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.
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.
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.
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.
Here, we can see that the shape of the graph is a parabola. Therefore, the given graph is of a Quadratic Polynomial.
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.
Here, we can see that the shape of the graph is parabola. Therefore, the given graph is of a Quadratic Polynomial.
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.
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.
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.
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.
(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.
Sol :
(ii) Here, the graph of y = p(x) intersects the x – axis at three points. So, the number of zeroes is 3.
Sol :
(iii) Here, the graph of y = p(x) intersects the x – axis at one point only. So, the number of zeroes is 1.
Sol :
(iv) Here, the graph of y = p(x) intersects the x – axis at exactly one point. So, the number of zeroes is 1.
Sol :
(v) Here, the graph of y = p(x) intersects the x – axis at two points. So, the number of zeroes is 2.
Sol :
(vi) Here, the graph of y = p(x) intersects the x – axis at exactly one point. So, the number of zeroes is 1.
(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.
Sol :
(i) Here, the graph of y = p(x) intersect the x – axis at zero points. So, the number of zeroes is 0.
(i) Here, the graph of y = p(x) intersect the x – axis at zero points. So, the number of zeroes is 0.
Sol :
(ii) Here, the graph of y = p(x) intersects the x – axis at two points. So, the number of zeroes is 2.
Sol :
(iii) Here, the graph of y = p(x) intersects the x – axis at four points. So, the number of zeroes is 4.
Sol :
(iv) Here, the graph of y = p(x) does not intersects the x – axis. So, the number of zeroes is 0.
Sol :
(v) Here, the graph of y = p(x) intersects the x – axis at two points. So, the number of zeroes is 2.
Sol :
(vi) Here, the graph of y = p(x) intersects the x – axis at two points. So, the number of zeroes is 2.
S.no | Chapters | Links |
---|---|---|
1 | Real numbers | Exercise 1.1 Exercise 1.2 Exercise 1.3 Exercise 1.4 |
2 | Polynomials | Exercise 2.1 Exercise 2.2 Exercise 2.3 |
3 | Pairs of Linear Equations in Two Variables | Exercise 3.1 Exercise 3.2 Exercise 3.3 Exercise 3.4 Exercise 3.5 |
4 | Trigonometric Ratios and Identities | Exercise 4.1 Exercise 4.2 Exercise 4.3 Exercise 4.4 |
5 | Triangles | Exercise 5.1 Exercise 5.2 Exercise 5.3 Exercise 5.4 Exercise 5.5 |
6 | Statistics | Exercise 6.1 Exercise 6.2 Exercise 6.3 Exercise 6.4 |
7 | Quadratic Equations | Exercise 7.1 Exercise 7.2 Exercise 7.3 Exercise 7.4 Exercise 7.5 |
8 | Arithmetic Progressions (AP) | Exercise 8.1 Exercise 8.2 Exercise 8.3 Exercise 8.4 |
9 | Some Applications of Trigonometry: Height and Distances | Exercise 9.1 |
10 | Coordinates Geometry | Exercise 10.1 Exercise 10.2 Exercise 10.3 Exercise 10.4 |
11 | Circles | Exercise 11.1 Exercise 11.2 |
12 | Constructions | Exercise 12.1 |
13 | Area related to Circles | Exercise 13.1 |
14 | Surface Area and Volumes | Exercise 14.1 Exercise 14.2 Exercise 14.3 Exercise 14.4 |
15 | Probability | Exercise 15.1 |
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