Polynomials

Class 10 > Math > NCERT > Polynomials

Case Based Questions for Ch 2 Polynomials Class 10 Math

Important Questions

1

Construction of Humps
ABC construction company got the contract of making speed humps on roads. Speed humps are parabolic in shape and prevents overspeeding, minimise accidents and gives a chance for pedestrians to cross the road. The mathematical representation of a speed hump is shown in the given graph.

Based on the above information, answer the following questions.


(i) The polynomial represented by the graph can be _____ polynomial.
(a) Linear
(b) Quadratic
(c) Cubic
(d) Zero


(ii) The zeroes of the polynomial represented by the graph are
(a) 1, 5
(b) 1, -5
(c) -1, 5
(d) -1, -5


(iii) The sum of zeroes of the polynomial represented by the graph are
(a) 4
(b) 5
(c) 6
(d) 7


(iv) If Ξ± and Ξ² are the zeroes of the polynomial represented by the graph such that Ξ² > Ξ±, then |8Ξ± + Ξ²| =
(a) 1
(b) 2
(c) 3
(d) 4


(v) The expression of the polynomial represented by the graph is
(a) –x2 – 4x – 5
(b) x2 + 4x + 5
(c) x2 + 4x – 5
(d) –x2 + 4x + 5

Answer

(i) (b) Since, the given graph is parabolic is shape, therefore it will represent a quadratic polynomial.
[∡Graph of quadratic polynomial is parabolic in shape]
(ii) (c) Since, the graph cuts the x-axis at -1, 5. So the polynomial has 2 zeroes i.e., -1 and 5.
(iii) (a) Sum of zeroes = -1 + 5 = 4
(iv) (c) Since Ξ± and Ξ² are zeroes of the given polynomial and Ξ² > Ξ±
∴ α = -1 and β = 5.
∴|8Ξ± + Ξ²| = |8(βˆ’1) + 5| = |βˆ’8 + 5| = |βˆ’3| = 3.
(v) (d) Since the zeroes of the given polynomial are - 1 and 5.
∴ Required polynomial p(x)
= k{x2 -(βˆ’1 + 5)x + (βˆ’1)(5)} = k(x2 – 4x – 5)
For k = -1, we get
p(x) = βˆ’ x2 + 4x + 5, which is the required polynomial.

2

Honeycomb
While playing in garden, Sahiba saw a honeycomb and asked her mother what is that. She replied that it's a honeycomb made by honey bees to store honey. Also, she told her that the shape of the honeycomb formed is parabolic. The mathematical representation of the honeycomb structure is shown in the graph.

Based on the above information, answer the following questions.


(i) Graph of a quadratic polynomial is _____ in shape.
(a) straight line
(b) parabolic
(c) circular
(d) None of these


(ii) The expression of the polynomial represented by the graph is
(a) x2 – 49
(b) x2 – 64
(c) x2 – 36
(d) x2 – 81


(iii) Find the value of the polynomial represented by the graph when x = 6.
(a) -2
(b) -1
(c) 0
(d) 1


(iv) The sum of zeroes of the polynomial x2 + 2x – 3 is
(a) -1
(b) -2
(c) 2
(d) 1


(v) If the sum of zeroes of polynomial at2 + 5t + 3a is equal to their product, then find the value of a.
(a) -5
(b) -3
(c) 5/3
(d) -5/3

Answer

(i) (b) Graph of a quadratic polynomial is a parabolic in shape.
(ii) (c) Since the graph of the polynomial cuts the x – axis at (-6, 0) and (6, 0). So, the zeroes of polynomial are -6 and 6.
∴ Required polynomial is
p(x) = x2 – (-6 + 6)x + (-6)(6) = x2 – 36
(iii) (c) We have, p(x) = x2 – 36
Now, p(6) = 62 – 36 = 36 – 36 = 0
(iv) (b) Let f(x) = x2 + 2x – 3. Then,
Sum of zeroes = -(coefficient of x)/(coefficient of x2) = -(2)/1 = -2
(v) (d) The given polynomial is at2 + 5t + 3a
Given, sum of zeroes = product of zeroes
β‡’ -5/a = 3a/a β‡’ a = -5/3

3

Just before the morning assembly a teacher of kindergarten school observes some clouds in the sky and so she cancels the assembly. She also observes that the clouds has a shape of the polynomial. The mathematical representation of a cloud is shown in the figure.


(i) Find the zeroes of the polynomial represented by the graph.
(a) -1/2, 7/2
(b) Β½, -7/2
(c) -1/2, -7/2
(d) 1/2, 7/2


(ii) What will be the expression for the polynomial represented by the graph?
(a) p(x) = 12x2 – 4x – 7
(b) p(x) = -x2 – 12x + 3
(c) p(x) = 4x2 + 12x + 7
(d) p(x) = -4x2 – 12x + 7


(iii) What will be the value of polynomial represented by the graph, when x = 3?
(a) 65
(b) -65
(c) 68
(d) -68


(iv) If Ξ± and Ξ² are the zeroes of the polynomial f(x) = x2 + 2x – 8, then Ξ±4 + Ξ²4 =
(a) 262
(b) 252
(c) 272
(d) 282


(v) Find a quadratic polynomial where sum and product of its zeroes are 0, √7 respectively.
(a) k(x2 + √7)
(b) k(x2 – √7)
(c) k(x2 + √5)
(d) none of these

Answer

(i) (b) Since the graph of the polynomial intersect the x – axis at x = Β½ , -7/2, therefore required zeroes of the polynomial are Β½ and -7/2.
(ii) (d) ∡ ½ and -7/2 are the zeroes of the polynomial.
So, at x = Β½, -7/2, the value of the polynomial will be 0.
From options, required polynomial is p(x) = -4x2 – 12x + 7.
(iii) (b) We have, p(x) = -4x2 – 12x + 7
∴ p(3) = -4(3)2 – 12(3) + 7 = -36 – 36 + 7 = -65
(iv) (c) Here f(x) = x2 + 2x – 8 and Ξ±, Ξ² are its zeroes.
∴ α + β = -2 and αβ = -8
Now, Ξ±4 + Ξ²4 = (Ξ±2 + Ξ²2)2 – 2Ξ±2Ξ²2
= ((Ξ±+Ξ²)2 – 2Ξ±Ξ²)2 – 2(Ξ±Ξ²)2
= [(-2)2 – 2(-8)]2 – 2(-8)2
= [4 + 16]2 – 2(-8)2 = (20)2 – 2(64)
= 400 – 128 = 272
(v) (a) We have sum of zeroes = 0 and product of zeroes = √7
So, required polynomial = k(x2 – 0.x + √7)
= k(x2 + √7)

4

Avocado
Pankaj’s father gave him some money to buy avocado from the market at the rate of p(x) = x2 – 24x + 128. Let Ξ±, Ξ² are the zeroes of p(x).

Based on the above information, answer the following questions.


(i) Find the value of Ξ± and Ξ², where Ξ± < Ξ².
(a) -8, -16
(b) 8, 16
(c) 8, 15
(d) 4, 9


(ii) Find the value of Ξ± + Ξ² + Ξ±Ξ².
(a) 151
(b) 158
(c) 152
(d) 155


(iii) The value of p(2) is
(a) 80
(b) 81
(c) 83
(d) 84


(iv) If Ξ± and Ξ² are zeroes of x2 + x – 2 , then 1/Ξ± + 1/Ξ² =
(a) Β½
(b) 1/3
(c) ΒΌ
(d) 1/5


(v) If sum of zeroes of q(x) = kx2 + 2x 3k is equal to their product, then k =
(a) 2/3
(b) 1/3
(c) -2/3
(d) -1/3

Answer

(i) (b) Given, Ξ± and Ξ² are the zeroes of p(x) = x2 – 24x + 128.
Putting p(x) = 0, we get
x2 – 8x – 16x + 128 = 0
β‡’ x(x – 8) – 16(x – 8) = 0
β‡’ (x – 8)(x – 16) = 0 β‡’ x = 8 or x = 16
∴ α = 8, β = 16
(ii) (c) Ξ± + Ξ² + Ξ±Ξ² = 8 + 16 + (8)(16)
= 24 + 128 = 152
(iii) (d) p(2) = 22 – 24(2) + 128 = 4 – 48 + 128 = 84
(iv) (a) Since Ξ± and Ξ² are zeroes of x2 + x – 2.
∴ α + β = -1 and αβ = -2
Now, 1/Ξ± + 1/Ξ² = (Ξ² + Ξ±)/(Ξ±Ξ²) = -1/-2 = Β½
(v) (c) Sum of zeroes = -2/k
Product of zeroes = 3k/k = 3
According to question, we have -2/k = 3
β‡’ k = -2/3

5

In a soccer match, the path of the soccer ball in a kick is recorded as shown in the following graph.

Based on the above information, answer the following questions.


(i) The shape of path of the soccer ball is a
(a) Circle
(b) Parabola
(c) Line
(d) None of these


(ii) The axis of symmetry of the given parabola is
(a) y – axis
(b) x – axis
(c) line parallel to y – axis
(d) line parallel to x – axis


(iii) The zeroes of the polynomial, represented in the given graph, are
(a) -1, 7
(b) 5, -2
(c) -2, 7
(d) -3, 8


(iv) Which of the following polynomial has -2 and -3 as its zeroes ?
(a) x2 – 5x – 5
(b) x2 + 5x – 6
(c) x2 + 6x – 5
(d) x2 + 5x + 6


(v) For what value of β€˜x’ , the value of the polynomial f(x) = (x – 3)2 + 9 is 9 ?
(a) 1
(b) 2
(c) 3
(d) 4

Answer

(i) (b) The shape of the path of the soccer ball is a parabola.
(ii) (c) The axis of symmetry of the given curve is a line parallel to y – axis.
(iii) (a) The zeroes of the polynomial, represented in the given graph, are -2 and 7, since the curve cuts the x – axis at these points.
(iv) (d) A polynomial having zeroes -2 and -3 is
p(x) = x2 – (-2 – 3)x + (-2)(-3) = x2 + 5x + 6
(v) (c) We have, f(x) = (x – 3)2 + 9
Now, 9 = (x – 3)2 + 9
β‡’ (x – 3)2 = 0 β‡’ x – 3 = 0 β‡’ x = 3

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