KC Sinha Mathematics Solution Class 10 Chapter 2 Polynomials Exercise 2.1


Exercise 2.1
Exercise 2.2
Exercise 2.3

Exercise 2.1


Question 1 

Examine, seeing the graph of the polynomials given below, whether they are a linear or quadratic polynomial or neither linear nor quadratic polynomial:
(i)

 








Sol :
(i) In general, we know that for a linear polynomial ax + b, a≠0, the graph of y = ax + b is a straight line which intersects the x – axis at exactly one point.
And here, we can see that the graph of y = p(x) is a straight line and intersects the x – axis at exactly one point. Therefore, the given graph is of a Linear Polynomial.

(ii) 












Sol :
(ii) Here, the graph of y = p(x) is a straight line and parallel to the x – axis . Therefore, the given graph is of a Linear Polynomial.

(iii) 










Sol :
(iii) For any quadratic polynomial ax2 + bx + c, a ≠ 0, the graph of the corresponding equation y = a x2 + bx + c has one of the two shapes either open upwards likeor open downwards likedepending on whether a > 0 or a < 0. (These curves are called parabolas.)
Here, we can see that the shape of the graph is a parabola. Therefore, the given graph is of a Quadratic Polynomial.

(iv) 










Sol :
(iv) For any quadratic polynomial ax2 + bx + c, a ≠ 0, the graph of the corresponding equation y = ax2 + bx + c has one of the two shapes either open upwards like or open downwards like depending on whether a > 0or a < 0. (These curves are called parabolas.)
Here, we can see that the shape of the graph is parabola. Therefore, the given graph is of a Quadratic Polynomial.

(v) 












Sol :
(v) The shape of the graph is neither a straight line nor a parabola. So, the graph is not of a linear polynomial nor a quadratic polynomial.

(vi) 











Sol :
(vi) The given graph have a straight line but it doesn’t intersect at x – axis and the shape of the graph is also not a parabola. So, it is not a graph of a quadratic polynomial. Therefore, it is not a graph of linear polynomial or quadratic polynomial.

(vii) 












Sol :
(vii) The shape of the graph is neither a straight line nor a parabola. So, the graph is not of a linear polynomial or a quadratic polynomial.


(viii) 











Sol :
(viii) The shape of the graph is neither a straight line nor a parabola. So, the graph is not of a linear polynomial or a quadratic polynomial.


Question 2 

The graphs of y – p(x) are given in the figures below, where p(x) is a polynomial. Find the number of zeros in each case.
(i) 










Sol :
(i) Here, the graph of y = p(x) intersects the x – axis at two points. So, the number of zeroes is 2.


(ii) 










Sol :
(ii) Here, the graph of y = p(x) intersects the x – axis at three points. So, the number of zeroes is 3.


(iii) 

 








Sol :
(iii) Here, the graph of y = p(x) intersects the x – axis at one point only. So, the number of zeroes is 1.


(iv) 










Sol :
(iv) Here, the graph of y = p(x) intersects the x – axis at exactly one point. So, the number of zeroes is 1.


(v) 









Sol :
(v) Here, the graph of y = p(x) intersects the x – axis at two points. So, the number of zeroes is 2.


(vi) 









Sol :
(vi) Here, the graph of y = p(x) intersects the x – axis at exactly one point. So, the number of zeroes is 1.

Question 3 

The graphs of y = p(x) are given in the figures below, where p(x) is a polynomial Find the number of zeroes in each case.
(i) 










Sol :
(i) Here, the graph of y = p(x) intersect the x – axis at zero points. So, the number of zeroes is 0.


(ii) 










Sol :
(ii) Here, the graph of y = p(x) intersects the x – axis at two points. So, the number of zeroes is 2.

(iii) 









Sol :
(iii) Here, the graph of y = p(x) intersects the x – axis at four points. So, the number of zeroes is 4.

(iv) 









Sol :
(iv) Here, the graph of y = p(x) does not intersects the x – axis. So, the number of zeroes is 0.

(v) 










Sol :
(v) Here, the graph of y = p(x) intersects the x – axis at two points. So, the number of zeroes is 2.

(vi) 










Sol :
(vi) Here, the graph of y = p(x) intersects the x – axis at two points. So, the number of zeroes is 2.

S.noChaptersLinks
1Real numbersExercise 1.1
Exercise 1.2
Exercise 1.3
Exercise 1.4
2PolynomialsExercise 2.1
Exercise 2.2
Exercise 2.3
3Pairs of Linear Equations in Two VariablesExercise 3.1
Exercise 3.2
Exercise 3.3
Exercise 3.4
Exercise 3.5
4Trigonometric Ratios and IdentitiesExercise 4.1
Exercise 4.2
Exercise 4.3
Exercise 4.4
5TrianglesExercise 5.1
Exercise 5.2
Exercise 5.3
Exercise 5.4
Exercise 5.5
6StatisticsExercise 6.1
Exercise 6.2
Exercise 6.3
Exercise 6.4
7Quadratic EquationsExercise 7.1
Exercise 7.2
Exercise 7.3
Exercise 7.4
Exercise 7.5
8Arithmetic Progressions (AP)Exercise 8.1
Exercise 8.2
Exercise 8.3
Exercise 8.4
9Some Applications of Trigonometry: Height and DistancesExercise 9.1
10Coordinates GeometryExercise 10.1
Exercise 10.2
Exercise 10.3
Exercise 10.4
11CirclesExercise 11.1
Exercise 11.2
12ConstructionsExercise 12.1
13Area related to CirclesExercise 13.1
14Surface Area and VolumesExercise 14.1
Exercise 14.2
Exercise 14.3
Exercise 14.4
15ProbabilityExercise 15.1

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