroes = c/a.
(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

10. Assertion
(a) : The polynomial p(x) = 4x² - 1 has two distinct real zeroes. Reason (R): A quadratic polynomial can have at most two zeroes.
(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

SECTION B - Short Answer Questions (2 marks each)

11. Find the zeroes of the polynomial x² + 7x + 12 and verify the relationship between zeroes and coefficients.

12. If α and β are the zeroes of the polynomial t² - 4t + 3, find the value of α² + β².

13. Find a quadratic polynomial whose zeroes are 2 + √3 and 2 - √3.

14. If one zero of the polynomial (a² + 9)x² + 13x + 6a is reciprocal of the other, find the value of a.

SECTION C - Short Answer Questions (3 marks each)

15. If α and β are the zeroes of the polynomial 2x² + 7x + 5, find the value of α² + β² + αβ.

16. Find a quadratic polynomial whose zeroes are 1/(2α) and 1/(2β), where α and β are the zeroes of the polynomial 4x² - 5x + 1.

17. If α and β are the zeroes of the polynomial x² - 6x + k such that α² + β² = 28, find the value of k.

SECTION D - Long Answer Question (5 marks)

18. If α and β are the zeroes of the polynomial 6x² + x - 2, find: (i) α + β (ii) αβ (iii) α² + β² (iv) 1/α + 1/β (v) α²β + αβ²

SECTION E - Case Study Based Questions (4 marks each)

19. Case Study-1: Garden Design A landscape architect is designing a rectangular garden. The length of the garden (in meters) is represented by one zero of a quadratic polynomial, and the width is represented by the other zero. The polynomial describing the dimensions is p(x) = x² - 12x + 35. (i) Find the zeroes of the polynomial p(x) = x² - 12x + 35. (1 mark) (ii) What are the dimensions of the garden? (1 mark) (iii) Find the perimeter of the garden. (2 marks) OR (iii) If the cost of fencing is ₹150 per meter, find the total cost of fencing the garden. (2 marks)

20. Case Study-2: Projectile Motion The height h (in meters) of a ball thrown upward is given by the polynomial h(t) = -5t² + 20t + 1, where t is the time in seconds. (i) Is this polynomial quadratic? Give reason. (1 mark) (ii) What is the value of the discriminant of this polynomial? (1 mark) (iii) Find the sum and product of the zeroes of the polynomial. (2 marks) OR (iii) At what time(s) will the ball be at a height of 16 meters? (2 marks) DETAILED ANSWER KEY - PAPER 01

SECTION A - Answers to MCQs

1.
(b) 1/2

Solution:

Given: f(x) = x² + x - 2 Sum of zeroes (α + β) = -b/a = -1/1 = -1 Product of zeroes (αβ) = c/a = -2/1 = -2 1/α + 1/β = (α + β)/(αβ) = (-1)/(-2) = 1/2

2.
(a) ±√(2/3)

Solution:

p(x) = 3x² - 2 = 0 3x² = 2 x² = 2/3 x = ±√(2/3)

3.
(a) 1

Solution:

If x = -1 is a zero, then p(-1) = 0 2(-1)² + 3(-1) + k = 0 2 - 3 + k = 0 k = 1

4.
(b) x² + x - 12

Solution:

If zeroes are α = 3 and β = -4 Sum (α + β) = 3 + (-4) = -1 Product (αβ) = 3 × (-4) = -12 Polynomial = x² - (sum)x + product = x² - (-1)x + (-12) = x² + x - 12

5.
(c) 9

Solution:

Sum of zeroes = -b/a = -(-k)/3 = k/3 Given: k/3 = 3 k = 9

6.
(c) 2

Solution:

The parabola intersects the x-axis at two points, so there are 2 zeroes.

7.
(a) 13

Solution:

f(x) = x² - 5x + 6 α + β = 5 and αβ = 6 α² + β² = (α + β)² - 2αβ = (5)² - 2(6) = 25 - 12 = 13

8.
(c) 3, -3

Solution:

x² - 9 = 0 (x + 3)(x - 3) = 0 x = -3 or x = 3

9.
(a) Both A and R are true and R is the correct explanation of A

Solution:

For x² - 2x + k: Sum = -(-2)/1 = 2 Product = k/1 = k Given: 2 = k/2 k = 4 Both assertion and reason are correct, and R explains A.

10.
(b) Both A and R are true but R is not the correct explanation of A

Solution:

4x² - 1 = 0 gives x = ±1/2 (two distinct real zeroes) A quadratic polynomial can have at most two zeroes (true) However, R doesn't specifically explain why THIS polynomial has two distinct zeroes.

SECTION B - Answers to Short Answer Questions

11.

Solution:

x² + 7x + 12 = 0 x² + 3x + 4x + 12 = 0 x(x + 3) + 4(x + 3) = 0 (x + 3)(x + 4) = 0 Zeroes: α = -3, β = -4 Verification: Sum of zeroes = α + β = -3 + (-4) = -7 = -b/a = -7/1 ✓ Product of zeroes = αβ = (-3)(-4) = 12 = c/a = 12/1 ✓ 12.

Solution:

For polynomial t² - 4t + 3: Sum of zeroes (α + β) = -(-4)/1 = 4 Product of zeroes (αβ) = 3/1 = 3 α² + β² = (α + β)² - 2αβ = (4)² - 2(3) = 16 - 6 = 10 13.

Solution:

Given zeroes: α = 2 + √3 and β = 2 - √3 Sum = α + β = (2 + √3) + (2 - √3) = 4 Product = αβ = (2 + √3)(2 - √3) = 4 - 3 = 1 Required polynomial = x² - (sum)x + product = x² - 4x + 1 14.

Solution:

If one zero is reciprocal of other, then product of zeroes = 1 Product of zeroes = c/a = 6a/(a² + 9) 6a/(a² + 9) = 1 6a = a² + 9 a² - 6a + 9 = 0 (a - 3)² = 0 a = 3

SECTION C - Answers to Short Answer Questions

15.

Solution:

For polynomial 2x² + 7x + 5: α + β = -b/a = -7/2 αβ = c/a = 5/2 α² + β² = (α + β)² - 2αβ = (-7/2)² - 2(5/2) = 49/4 - 10/2 = 49/4 - 20/4 = 29/4 α² + β² + αβ = 29/4 + 5/2 = 29/4 + 10/4 = 39/4 16.

Solution:

For 4x² - 5x + 1: α + β = 5/4 and αβ = 1/4 New zeroes are 1/(2α) and 1/(2β) Sum of new zeroes = 1/(2α) + 1/(2β) = (α + β)/(2αβ) = (5/4)/(2 × 1/4) = (5/4)/(1/2) = 5/2 Product of new zeroes = 1/(2α) × 1/(2β) = 1/(4αβ) = 1/(4 × 1/4) = 1 Required polynomial = x² - (sum)x + product = x² - (5/2)x + 1 = 2x² - 5x + 2 17.

Solution:

For x² - 6x + k: α + β = 6 αβ = k Given: α² + β² = 28 (α + β)² - 2αβ = 28 (6)² - 2k = 28 36 - 2k = 28 2k = 8 k = 4

SECTION D - Answer to Long Answer Question

18.

Solution:

For polynomial 6x² + x - 2: (i) α + β = -b/a = -1/6 (ii) αβ = c/a = -2/6 = -1/3 (iii) α² + β² = (α + β)² - 2αβ = (-1/6)² - 2(-1/3) = 1/36 + 2/3 = 1/36 + 24/36 = 25/36 (iv) 1/α + 1/β = (α + β)/(αβ) = (-1/6)/(-1/3) = (-1/6) × (-3/1) = 1/2 (v) α²β + αβ² = αβ(α + β) = (-1/3)(-1/6) = 1/18

SECTION E - Answers to Case Study Based Questions

19.

Solution:

(i) p(x) = x² - 12x + 35 x² - 7x - 5x + 35 = 0 x(x - 7) - 5(x - 7) = 0 (x - 5)(x - 7) = 0 Zeroes: x = 5 and x = 7 (ii) Dimensions of garden: Length = 7 meters Width = 5 meters (iii) Perimeter = 2(length + width) = 2(7 + 5) = 2(12) = 24 meters OR (iii) Cost of fencing = Perimeter × Rate Perimeter = 24 meters Cost = 24 × ₹150 = ₹3,600 20.

Solution:

Given: h(t) = -5t² + 20t + 1 (i) Yes, this is a quadratic polynomial because the highest degree of the variable t is 2. (ii) Discriminant = b² - 4ac Here a = -5, b = 20, c = 1 D = (20)² - 4(-5)(1) D = 400 + 20 D = 420 (iii) Sum of zeroes = -b/a = -20/(-5) = 4 Product of zeroes = c/a = 1/(-5) = -1/5 OR (iii) When h = 16: -5t² + 20t + 1 = 16 -5t² + 20t - 15 = 0 t² - 4t + 3 = 0 (t - 1)(t - 3) = 0 t = 1 second or t = 3 seconds The ball will be at 16 meters height at 1 second and 3 seconds.