Chapter 2 Polynomials NCERT Exemplar Solutions Exercise 2.2 Class 10 Maths

Chapter 2 Polynomials NCERT Exemplar Solutions Exercise 2.2 Class 10 Maths

Chapter Name

NCERT Maths Exemplar Solutions for Chapter 2 Polynomials Exercise 2.2

Book Name

NCERT Exemplar for Class 10 Maths

Other Exercises

  • Exercise 2.1
  • Exercise 2.3
  • Exercise 2.4

Related Study

NCERT Solutions for Class 10 Maths

Exercise 2.2 Solutions

Short Answer Questions with Reasoning: 

1. Answer the following and justify: 
(i) Can x2 – 1 be the quotient on division of x6 + 2x3 + x - 1 by a polynomial in x of degree 5 ? 
(ii) What will the quotient and remainder be on division of ax2 + bx + c by px3 + qx2 + rx + s, p ≠ 0 ?
(iii) If on division of a polynomial p(x) by a polynomial g(x), the quotient is zero, what is the relation between the degrees of p (x) and g (x)?
(iv) If on division of a non-zero polynomial p (x) by a polynomial g (x), the remainder is zero, what is the relation between the degrees of p (x) and g (x)?
(v) Can the quadratic polynomial x2 + kx + k have equal zeroes for some odd integer k ? 1 ? 

Solution
(i) No, x2 - 1 cannot be the quotient  on division of x6 + 2x3 + x - 1 by a polynomial in x of  degree 5.
Explanation : 
When a polynomial with degree 6 is divided by degree 5 polynomial, the quotient will be of degree 1. 
Assuming that (x2 - 1) divides the degree 6 polynomial and the quotient obtained is degree 5 polynomial. 
As a = bq + r, so, 
(Degree 6 polynomial) = (x2 - 1) (degree 5 polynomial) + r(x) 
= (degree 7 polynomial) + r(x) 
[As, (x2 term × x5 term = x7 term)]
= (degree 7 polynomial)
So, this contradict our assumption. 
Hence, x2 - 1 cannot be the quotient on division of x6 + 2x3 + x - 1 by a polynomial in x of degree 5.

(ii) Degree of the polynomial px3 + qx2 + rx + s = 3 
Degree of the polynomial ax2 + bx + c = 2 
Here, Degree of px3 + qx2 + rx + s is greater than degree of the ax2 + bx + c
Hence, the quotient would be zero, 
Therefore, the remainder would be the dividend = ax2 + bx + c. 

(iii) We have, p(x) = g(x) × q(x) + r(x) 
As given in the question, 
q(x) = 0 
When q(x) = 0, 
r(x) is also = 0 
Here, when we divide p(x) by g(x), 
Then, p(x) should be = 0 
Therefore, the relation between the degrees of p(x) and g(x) is the degree p(x) < degree g(x). 

(iv) To divide p(x) by g(x) 
We have, 
Degree of p(x) > degree of g(x) 
or 
Degree of p(x) = degree of g(x) 
Hence, the relation between the degrees of p(x) and g(x) is degree of p(x) > degree of g(x) 

(v) A Quadratic Equation has equal roots when : 
b2 - 4ac = 0 
Given, 
x2 + kx + k = 0 
a = 1, 
b = k, 
x = k
Putting values in the equation we get, 
k2 - 4(1)(k) = 0 
⇒ k3 - 4k = 0 
⇒ k(k - 4) = 0 
⇒ k = 0, 
⇒ k = 4 
Here, it is given that k is greater than I, So, the value of k is 4 if the equation has common roots. 
If k = 4, then the equation (x2 + kx + k) will have equal roots.


2. Are the following statements ‘True’ or ‘False’? Justify your answers.
(i) If the zeroes of a quadratic polynomial ax2 + bx + c are both positive, then a, b and c all have the same sign. 
(ii) If the graph of a polynomial intersects the x - axis at only one point, it cannot be a quadratic polynomial. 
(iii) If the graph of a polynomial intersects the x - axis at exactly two points, it need not be a quadratic polynomial. 
(iv) If two of the zeroes of a cubic polynomial are zero, then it does not have linear and constant terms. 
(v) If all the zeroes of a cubic polynomial are negative, then all the coefficients and the constant term of the polynomial have the same sign. 
(vi) If all three zeroes of a cubic polynomial x3 + ax2 - bx + c are positive, then at least one of a, b and c is non - negative. 
(vii) The only value of k for which the quadratic polynomial kx2 + x + k has equal zeros is 1/2. 

Solution

(i) False
Taking α and β as the roots of he quadratic polynomial. If α and β are positive then α + β = -b/a. For making sum of roots positive either b or a must be negative. 

(ii) False 
The statement is false, because when two zeroes of a quadratic polynomial are equal, then two intersecting points coincide to become one point. 

(iii) True 
If a polynomial of degree more than two has two zeroes and other zeroes are not real or are imaginary, then graph of the polynomial will intersect at two points on x-axis. 

(iv) True
Taking, 
β = 0, 
γ = 0 
f(x) = (x - α)(x - β)(x - γ)
= (x - α) x . x 
f(x) = x3 - ax2 
So, it has no linear (coefficient of x) and constant terms. 

(v) True 
α, β and γ are all negative for cubic polynomial ax3 + bx2 + cx + d. 
Here, d and a will have same sign. 
So, sign of b, c, d are same as of a. 
Signs of a, b, c, d will be same either positive or negative. 

(vi) False: As all zeroes of cubic polynomial are positive 
let f(x) = x3 +  ax2 - bx + c 
α + β + γ = positive, (say + x)

a and d have opposite signs. 
Therefore, we can conclude that, 
From (i) if a is positive, then b is negative. 
From (ii) if a is positive, then c is also positive. 
From (iii) if a is positive, then d is negative. 
So, if zeroes α, β, γ of cubic polynomial are positive then out of a, b, c at least one is negative. 

(vii) False

So, the values of k are ± 1/2 so that the given equation has equals roots.

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