KC Sinha: Exercise 2.1 - Mathematics Solution Class 10 Chapter 2 Real numbers

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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.

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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.

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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.