NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Exercise 2.1 Solutions

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NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Exercise 2.1 Solutions are based on Polynomials and their applications.

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Read More: NCERT Solutions For Class 9 Maths Chapter 2 Polynomials

Exercise Solutions of Class 9 Maths Chapter 2 Polynomials

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Chapter Related Topics
Polynomials	Remainder Theorem	Degree of polynomial
Polynomial Important Questions	Polynomials in One Variable	Synthetic Division
Remainder	Algebraic Identities	Cubic Polynomials

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CBSE X Related Questions

1.
Prove that: \(\frac{\sec^3 \theta}{\sec^2 \theta - 1} + \frac{\csc^3 \theta}{\csc^2 \theta - 1} = \sec \theta \cdot \csc \theta (\sec \theta + \csc \theta)\)
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2.
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.
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3.
The graph of \(y = f(x)\) is given. The number of zeroes of \(f(x)\) is :
- 0
- 1
- 3
- 2
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4.
Solve the linear equations \(3x + y = 14\) and \(y = 2\) graphically.
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5.
If \(PQ\) and \(PR\) are tangents to the circle with centre \(O\) and radius \(4 \text{ cm}\) such that \(\angle QPR = 90^{\circ}\), then the length \(OP\) is
- \(4 \text{ cm}\)
- \(4\sqrt{2} \text{ cm}\)
- \(8 \text{ cm}\)
- \(2\sqrt{2} \text{ cm}\)
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6.
Arc \(PQ\) subtends an angle \(\theta\) at the centre of the circle with radius \(6.3 \text{ cm}\). If \(\text{Arc } PQ = 11 \text{ cm}\), then the value of \(\theta\) is
- \(10^{\circ}\)
- \(60^{\circ}\)
- \(45^{\circ}\)
- \(100^{\circ}\)
View Solution

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