NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Exercise 2.1

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NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Exercise 2.1 is given in this article with a step by step explanation. Chapter 2 Polynomials Exercise 2.1 covers different cases of the geometrical meaning of the zeroes of a polynomial. The exercise has 1 question with 6 cases and includes finding the zeroes of a polynomial through the graphical method.

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Check out the solutions of Class 10 Maths NCERT solutions chapter 2 Polynomials Exercise 2.1

Read More: NCERT Solutions For Class 10 Maths Polynomials

Check out other exercise solutions of Class 10 Maths Chapter 2 Polynomials

Class 10 Chapter 2 Polynomials Topics:

Degree of Polynomial	Parabola Graph	Polynomials Formula
Polynomials Important Questions	Polynomials Revision Notes	Cubic Polynomial
Roots of Polynomials	Vietas Formula	Difference between zeroes and coefficients

CBSE Class 10 Maths Study Guides:

Calculus	Math Formula	Math MCQs
Difference between in Maths	Math Study Notes	NCERT Solutions for Class 10 Maths
Mensuration	Arithmetic	Trigonometry

CBSE X Related Questions

1.
A conical cavity of maximum volume is carved out from a wooden solid hemisphere of radius 10 cm. Curved surface area of the cavity carved out is (use \(\pi = 3.14\))
- \(314 \sqrt{2}\) \(\text{cm}^{2}\)
- \(314\) \(\text{cm}^{2}\)
- \(\frac{3140}{3}\) \(\text{cm}^{2}\)
- \(3140 \sqrt{2}\) \(\text{cm}^{2}\)
View Solution
2.
The graph of \(y = f(x)\) is given. The number of zeroes of \(f(x)\) is :
- 0
- 1
- 3
- 2
View Solution
3.
The dimensions of a window are 156 cm \(\times\) 216 cm. Arjun wants to put grill on the window creating complete squares of maximum size. Determine the side length of the square and hence find the number of squares formed.
View Solution
4.
Solve the linear equations \(3x + y = 14\) and \(y = 2\) graphically.
View Solution
5.
Three tennis balls are just packed in a cylindrical jar. If radius of each ball is \(r\), volume of air inside the jar is
- \(2\pi r^3\)
- \(3\pi r^3\)
- \(5\pi r^3\)
- \(4\pi r^3\)
View Solution
6.
Prove that: \(\frac{\sec^3 \theta}{\sec^2 \theta - 1} + \frac{\csc^3 \theta}{\csc^2 \theta - 1} = \sec \theta \cdot \csc \theta (\sec \theta + \csc \theta)\)
View Solution

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