1. If one zero of the quadratic polynomial x² + 3x + k is 2, then the value of k is

(a) 10

(b) -10

(c) 5

(d) -5

Answer: b

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2. Given that two of the zeroes of the cubic poly-nomial ax^{3} + bx² + cx + d are 0, the third zero is

Answer: a

3. If one of the zeroes of the quadratic polynomial (k – 1) x² + kx + 1 is – 3, then the value of k is

Answer: a

4. A quadratic polynomial, whose zeroes are -3 and 4, is

(a) x²- x + 12

(b) x² + x + 12

(c) \(\frac{x^{2}}{2}-\frac{x}{2}-6\)

(d) 2x² + 2x – 24

Answer: c

5. If the zeroes of the quadratic polynomial x2 + (a + 1) x + b are 2 and -3, then

(a) a = -7, b = -1

(b) a = 5, b = -1

(c) a = 2, b = -6

(d) a – 0, b = -6

Answer: d

6. The number of polynomials having zeroes as -2 and 5 is

(a) 1

(b) 2

(c) 3

(d) more than 3

Answer: d

7. Given that one of the zeroes of the cubic polynomial ax^{3} + bx² + cx + d is zero, the product of the other two zeroes is

Answer: b

8. If one of the zeroes of the cubic polynomial x^{3} + ax² + bx + c is -1, then the product of the

other two zeroes is

(a) b – a + 1

(b) b – a – 1

(c) a – b + 1

(d) a – b – 1

Answer: a

9. The zeroes of the quadratic polynomial x2 + 99x + 127 are

(a) both positive

(b) both negative

(c) one positive and one negative

(d) both equal

Answer: b

10. The zeroes of the quadratic polynomial x² + kx + k, k? 0,

(a) cannot both be positive

(b) cannot both be negative

(c) are always unequal

(d) are always equal

Answer: a

11. If the zeroes of the quadratic polynomial ax² + bx + c, c # 0 are equal, then

(a) c and a have opposite signs

(b) c and b have opposite signs

(c) c and a have the same sign

(d) c and b have the same sign

Answer: c

12. If one of the zeroes of a quadratic polynomial of the form x² + ax + b is the negative of the other, then it

(a) has no linear term and the constant term is negative.

(b) has no linear term and the constant term is positive.

(c) can have a linear term but the constant term is negative.

(d) can have a linear term but the constant term is positive.

Answer: a

13. Which of the following is not the graph of quadratic polynomial?

Answer: d

14. The number of polynomials having zeroes as 4 and 7 is

(a) 2

(b) 3

(c) 4

(d) more than 4

Answer: d

15. A quadratic polynomial, whose zeores are -4 and -5, is

(a) x²-9x + 20

(b) x² + 9x + 20

(c) x²-9x- 20

(d) x² + 9x- 20

Answer: b

16. The zeroes of the quadratic polynomial x² + 1750x + 175000 are

(a) both negative

(b) one positive and one negative

(c) both positive

(d) both equal

Answer: a

17. The zeroes of the quadratic polynomial x² – 15x + 50 are

(a) both negative

(b) one positive and one negative

(c) both positive

(d) both equal

Answer: c

18. The zeroes of the quadratic polynomial 3x² – 48 are

(a) both negative

(b) one positive and one negative

(c) both positive

(d) both equal

Answer: b

19. The zeroes of the quadratic polynomial x² – 18x + 81 are

(a) both negative

(b) one positive and one negative

(c) both positive and unequal

(d) both equal and positive

Answer: d

20. The zeroes of the quadratic polynomial x² + px + p, p ≠ 0 are

(a) both equal

(b) both cannot be positive

(c) both unequal

(d) both cannot be negative

Answer: b

21. If one of the zeroes of the quadratic polynomial (p – l)x² + px + 1 is -3, then the value of p is

Answer: b

22. If the zeroes of the quadratic polynomial Ax² + Bx + C, C # 0 are equal, then

(a) A and B have the same sign

(b) A and C have the same sign

(c) B and C have the same sign

(d) A and C have opposite signs

Answer: b

23. If x^{3} + 1 is divided by x² + 5, then the possible degree of quotient is

(a) 0

(b) 1

(c) 2

(d) 3

Answer: b

24. If x^{3} + 11 is divided by x² – 3, then the possible degree of remainder is

(a) 0

(b) 1

(c) 2

(d) less than 2

Answer: d

25. If x^{4} + 3x² + 7 is divided by 3x + 5, then the possible degrees of quotient and remainder are:

(a) 3, 0

(b) 4, 1

(c) 3, 1

(d) 4, 0

Answer: a

26. If x^{5} + 2x^{4} + x + 6 is divided by g(x), and quotient is x² + 5x + 7, then the possible degree of g(x) is:

(a) 4

(b) 2

(c) 3

(d) 5

Answer: c

27. If x^{5} + 2x^{4} + x + 6 is divided by g(x) and quo-tient is x² + 5x + 7, then the possible degree of remainder is:

(a) less than 1

(b) less than 2

(c) less than 3

(d) less than 4

Answer: c

28. What is the number of zeroes that a linear poly-nomial has/have:

(a) 0

(b) 1

(c) 2

(d) 3

Answer: b

29. What is the number(s) of zeroes that a quadratic polynomial has/have:

(a) 0

(b) 1

(c) 2

(d) 3

Answer: c

30. What is the number(s) of zeores that a cubic polynomial has/have:

(a) 0

(b) 1

(c) 2

(d) 3

Answer: d

31. If one of the zeroes of the cubic polynomial x^{3} + px² + qx + r is -1, then the product of the other two zeroes is

(a) p + q + 1

(b) p-q- 1

(c) q – p + 1

(d) q – p – 1

Answer: c

32. If one zero of the quadratic polynomial x² + 3x + b is 2, then the value of b is

(a) 10

(b) -8

(c) 9

(d) -10

Answer: d

33. If 1 is one of the zeroes of the polynomial x² + x + k, then the value of k is:

(a) 2

(b) -2

(c) 4

(d) -4

Answer: b

34. If p(x) is a polynomial of at least degree one and p(k) = 0, then k is known as

(a) value of p(x)

(b) zero of p(x)

(c) constant term of p{x)

(d) none of these

Answer: b

Explaination:

Zero of p(x)

Let p(x) = ax + b

Put x = k

p(k) = ak + b = 0

∴ k is zero of p(x).

35. If one of the zeroes of the quadratic polynomial (k -1)x² + kx + 1 the value of k is [NCERT Exemplar Problems]

Answer: c

Explaination:

(k – 1)² + kx + 1

One zero is – 3, so it must satisfy the equation and make it zero.

∴ (k- 1) (- 3)² + k(-3) + 1 =0

⇒ 9k – 9 – Ik + 1 = 0

⇒ 6k – 8 = 0

⇒ k = \(\frac{8}{6}\) = \(\frac{4}{3}\)

36. If the zeroes of the quadratic polynomial x² + (a + 1) x + b are 2 and -3, then [NCERT Exemplar Problem, CBSE 2011]

(a) a = -7, b = -1

(b) a = 5, b = -1

(c) a = 2, b = -6

(d) a = 0, b = -6

Answer: d

Explaination:

x² + (a + 1)x + b

∵ x = 2 is a zero

and x = – 3 is another zero

∴ (2)² + (a + 1)² + 6 = 0

and (- 3)² + (a + 1) (- 3) + b = 0

⇒ 4 + 2cr + 2 + & = 0

and 9 – 3a – 3 + b = 0

⇒ 2a + b = – 6 …(i)

and – 3a + b = – 6 …(ii)

Solving (i) and (ii), we get 5a = 0

⇒ a = 0 and b = – 6.

37. Which of the following is not the graph of a quadratic polynomial?[NCERT Exemplar Problems]

Answer: a

Explaination:

∵ It crosses the x-axis in three points.

38. Zeroes of a polynomial can be determined graphically. No. of zeroes of a polynomial is equal to no. of points where the graph of polynomial

(a) intersects y-axis

(b) intersects x-axis

(c) intersects y-axis or intersects x-axis

(d) none of these

Answer: b

Explaination:

(b) Intersects x-axis.

39. If graph of a polynomial does not intersects the x-axis but intersects y-axis in one point, then no. of zeroes of the polynomial is equal to

(a) 0

(b) 1

(c) 0 or 1

(d) none of these

Answer: a

Explaination:

(a) Zero

40. A polynomial of degree n has

(a) only 1 zero

(b) at least n zeroes

(c) atmost n zeroes

(d) more than n zeroes

Answer: c

Explaination:

Maximum number of zeroes of a

polynomial = degree of the polynomial.

41. If p(x) = axr + bx + c, then –\(\frac{b}{a}\) is equal to

(a) 0

(b) 1

(c) product of zeroes

(d) sum of zeroes

Answer: d

Explaination:

(d) Sum of zeroes = –\(\frac{b}{a}\)

42. If p(x) = ax² + bx + c one zero is and a + b + c = 0, then one zero is

(a) \(\frac{-b}{a}\)

(b) \(\frac{c}{a}\)

(c) \(\frac{b}{c}\)

(d) none of these

Answer: b

Explaination:

p(1) = 0; a(1)² + b(1) + c = 0

⇒ a + b + c = 0

∴ one zero (α) = 1

αβ = product of zeroes = \(\frac{c}{a}\)

⇒ 1.β = \(\frac{c}{a}\)

⇒ β = \(\frac{c}{a}\)

∴ zeroes are 1 and \(\frac{c}{a}\)

43. If p{x) = ax2 + bx + one of the zeroes is c and a + c = b, then

Answer: c

Explaination:

p(-1) = 0; a(-1)² + b(-1) + c = 0

⇒ a – b + C = 0,

∴ One zero (a) = -1

αα = product of zeroes = \(\frac{c}{a}\)

⇒ (-1).β = \(\frac{c}{a}\)

⇒ β = \(\frac{-c}{a}\)

44. The number of polynomials having zeroes as -2 and 5 is [NCERT Exemplar Problems]

(a) 1

(b) 2

(c) 3

(d) more than 3

Answer: d

Explaination:

∵ x² – 3x – 10, 2x² – 6x – 20,

\(\frac{1}{2}\)x² – \(\frac{3}{2}\)x – 5, 3x² – 9x – 30 etc.,

have zeroes – 2 and 5.

45. Given that one of the zeroes of the cubic polynomial ax3 + bx2 + cx + d is zero, the product of the other two zeroes is [NCERT Exemplar Problems]

Answer: b

Explaination:

∵ αβ + βγ + γα = \(\frac{c}{a}\)

Let, α = 0

So, 0 + βγ + 0 = \(\frac{c}{a}\)

⇒ βγ= \(\frac{c}{a}\)

46. If one of the zeroes of a quadratic polynomial of the form x² + ax + b is the negative of the other, then it

(a) has no linear term and the constant term is negative.

(b) has no linear term and the constant term is positive.

(c) can have a linear term but the constant term is negative.

(d) can have a linear term but the constant term is positive.

Answer: a

Explaination:

f(x) = x² + ax + b

Given: zeroes are α and – α

Sum of zeroes = α – α = 0

∴ f(x) = x² + b, which is not linear

Product of zeroes = α(-α) = – α² = \(\frac{b}{1}\)

⇒ -α² = b

It is possible when, b < 0.

Hence, it has no linear term and the constant term is negative.

47. If 4x² – 6x – m is divisible by x – 3, the value of m is exact divisor of

(a) 9

(b) 45

(c) 20

(d) 36

Answer: a

Explaination:

Here p(3) = 0

⇒ 4(3)² – 6 × 3-m = 0

⇒ 36 – 18 – m = 0

⇒ m=18

∴ Value of m is exactly divisible by 9.

48. Which one of the following statements is correct

(a) if x^{6} + 1 is divided by x + 1, then the remainder is -2.

(b) if x^{6} + 1 is divided by x – 1, then the remainder is 2.

(c) if x^{6} + 1 is divided by x + 1, then the remainder is 1.

(d) if x^{6} + 1 is divided by x – 1, then the remainder is -1.

Answer: b

Explaination:

p(x) = x^{6} + 1

when divided by x – 1, then remainder = p(1)

∴ p(1) = 1^{6} + 1 = 2

49. Consider the following statements

(i) x – 2 is a factor of x^{3} – 3x² + 4x – 4.

(ii) x + 1 is a factor of 2x^{3} + 4x + 6.

(iii) x – 1 is a factor of x^{5} + x^{4} – x^{3} + x² -x + 1.

In these statements

(a) 1 and 2 are correct

(b) 1, 2 and 3 are correct

(c) 2 and 3 are correct

(d) 1 and 3 are correct

Answer: a

Explaination:

x – 2 is a factor of x^{3} – 3x² + 4x – 4

∵ remainder is zero Similarly x + 1 is a factor of 2x^{3} + 4x + 6

but x – 1 is not a factor of x^{5} + x^{4} – x^{3} + x² – x + 1

∵ remainder is not zero

∴ Statements 1 and 2 are correct.

50. If f(x) = 5x – 10 is divided by x – √2, then the remainder will be

(a) non zero rational number

(b) an irrational number

(c) 0

(d) \(f\left(\frac{1}{\sqrt{2}}\right)\)

Answer: b

Explaination:

Remainder = f(√2) = 5 × √2 – 10

= an irrational number

51. Zeroes of p(z) = z² – 27 are ______ and ______ .

Answer:

Explaination:

For zeroes z² – 27 = 0

⇒ z² = 27

⇒ z = ±√27

⇒ z = ± 3√3

52. Verify that x = 3 is a zero of the polynomial. p(x) = 2x^{3} – 5x² – 4x + 3

Answer:

Explaination:

Here p(x) = 2x^{3} – 5x² – 4x + 3

∴ p{3) = 2(3)^{3} – 5 x (3)² -4 x 3 + 3

= 54 – 45- 12 + 3 = 0

∵ P(3) = 0

∴ x = 3 is a zero of p(x)

53. The graph of y =f(x) is given below. How many zeroes are there of f(x)?

Answer:

Explaination:

Graph of y = f(x) intersect x-axis in one point only.

Therefore number of zeroes of f(x) is one.

54. The graph of y = f(x) is given, how many zeroes are there of f(x)?

Answer:

Explaination:

∵ Graph y =f(x) does not intersect x-axis.

∴ f(x) has no zeroes.

55. The graph of y = f(x) is given below, for some polynomial f(x). Find the number of zeroes of f(x).

Answer:

Explaination:

∵ Graph of f(x) intersects x-axis at three different points.

∴ Number of zeroes of f(x) = 3.

56. The graph of x = p(y) is given below, for some polynomial p(y). Find the number of zeroes of p(y).

Answer:

Explaination:

∵ Graph of p(y) intersects y-axis in four different points.

∴ Numbers of zeroes = 4

57.

Graph of the polynomial p(x) =px² + 4x – 4 is given as above. Find the value of p.

Answer:

Explaination:

Graph of p(x) touches the x-axis at (2, 0)

∴ x = 2 is a zero of the p(x)

⇒ p(2) = 0

⇒ p(2)² + 4 × 2 – 4 = 0

4p + 4 = 0

⇒ p = -1

58. If the product of the zeroes of x2 – 3kx + 2k1 – 1 is 7, then values of k are _____ and _____ .

Answer:

Explaination:

Product of zeroes = 7

⇒ 2k² – 1 = 7

⇒ 2k² = 8

⇒ k² = 4

⇒ k = ± 2

59. If zeroes ofp(x) = 2x² -Ix + k are reciprocal of each other, then value of k is _____ .

Answer:

Explaination:

∵ Zeroes are reciprocal of each other

∴ Product of zeroes = 1

⇒ \(\frac{k}{2}\) = 1

⇒ k = 2

60. Find the product of the zeroes of – 2x² + kx + 6.

Answer:

Explaination:

Here a = – 2, b = k, c = 6

Product of zeroes = \(\frac{c}{a}\)

i. e., α × β = \(\frac{6}{-2}\) = -3

61. Find the sum of the zeroes of the given quadratic polynomial -3x² + k.

Answer:

Explaination:

Since polynomial is -3x² + 0x + k

∴ a = -3, b = 0, c = k

and sum of zeroes = \(\frac{-b}{a}\)

i. e., α + β = \(\frac{-b}{a}\)

⇒ α + β = \(\frac{0}{-3}\) = 0

62. If one zero of the polynomial x² -4x+ 1 is 2 + √3, write the other zero.

Answer:

Explaination:

Let other zero be α ,

63. Write the polynomial, the product and sum of whose zeroes are –\(\frac{9}{2}\) and –\(\frac{3}{2}\) respectively.

Answer:

Explaination:

64. The value of m, in order that x² – mx – 2 is the quotient where x^{3} + 3x² – 4 is divided by x + 2 is ____ .

Answer:

Explaination:

Since quotient = x² – mx – 2

∴ x² – mx – 2 = x² + x – 2

On comparison of coefficient of x, we get m = – 1.

65. If one factor of x^{3} + 7kx² – 4kx + 12 is (x + 3), then the value of k is ______ .

Answer:

Explaination:

p(x) = x^{3} + Ike – 4kx + 12

p(-3) = (-3)^{3} + 7k(-3)² – 4k (- 3) + 12

∵ x + 3 is a factor

∴ p(-3) = 0

⇒ -27 + 63k + 12k+ 12 = 0

⇒ 75k – 15 = 0

⇒ k = \(\frac{1}{2}\)

66. A polynomial of degree five is divided by a quadratic polynomial. If it leaves a remainder, then find the degree of remainder.

Answer:

Explaination:

Degree of remainder is always less than divisor.

∴ Degree of remainder will be less than 2.

Hence, degree is 1 or 0.

67. Check whether 3x – 7 is a factor of polynomial 6x^{3} + x² – 26x – 25?

Answer:

Explaination:

68. If x^{3} + x² – ax + b is divisible by x² – x, write the values of a and b.

Answer:

Explaination:

∵ p(x) = x3+xl-ax + b is divisible by x² -x.

x (x – 1) is a factor of p(x).

⇒ x = 0 and x = 1 are zeroes of p(x).

⇒ p(0) = 0

⇒ (0)^{3} + (0)² – a × 0 + b = 0

⇒ b = 0 and p(1) = 0

⇒ 1^{3} + (1)² – a × 1 + b = 0

⇒ 2 – a + 0 — 0

⇒ a = 2