NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Exercise 2.2 Solutions

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NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Exercise 2.2 Solutions are based on the concept of zero of the polynomial.

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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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Difference Between Topics in Maths	Math Study Material	Algebra
Trigonometry	Number Systems	Mensuration
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CBSE X Related Questions

1.
In an A.P., if $a_{14} - a_8 = 24$, then the common difference of the A.P. is
- $6$
- $4$
- $\pm 4$
- $3$
View Solution
2.
If for a data, median is 5 and mode is 4, then mean is equal to :
- 7
- 11
- $\frac{11}{2}$
- $\frac{14}{3}$
View Solution
3.
If the zeroes of the polynomial $p(x) = 2x^2 - 7x + 6$ are $\alpha$ and $\beta$, then the value of $\frac{1}{\alpha} + \frac{1}{\beta}$ is
- $\frac{7}{2}$
- $\frac{6}{7}$
- $\frac{-7}{6}$
- $\frac{7}{6}$
View Solution
4.
The value of k for which sum of the zeroes of the polynomial $p(x) = 3x^2 - kx + 6$ is 2, is
- 2
- $-6$
- $-2$
- 6
View Solution
5.
In the given figure, $PQ$ is tangent to the circle with centre $O$. $S$ is a point on the circle such that $\angle SQT = 55^\circ$. The $m\angle QPS$ is
- $55^\circ$
- $20^\circ$
- $35^\circ$
- $70^\circ$
View Solution
6.
For an acute angle $\theta$, if $\sin \theta = \frac{1}{9}$, then value of $\frac{9 \csc \theta + 1}{9 \csc \theta - 1}$ is
- $0$
- $\frac{80}{81}$
- $1$
- $\frac{82}{80}$
View Solution

View All

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NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Exercise 2.2 Solutions

Exercise Solutions of Class 9 Maths Chapter 2 Polynomials

CBSE X Related Questions

1.
In an A.P., if $a_{14} - a_8 = 24$, then the common difference of the A.P. is

2.
If for a data, median is 5 and mode is 4, then mean is equal to :

3.
If the zeroes of the polynomial \(p(x) = 2x^2 - 7x + 6\) are \(\alpha\) and \(\beta\), then the value of \(\frac{1}{\alpha} + \frac{1}{\beta}\) is

4.
The value of k for which sum of the zeroes of the polynomial \(p(x) = 3x^2 - kx + 6\) is 2, is

5.
In the given figure, $PQ$ is tangent to the circle with centre $O$. $S$ is a point on the circle such that $\angle SQT = 55^\circ$. The $m\angle QPS$ is

6.
For an acute angle \(\theta\), if \(\sin \theta = \frac{1}{9}\), then value of \(\frac{9 \csc \theta + 1}{9 \csc \theta - 1}\) is

Similar Mathematics Concepts

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NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Exercise 2.2 Solutions

Exercise Solutions of Class 9 Maths Chapter 2 Polynomials

CBSE X Related Questions

1.In an A.P., if $a_{14} - a_8 = 24$, then the common difference of the A.P. is

2.If for a data, median is 5 and mode is 4, then mean is equal to :

3.If the zeroes of the polynomial \(p(x) = 2x^2 - 7x + 6\) are \(\alpha\) and \(\beta\), then the value of \(\frac{1}{\alpha} + \frac{1}{\beta}\) is

4.The value of k for which sum of the zeroes of the polynomial \(p(x) = 3x^2 - kx + 6\) is 2, is

5.In the given figure, $PQ$ is tangent to the circle with centre $O$. $S$ is a point on the circle such that $\angle SQT = 55^\circ$. The $m\angle QPS$ is

6.For an acute angle \(\theta\), if \(\sin \theta = \frac{1}{9}\), then value of \(\frac{9 \csc \theta + 1}{9 \csc \theta - 1}\) is

Similar Mathematics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
In an A.P., if $a_{14} - a_8 = 24$, then the common difference of the A.P. is

2.
If for a data, median is 5 and mode is 4, then mean is equal to :

3.
If the zeroes of the polynomial \(p(x) = 2x^2 - 7x + 6\) are \(\alpha\) and \(\beta\), then the value of \(\frac{1}{\alpha} + \frac{1}{\beta}\) is

4.
The value of k for which sum of the zeroes of the polynomial \(p(x) = 3x^2 - kx + 6\) is 2, is

5.
In the given figure, $PQ$ is tangent to the circle with centre $O$. $S$ is a point on the circle such that $\angle SQT = 55^\circ$. The $m\angle QPS$ is

6.
For an acute angle \(\theta\), if \(\sin \theta = \frac{1}{9}\), then value of \(\frac{9 \csc \theta + 1}{9 \csc \theta - 1}\) is