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In this section, we delve into Chapter 2 of the Class 10 RD Sharma textbook, focusing on Polynomials. Exercise 2.1 is designed to help students understand the fundamental concepts of polynomials, including their types, properties, and various operations performed on them.
This section provides comprehensive solutions for Exercise 2.1 from Chapter 2 of the Class 10 RD Sharma textbook. These solutions aim to assist students in grasping the core principles of polynomials, ensuring they build a solid mathematical foundation for more advanced topics.
Solution:
Given that,
α and β are the zeroes of the quadratic polynomial f(x) = 6x2 + x – 2.
therefore,
Sum of the zeroes = α + β = -1/6,
Product of the zeroes =α × β = -1/3.
Now,
(α/β) +(β/α) = (α2 + β2) - 2αβ / αβ
Now substitute the values of the sum of zeroes and products of the zeroes and we will get,
= -25/12
Hence the value of (α/β) +(β/α) is -25/12.
Solution:
Given that,
α and β are the zeroes of the quadratic polynomial f(x) = 6x2 + x – 2.
therefore,
Sum of the zeroes = α + β = 6/3
Product of the zeroes = α × β = 4/3
Now,
α/β + 2(1/α + 1/β) + 3αβ = [(α2 + β2) / αβ] + 2(1/α + 1/β) + 3αβ
[ ((α + β)2 - 2αβ) / αβ] + 2(1/α + 1/β) + 3αβ
Now substitute the values of the sum of zeroes and products of the zeroes and we will get,
α/β + 2(1/α + 1/β) + 3αβ = 8
Hence the value of α/β + 2(1/α + 1/β) + 3αβ is 8.
Solution:
Let as assume that the two zeroes of the polynomial are α and β.
Given that,
f(x) = x2 + px + 45
Now,
Sum of the zeroes = α + β = – p
Product of the zeroes = α × β = 45
therefore,
(α + β)2 - 4αβ = (-p)2 - 4 x 45 = 144
(-p)2 = 144 + 180 = 324
p = √324
Hence the value of p will be either 18 or -18.
Solution:
Given that,
α and β are the roots of the quadratic polynomial.
f(x) = x2 – px + q
Now,
Sum of the zeroes = p = α + β
Product of the zeroes = q = α × β
therefore,
LHS = [(α2 / β2) + (β2 / α2)]
= [(α^4 + β4) / α2.β2]
= [((α+ β)^2 - 2αβ)2 + 2(αβ)2] / (αβ)2
= [((p)2 - 2q)2 + 2(q)2] / (q)2
= [(p4 + 4q2 - 4pq2) - 2q2] / q2
= (p4 + 2q2 - 4pq2) / q2 = (p/q)2 - (4p2/q) + 2
LHS = RHS
Hence, proved.
Solution:
Given that,
α and β are the zeroes of the quadratic polynomial
f(x) = x2 – p(x + 1)– c
Now,
Sum of the zeroes = α + β = p
Product of the zeroes = α × β = (- p – c)
therefore,
(α + 1)(β + 1)
= αβ + α + β + 1
= αβ + (α + β) + 1
= (− p – c) + p + 1
= 1 – c = RHS
therefore, LHS = RHS
Hence proved.
Solution:
Given that,
α + β = 24 ------(i)
α – β = 8 ------(ii)
By solving the above two equations, we will get
2α = 32
α = 16
put the value of α in any of the equation.
Let we substitute it in (ii) and we will get,
β = 16 – 8
β = 8
Now,
Sum of the zeroes of the new polynomial = α + β = 16 + 8 = 24
Product of the zeroes = αβ = 16 × 8 = 128
Then, The quadratic polynomial = x2– (sum of the zeroes)x + (product of the zeroes) = x2 – 24x + 128
Hence, the required quadratic polynomial is f(x) = x2 + 24x + 128
Solution:
Given that,
f(x) = x2 – 1
Sum of the zeroes = α + β = 0
Product of the zeroes = αβ = – 1
therefore,
Sum of the zeroes of the new polynomial
= [(2α2 + 2β2)] / αβ
= [2(α2 + β2)] / αβ
= [2((α + β)2 - 2αβ)] / αβ = 4/(-1)
After substituting the value of the sum and products of the zeroes we will get,
As given in the question,
Product of the zeroes
= (2α)(2β) / αβ = 4
Hence, the quadratic polynomial is
x2 – (sum of the zeroes)x + (product of the zeroes)
= kx2 – (−4)x + 4x2 –(−4)x + 4
Hence, the required quadratic polynomial is f(x) = x2 + 4x + 4
Solution:
Given that,
f(x) = x2 – 3x – 2
Sum of the zeroes = α + β = 3
Product of the zeroes = αβ = – 2
therefore,
Sum of the zeroes of the new polynomial
= 1/(2α + β) + 1/(2β + α)
= (2α + β + 2β + α) / (2α + β)(2β + α)
= (3α + 3β) / (2(α2 + β2) + 5αβ)
= (3 x 3) / 2[2(α + β)2 - 2αβ + 5 x (-2)]
= 9 / 2[9-(-4)]-10 = 9/16
Product of zeroes = 1/(2α + β) x 1/(2β + α)
= 1 / (4αβ + 2α2 + 2β2 + αβ)
= 1 / [5αβ + 2((α + β)2 - 2αβ)]
= 1 / [5 x (-2) + 2((3)2 - 2 x (-2))] = 1/16
therefore, the quadratic polynomial is,
x2- (sum of the zeroes)x + (product of the zeroes)
= (x2 + (9/16)x +(1/16))
Hence, the required quadratic polynomial is (x2 + (9/16)x +(1/16)).
Solution:
Given that,
f(x) = x2 + px + q
Sum of the zeroes = α + β = -p
Product of the zeroes = αβ = q
therefore,
Sum of the zeroes of new polynomial = (α + β)2 + (α – β)2
= (α + β)2 + α2 + β2 – 2αβ
= (α + β)2 + (α + β)2 – 2αβ – 2αβ
= (- p)2 + (- p)2 – 2 × q – 2 × q
= p2 + p2 – 4q = p2 – 4q
Product of the zeroes of new polynomial = (α + β)2 x (α – β)2
= (- p)2((- p)2 - 4q)
= p2 (p2–4q)
therefore, the quadratic polynomial is,
x2 – (sum of the zeroes)x + (product of the zeroes)
= x2 – (2p2 – 4q)x + p2(p2 – 4q)
Hence, the required quadratic polynomial is f(x) = k(x2 – (2p2 –4q) x + p2(p2 - 4q)).
(i) α + 2, β + 2
(ii) [α-1] / [α+1], [β-1] / [β+1]
Solution:
Given that,
f(x) = x2 – 2x + 3
Sum of the zeroes = α + β = 2
Product of the zeroes = αβ = 3
(i) Sum of the zeroes of new polynomial = (α + 2) + (β + 2)
= α + β + 4 = 2 + 4 = 6
Product of the zeroes of new polynomial = (α + 1)(β + 1)
= αβ + 2α + 2β + 4
= αβ + 2(α + β) + 4 = 3 + 2(2) + 4 = 11
therefore, quadratic polynomial is :
x2 – (sum of the zeroes)x + (product of the zeroes)
= x2 – 6x +11
Hence, the required quadratic polynomial is f(x) = k(x2 – 6x + 11).
(ii) Sum of the zeroes of new polynomial :
= [(α-1)/(α+1)] + [(β-1)/(β+1)]
= [(α-1)(β+1) + (β-1)(α+1)] / (α+1)(β+1)
= [αβ + α - β - 1 + αβ - α + β - 1)] / (α+1)(β+1)
= (3-1+3-1) / (3+1+2) = 2/3
Product of the zeroes of new polynomial :
= [(α-1)/(α+1)] + [(β-1)/(β+1)]
= 26 = 13(2/6) = 1/3
therefore, the quadratic polynomial is,
x2 – (sum of the zeroes)x + (product of the zeroes)
= x2 - (2/3)x + (1/3)
Hence, the required quadratic polynomial is f(x) = k(x2 – (2/3)x + (1/3))
(i) α – β
(ii) 1/α - 1/β
(iii) 1/α + 1/β - 2αβ
(iv) α2β + αβ2
(v) α4 + β4
(vi) 1/(aα + b) + 1/(aβ + b)
(vii) β/(aα + b) + α/(aβ + b)
(viii) [(α2/β) + (β2/α)] + b[α/a + β/a]
Solution:
Given that,
f(x) = ax2 + bx + c
Sum of the zeroes of polynomial = α + β = -b/a
Product of zeroes of polynomial = αβ = c/a
Since, α + β are the zeroes of the given polynomial therefore,
(i) α – β
The two zeroes of the polynomials are :
= [√-b+b2-4ac]/2a - ([-b+√(b2-4ac)]/2a)
= [-b+√(b2-4ac) + b+√(b2-4ac)] / 2a
= √(b2-4ac) / a
(ii) 1/α - 1/β
= (β-1) / αβ = -(α-β)/αβ -------(1)
From above question as we know that,
α-β = √(b2-4ac) / a
and,
αβ = c/a
Put the values in (i) and we will get,
= -[(√(b2-4ac))/c]
(iii) (1/α) + (1/β) - 2αβ
= (α+β)/αβ - 2αβ ---------- (i)
Since,
Sum of the zeroes of polynomial = α + β = – b/a
Product of zeroes of polynomial = αβ = c/a
After putting it in (i), we will get
= (-b/a x a/c - 2c/a) = -[b/c + 2c/a]
(iv) α2β + αβ2
= αβ(α + β) --------(i)
Since,
Sum of the zeroes of polynomial = α + β = – b/ a
Product of zeroes of polynomial = αβ = c/a
After putting it in (i), we will get
= c/a(-b/a) = -bc/a^2
(v) α4 + β4
= (α2 + β2)2 – 2α2β2
= ((α + β)2 – 2αβ)2 – (2αβ)2 ---------(i)
Since,
Sum of the zeroes of polynomial = α + β = – b/a
Product of zeroes of polynomial = αβ = c/a
After substituting it in (i), we will get
= [(-b/a) -2(c/a)]2 - [2(c/a)2]
= [(b2 -2(ac)) / a2]2 - [2(c/a)2]
= [(b2 - 2ac)2 - 2a2 c2] / a4
(vi) 1/(aα + b) + 1/(aβ + b)
= (aβ + b + aα + b) / (aα + b)(aβ + b)
= (a(α + β) + 2b) / (a2 x αβ + abα + abβ + b2)
Since,
Sum of the zeroes of polynomial = α + β = – b/a
Product of zeroes of polynomial = αβ = c/a
After putting it, we will get
= b / (ac - b2 + b2) = b/ac
(vii) β/(aα + b) + α/(aβ + b)
= [β(aβ + b) + α(aα + b)] / (aβ + b)(aα + b)
= [aα2 + bβ2 + bα + bβ] / (a2 x (c/a) + ab(α+β) + b2)
Since,
Sum of the zeroes of polynomial = α + β = – b/a
Product of zeroes of polynomial = αβ = c/a
After putting it, we will get
= a[(α+β)2 - b(α+β)] / ac
= a[b2/a - 2c/a] - b2/a
= a[(b2 - 2c - b2)/a] / ac
= (b2 - 2c - b2) / ac = -2/a
(viii) [(α2/β) + (β2/α)] + b[α/a + β/a]
= a[(α2 + β2) / αβ] + b[(α2+β2)/αβ]
= a[(α+β)2 - 2αβ] + b((α+β)2 - 2αβ) / αβ
Since,
Sum of the zeroes of polynomial= α + β = – b/a
Product of zeroes of polynomial= αβ = c/a
After putting it, we will get
= a[(-ba)2 - 3x(c/a)] + b((-b/a)2 - 2(c/a)) / (c/a)
= [(-b2a2/a2c)+(3bca2/a2)+(b/a)2 - (2bca2/a2c)] = b
Exercise 2.1 | Set 2 of RD Sharma's Class 10 Mathematics textbook focuses on the fundamental concepts of polynomials. This section covers topics such as identifying polynomials, determining their degrees, finding zeros, and evaluating polynomials for given values of variables. Students learn to classify polynomials based on their degrees and number of terms, understand the relationship between factors and zeros, and solve problems involving the behavior of polynomial functions. The exercise also includes questions on finding the values of unknown coefficients in polynomials satisfying certain conditions.
