Class 10 Maths Chapter 2 Polynomials Objective Questions

 

Basics Revisited

1. 1. Write the coefficient of  x2 in each of the following.

Answer: C. 1, -1, π/2, 0

Solution: The constant multiplied by x2 is the coefficient of x2

(1) 2 + x2 + x → coefficient of x2 = 1

(2) 2 – x2 + x3 → coefficient of x2 = -1

(3) (π/2) x2 +x → coefficient of x2 = π/2

(4) √2 x−1 → coefficient of x2 = 0

2. 2. The polynomial p(x) = x−323 is a ___.
      1. Constant Polynomial
      2. Cubic Polynomial
      3. Quadratic Polynomial
      4. Linear Polynomial

Answer: D. Linear Polynomial

Solution: The polynomial of degree one is called a linear polynomial.

Therefore, x−323 is a linear polynomial.

Graphical Representations

3. 3. Three curves, i.e. a) y = x2b) y = x4c) y = x6, are depicted in the graph shown below. Which of the polynomials does graph 3 represent?

3. 3. 1. y = x4
      2. y = x6
      3. y = x2
      4. Cannot be determined

Answer: B. y = x6

Solution: Consider the polynomial where n is a positive even integer.

As the value of n increases, then the curve goes closer to the positive y-axis.

Thus, graph 3 represents the polynomial x6

4. 4. Three curves, i.e.,

a) Y = −x2

b) y = −x3

c) y = −x7

are depicted in the following graph and are numbered from 1 to 3.

Identify the correct relation.

4. 4. 1. (a)-(1) , (b)-(2), (c)-(3)
      2. (a)-(3) , (b)-(2), (c)-(1)
      3. (a)-(1) , (b)-(3), (c)-(2)
      4. (a)-(2) , (b)-(3), (c)-(1)

Answer: A. (a)-(1), (b)-(2), (c)-(3)

Solutions: When a polynomial is of the form y = −xn, the graph of the polynomial is the mirror image of the graph of the polynomial y = xn.

Also, when the value of n increases, the graph draws closer to the y-axis.

Thus, graph 1 represents y = −x2, graph 2 represents y = −x3 and graph 3 represents y = −x7

Visualisation of a polynomial

5. 5. If x = 2,y = −1, then the value of x2+4xy+4y2 is
      1. 2
      2. -1
      3. 1
      4. 0

Answer: D. 0

Solution: Substituting the values,

X2+4xy+4y2

= (2)2 + 4(2) (-1) + 4(-1)2

= 4−8+4=0

6. 6. According to the graph below, the product of the zeroes of the polynomial will be

6. 6. 1. Cannot be determined
      2. Zero
      3. Negative
      4. Positive

Answer: C. Negative

Solution: One of the zeros of the polynomial lies on the positive x-axis. Thus, the abscissa or the x -coordinate, which is the corresponding zero, is positive.

The other zero lies on the negative x-axis. Thus, the abscissa or x -coordinate, which is the corresponding zero, is negative.

Thus, the product of zeroes is going to be positive negative = negative.

Zeroes of a Polynomial

7. 7. The number of polynomials having 3 and 7 as zeroes are
      1. More than 3
      2. 3
      3. 2
      4. 1

Answer: A. More than 3

Solution: (x-3)A(X-7)B, where a and b can take any natural number values.

Hence, infinite possibilities.

8. 8. If α, β and γ are the zeros of the polynomial

8. 8. 1. – (b/a)
      2. – (c/d)
      3. a/d
      4. c/d

Answer: B. –(c/d)

Solutions: If α, β and γ are the zeros of the polynomial

9. 9. If α, β are the zeros of the polynomial, x2-px +36 and α2 + β2 = 9, then what is the value of p?
      1. 6
      2. 3
      3. 9
      4. 8

Answer: C. 9

Solution: Here a = 1, b = -p, c = 36.

Factorisation of Polynomials

10. 10. What is the factorisation of 2x2−7x−15?
        1. (x+5) (2x-3)
        2. (x+3) (2x-5)
        3. (x-5) (2x+3)
        4. (x-3) (2x-5)

Answer: C. (x-5) (2x+3)

Solution: Find two numbers such that their product is -30 and the sum is -7.

11. 11. What is the factorisation of x2 −5x+6?
        1. (x+5) (x-3)
        2. (x-6) (x+1)
        3. (x-1) (x+5)
        4. (x-2) (x-3)

Answer: D. (x-2) (x-3)

12. 12. Which among the following options is one of the factors of x2+ x/6+ 1/6?
        1. 3x +1
        2. 2x + 1
        3. X – (1/5)
        4. X- (1/2)

Answer: B. 2x +1

Relationship between zeroes and coefficient

13. 13. Find the sum  and product of roots  for the given polynomial:

Answer: A. -(1/2), – (5/2)

Solution: We know that, for a quadratic equation

ax2 + bx+ c = 0 sum of roots = α+β & product of roots = αβ

Comparing 2x2+x−5=0 with ax2 + bx+ c=0, we get

a=2, b=1, c=−5

⇒α+β= – (1/2)

⇒αβ= – (5/2)

14. 14. If p, q & r are the zeroes of a cubic polynomial ax3+bx2+cx+d, then what will be p+q+r?
        1. c/a
        2. b/a
        3. –(c/a)
        4. –(b/a)

Answer: D. –(b/a)

Solution: We know that for a cubic polynomial ax3+bx2+cx+d Sum of zeroes = –(b/a)

Therefore, p+q+r= – (b/a).

Division Algorithm

15. 15. In the division algorithm, when should one stop the division process?

1. When the remainder is zero.

2. When the degree of the remainder is less than the degree of the divisor.

3. When the degree of the quotient is less than the degree of the divisor.

15. 15. 1. Statements 1 and 2 are correct
        2. Statements 2 and 3 are correct
        3. Statements 1 and 3 are correct
        4. Only statement 3 is correct

Answer: (A) Statements 1 and 2 are correct

Solution: We stop the division process when either the remainder is zero, or its degree is less than the degree of the divisor.

16. 16. If the remainder when x3+2x2+kx+3 is divided by x-3 is 21, find the zeroes of x3+2x2+kx−18.
        1. -2, 3, 3
        2. -3, 2, 3
        3. -3, -2,3
        4. -3,-3, 2

Answer: C. -3, -2, 3

Solution: P (3) = 48 + 3k = 21

⇒ K = -9

Hence, x3+2x2−9x+3= (x−3) x Quotient + 21

⇒x3+2x2−9x−18 =9x−3) x Quotient

Quotient = (x3+2x2−9x−18) / (x-3)

Factorising the quotient, x2+ 5x +6= x2+ 3x + 2x +6 = x(x+3) +2(x+3) =(x+2) (x+3)

Hence, the factors of x3+2x2−9x−18 are x−3, x+2 and x+3

⇒ the zeroes are -3,-2, 3.

Algebraic identities

17. 17. If x+x-1 = 10, (x≠0), then what will be x2+x-2?
        1. 100
        2. 10
        3. 98
        4. 102

Answer: C. 98

18. 18. f α and β are the zeros of the polynomial

18. 18. 1. –(45/8)
        2. 45/8
        3. -(8)/ 45
        4. 8/45

Answer: B. 45/8

19. 19. What term should be added to a2+2ab to make it a perfect square?
        1. 2ab
        2. a2
        3. b2

Answer: C. b2

Solution: To make (a2+2ab) a perfect square, b2 is to be added.

So (a2+2ab+b2) will become a perfect square using the identity