CBSE Class 10 Sample Paper for 2021 Boards - Maths Standard

Question 19 - CBSE Class 10 Sample Paper for 2021 Boards - Maths Standard

Last updated at Oct. 19, 2020 by Teachoo

Case Study Based- 3

Applications of Parabolas-Highway Overpasses/Underpasses A highway underpass is parabolic in shape.

Parabola

A parabola is the graph that results from
p(x)= ax
2
+ bx + c

Parabolas are symmetric about a vertical line known as the
Axis of Symmetry
. The Axis of Symmetry runs through the maximum or minimum point of the parabola which is called the

Question 19 Case Study Based- 3 Applications of Parabolas-Highway Overpasses/Underpasses A highway underpass is parabolic in shape.Parabola
A parabola is the graph that results from p(x)= ax2 + bx + c
Parabolas are symmetric about a vertical line known as the Axis of Symmetry.
The Axis of Symmetry runs through the maximum or minimum point of the parabola which is called the Vertex (a) If the highway overpass is represented by x2 – 2x – 8.
Then its zeroes are
(i) (2, –4) (ii) (4, –2) (iii) (–2, –2) (iv) (–4, –4)
Finding Zeroes
x2 − 2x − 8 = 0
Solving by Spitting the middle term
x2 − 4x + 2x − 8 = 0
x(x − 4) + 2(x − 4) = 0
(x − 4) (x + 2) = 0
Therefore,
x = 4, x = −2
So, (ii) is correct
Splitting the middle term method
We need to find two numbers whose
Sum = −2
Product = 1 × −8 = −8
(b) The highway overpass is represented graphically. Zeroes of a polynomial can be expressed graphically. Number of zeroes of polynomial is equal to number of points where graph of polynomial (i) Intersects x-axis (ii) Intersects y-axis (iii) Intersects y-axis or x-axis (iv) None of the aboveIntersects the x-axis
(c) Graph of a quadratic polynomial is a (i) straight line (ii) circle (iii) parabola (iv) ellipseParabola
(d) The representation of Highway Underpass whose one zero is 6 and sum of the zeroes is 0, is (i) x2 – 6x + 2 (ii) x2 – 36 (iii) x2 – 6 (iv) x2 – 3Given
Sum of Zeros = 0
First Zero + Second Zero = 0
6 + Second Zero = 0
Second Zero = −6
The Required polynomial is
p(x) = x2 − (Sum of Zeroes)x + Product of Zeroes
= x2 − 0x + (6 × −6)
= x2 − 36
(e) The number of zeroes that polynomial f(x) = (x – 2)2 + 4 can have is: (i) 1 (ii) 2 (iii) 0 (iv) 3 To find Zero,
We put f(x) = 0 and find value of x
Putting f(x) = 0
(x – 2)2 + 4 = 0
(x – 2)2 = −4
Since Square cannot be negative
This is not possible
Therefore, this Polynomial doesn’t have a Zero
Number of Zeros = 0

CBSE Class 10 Sample Paper for 2021 Boards - Maths Standard

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.