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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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.
Assertion (A) : H.C.F. \((36 m^{2}, 18 m) = 18 m\), where \(m\) is a prime number.
Reason (R) : H.C.F. of two numbers is always less than or equal to the smaller number.
- Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
- Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
- Assertion (A) is true, but Reason (R) is false.
- Assertion (A) is false, but Reason (R) is true.
View Solution
2.
Prove that: $\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \tan \theta + \cot \theta$
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3.
The natural number 1 is :
- a prime number.
- a composite number.
- prime as well as composite.
- neither prime nor composite.
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4.
The graph of \(y = f(x)\) is given. The number of zeroes of \(f(x)\) is :
- 0
- 1
- 3
- 2
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5.
\(ABCD\) is a parallelogram such that \(AF = 7 \text{ cm}\), \(FB = 3 \text{ cm}\) and \(EF = 4 \text{ cm}\), length \(FD\) equals
- \(\frac{21}{4} \text{ cm}\)
- \(\frac{28}{3} \text{ cm}\)
- \(\frac{12}{7} \text{ cm}\)
- \(5.5 \text{ cm}\)
View Solution
6.
A chord of a circle, of radius 14 cm, subtends an angle of $60^\circ$ at the centre. Find the area of the smaller sector and perimeter of the smaller segment.
View Solution

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