Class 10 Maths Chapter 2 Polynomials Practice Paper 1

Q1. 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

Answer: (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

Q2. The zeroes of the polynomial p(x) = 3x² - 2 are:
(a) ±√(2/3)
(b) ±√(3/2)
(c) ±√6
(d) ±2/3

Answer: (a) ±√(2/3)
Solution:
p(x) = 3x² - 2 = 0
3x² = 2
x² = 2/3
x = ±√(2/3)

Q3. 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

Answer: (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

Q4. 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

Answer: (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

Q5. 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

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

Q6. 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

Answer: (c) 2 Solution: The parabola intersects the x-axis at two points, so there are 2 zeroes.

Q7. 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

Answer: (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

Q8. The zeroes of the polynomial x² - 9 are:
(a) 3, 3
(b) -3, -3
(c) 3, -3
(d) No real zeroes

Answer: (c) 3, -3 Solution:
x² - 9 = 0
(x + 3)(x - 3) = 0
x = -3 or x = 3

Q9. 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

Answer: (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.

Q10. 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

Answer: (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.

Q11. Find the zeroes of the polynomial x² + 7x + 12 and verify the relationship between zeroes and coefficients.

Answer: 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 ✓

Q12. If α and β are the zeroes of the polynomial t² - 4t + 3, find the value of α² + β².

Answer: 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

Q13. Find a quadratic polynomial whose zeroes are 2 + √3 and 2 - √3.

Answer: 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

Q14. If one zero of the polynomial (a² + 9)x² + 13x + 6a is reciprocal of the other, find the value of a.

Answer: 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

Q15. If α and β are the zeroes of the polynomial 2x² + 7x + 5, find the value of α² + β² + αβ.

Answer: 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

Q16. 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.

Answer: 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

Q17. If α and β are the zeroes of the polynomial x² - 6x + k such that α² + β² = 28, find the value of k.

Answer: 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

Q18. 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

Answer: 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

Q19. 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)

Answer: 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

Q20. 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) Made with ❤ by Sumeet Sahu Website: uniquestudyonline.com

Answer: 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. Made with ❤ by Sumeet Sahu Website: uniquestudyonline.com