In this chapter, we provide RD Sharma Solutions for Class 10 Chapter 2 Polynomials Exercise 2.2 for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Solutions for Class 10 Chapter 2 Polynomials Exercise 2.1 pdf, free RD Sharma Solutions for Class 10 Chapter 2 Polynomials Exercise 2.2 book pdf download. Now you will get step by step solution to each question.
| Textbook | NCERT |
| Class | Class 10 |
| Subject | Maths |
| Chapter | Chapter 2 |
| Chapter Name | Polynomials |
| Exercise | 2.2 |
| Category | RD Sharma Solutions |
RD Sharma Solutions for Class 10 Chapter 2 Polynomials Ex 2.2 Download PDF
Chapter 2: Polynomials Exercise – 2.2
Question: 1
Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also, verify the relationship between the zeroes and coefficients in each of the following cases:
(i) f(x) = 2x3+ x2– 5x + 2; 1/2, 1, – 2
(ii) g(x) = x3– 4x2 + 5x – 2; 2, 1, 1
Solution:
(i) f(x) = 2x3 + x2 – 5x + 2; 1/2, 1, – 2
(a) By putting x = 1/2 in the above equation, we will get
(b) By putting x = 1 in the above equation, we will get
f(1) = 2(1)3 + (1)2 – 5(1) + 2
= 2 + 1 – 5 + 2 = 0
(c) By putting x = -2 in the above equation, we will get
f(−2) = 2(−2)3 + (−2)2 – 5(−2) + 2
= -16 + 4 + 10 + 2 = – 16 + 16 = 0
Now,
Sum of zeroes = α + β + γ = – b/a
Product of the zeroes = αβ + βγ + αγ = c/a
Hence, verified.
(ii) g(x) = x3 – 4x2 + 5x – 2; 2, 1, 1
(a) By putting x = 2 in the given equation, we will get
g(2) = (2)3 – 4(2)2 + 5(2) – 2
= 8 – 16 + 10 – 2 = 18 – 18 = 0
(b) By putting x = 1 in the given equation, we will get
g(1) = (1)3 – 4(1)2 + 5(1) – 2
= 1 – 4 + 5 – 2
= 0
Now,
Sum of zeroes= α + β + γ =-b/a
⇒ 2 + 1 + 1 = −(−4)
4 = 4
Product of the zeroes = αβ + βγ + αγ = c/a
2 × 1 + 1 × 1 + 1 × 2 = 5
2 + 1 + 2 = 5
5 = 5
αβγ = –(−2)
2 × 1 × 1 = 2
2 = 2
Hence, verified.
Question: 2
Find a cubic polynomial with the sum, sum of the product of its zeroes is taken two at a time, and product of its zeroes as 3, – 1 and – 3 respectively.
Solution:
Any cubic polynomial is of the form ax3 + bx2 + cx + d:
= x3 – (sum of the zeroes) x2 + (sum of the products of its zeroes) x – (product of the zeroes)
= x3 – 3x2 + (−1)x + (−3)
= k[x3 – 3x2 – x – 3]
k is any non-zero real numbers.
Question: 3
If the zeroes of the polynomial f(x) = 2x3 – 15x2 + 37x – 30, find them.
Solution:
Let, α = a – d, β = a and γ = a + d be the zeroes of the polynomial.
f(x) = 2x3 – 15x2 + 37x – 30
And, a (a2 + d2) = 15
Question: 4
Find the condition that the zeroes of the polynomial f(x) = x3 + 3px2 + 3qx + r may be in A.P.
Solution:
f(x) = x3 + 3px2 + 3qx + r
Let, a – d, a, a + d be the zeroes of the polynomial.
Then,
The sum of zeroes = – b/a
a + a – d + a + d = -3p 3a = -3p a = -p Since, a is the zero of the polynomial f(x),
Therefore, f(a) = 0
f(a) = a3 + 3pa2 + 3qa + r = 0
Therefore, f(a) = 0f(a) = 0
⇒ a3 + 3pa2 + 3qa + r = 0
= ⇒ (−p)3 + 3p(−p)2 + 3q(−p) + r = 0
= − p3 + 3p3 – pq + r = 0
= 2p3 – pq + r = 0
Question: 5
If zeroes of the polynomial f(x) = ax3 + 3bx2 + 3cx + d are tin A.P., prove that 2b3 – 3abc + a2d = 0.
Solution:
f(x) = x3 + 3px2 + 3qx + r
Let, a – d, a, a + d be the zeroes of the polynomial.
Then,
The sum of zeroes = – b/a
a + a – d + a + d = – 3b/a
Since, f(a) = 0
⇒ a(a2) + 3b(a)2 + 3c(a) + d = 0
⇒ a(a2) + 3b(a)2 + 3c(a) + d = 0
Question: 6
If the zeroes of the polynomial f(x) = x3 – 12x2 + 39x + k are in A.P., find the value of k.
Solution:
f(x) = x3 – 12x2 + 39x + k
Let, a-d, a, a + d be the zeroes of the polynomial f(x).
The sum of the zeroes = 12
3a = 12
a = 4
Now,
f(a) = 0
f(a) = a3 – 12a2 + 39a + k f(4) = 43 – 12(4)2 + 39(4) + k = 0
f(4) = 43 –12(4)2 + 39(4) + k = 0
64 – 192 + 156 + k = 0
k = – 28
All Chapter RD Sharma Solutions For Class10 Maths
I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment on the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.