RS Aggarwal Solutions Chapter 2 Polynomials MCQ Class 10 Maths
Chapter Name | RS Aggarwal Chapter 2 Polynomials |
Book Name | RS Aggarwal Mathematics for Class 10 |
Other Exercises |
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Related Study | NCERT Solutions for Class 10 Maths |
MCQ for Polynomials Class 10 Maths
1. Which of the following is a polynomial?
(a) x2 – 5x + 6√x + 3
(b) x3⁄2 – x + x1⁄2 + 1
(c) √x + 1/√x
(d) None of these
Solution
(d) none of these
A polynomial in x of degree n is an expression of the form p(x) = a0 + a1x + a2x2 + ...…+ anxn, where an ≠ 0.
2. Which of the following is not a polynomial?
(a) √3x2 - 2√3x + 5
(b) 9x2 – 4x + √2
(c) 3/2.x3 + 6x2 - 1√2.x – 8
(d) x + 3/x
Solution
(d) x + 3/x is not a polynomial.
It is because in the second term, the degree of x is –1 and an expression with a negative degree is not a polynomial.
3. The Zeroes of the polynomial x2– 2x – 3 are
(a) -3, 1
(b) -3, -1
(c) 3, -1
(d) 3, 1
Solution
(c) 3, –1
Let f(x) = x2 – 2x – 3= 0
= x2 – 3x + x – 3= 0
= x(x – 3) + 1(x – 3) = 0
= (x – 3) (x + 1) = 0
⇒ x = 3 or x = –1
4. The zeroes of the polynomial x2 - √2x – 12 are
(a) √2, -√2
(b) 3√2, -2√2
(c) -3√2, 2√2
(d) 3√2, 2√2
Solution
(b) 3√2, –2√2
Let f(x) = x2– √2x – 12 = 0
⇒ x2– 3√2x +2√2x – 12 = 0
⇒ x(x – 3√2) + 2√2(x – 3√2) = 0
⇒ (x – 3√2) (x + 2√2) = 0
⇒ x = 3√2 or x = –2√2
5. The zeroes of the polynomial 4x2 + 5√2x – 3 are
(a) -3√2, √2
(b) -3√2, √2/2
(c) (−3/√2), √2/4
(d) none of these
Solution
(c) –3√2, √2/4
Let f(x) = 4x2 + 5√2x – 3 = 0
⇒ 4x2 + 6√2x – √2x – 3 = 0
⇒ 2√2x(√2x + 3) –1 (√2x + 3) = 0
⇒ (√2x + 3) (2√2x – 1) = 0
⇒ x = –(3√2) or x = 1/(2√2)
⇒ x = –(3√2) or x = 1/(2√2) × √2/√2 = √2/4
6. The zeros of the polynomial x2 + 1/6.x – 2 are
(a) -3, 4
(b) −3/2, 4/3
(c) −4/3, 3/2
(d) none of these
Solution
(b) −3/2, 4/3
Let f(x) = x2 + 1/6.x – 2 = 0
⇒ 6x2 + x – 12 = 0
⇒ 6x2 + 9x – 8x – 12 = 0
⇒ 3x (2x + 3) – 4(2x + 3) = 0
⇒ (2x + 3) (3x – 4) = 0
∴ x = −3/2 or x = 4/3
7. The zeros of the polynomial 7x2 – 11/3.x – 2/3 are
(a) 2/7, −1/7
(b) 2/7, -1/3
(c) −2/3, 1/7
(d) none of these
Solution
(a) 2/3, −1/7
Let f(x) = 7x2 – 113x – 23 = 0
⇒ 21x2 – 11x – 2 = 0
⇒ 21x2 – 14x + 3x – 2 = 0
⇒ 7x (3x – 2) + 1(3x – 2) = 0
⇒ (3x – 2) (7x + 1) = 0
⇒ x = 2/3 or x = −1/7
8. The sum and product of the zeroes of a quadratic polynomial are 3 and -10 respectively. The quadratic polynomial is
(a) x2 – 3x + 10
(b) x2 + 3x – 10
(c) x2– 3x – 10
(d) x2 + 3x + 10
Solution
(c) x2 – 3x – 10
Given: Sum of zeroes, α + β = 3
Also, product of zeroes, αβ = –10
∴ Required polynomial = x2 – α + β = x2 – 3x – 10
9. A quadratic polynomial whose zeroes are 5 and -3, is
(a) x2 + 2x - 15
(b) x2 - 2x + 15
(c) x2 – 2x – 15
(d) none of these
Solution
(c) x2 – 2x – 15
Here, the zeroes are 5 and –3.
Let α = 5 and β = –3
So, sum of the zeroes, α + β = 5 + (–3) = 2
Also, product of the zeroes, αβ = 5 × (–3) = –15
The polynomial will be x2 – (α + β) x + αβ
∴ The required polynomial is x2 – 2x – 15.
10. A quadratic polynomial whose zeroes are 3/5 and −1/2, is
(a) 10x2 + x + 3
(b) 10x2 + x – 3
(c) 10x2 – x + 3
(d) x2 - 110x – 310
Solution
(d) x2 – 1/10.x – 3/10
Here, the zeroes are 3/5 and −1/2
Let α = 3/5 and β = −1/2
So, sum of the zeroes, α + β = 3/5 + (−1/2) = 1/10
Also, product of the zeroes, αβ = 3/5 × (−1/2) = −3/10
The polynomial will be x2 – (α + β)x + αβ
∴ The required polynomial is x2 – 1/10.x - 3/10.
11. The zeroes of the quadratic polynomial x2 + 88x + 125 are
(a) both positive
(b) both negative
(c) one positive and one negative
(d) both equal
Solution
(b) both negative
Let α and β be the zeroes of x2 + 88x + 125.
Then α + β = –88 and α × Î² = 125
This can only happen when both the zeroes are negative.
12. If α and β are the zeros of x2 + 5x + 8, then the value of (α + β) is
(a) 5
(b) -5
(c) 8
(d) -8
Solution
(b) –5
Given: α and β be the zeroes of x2 + 5x + 8.
If α + β is the sum of the roots and αβ is the product, then the required polynomial will be x2 – (α + β) x + αβ
∴ α + β = –5
13. If α and β are the zeroes of 2x2 + 5x - 9, then the value of αβ is
(a) −5/2
(b) 5/2
(c) −9/2
(d) 9/2
Solution
(c) −9/2
Given: α and β be the zeroes of 2x2 + 5x – 9.
If α + β are the zeroes, then x2 – (α + β)x + αβ is the required polynomial.
The polynomial will be x2 – 5/2.x – 9/2.
∴ αβ = −9/2.
14. If one zero of the quadratic polynomial kx2 + 3x + k is 2, then the value of k is
(a) 5/6
(b) −5/6
(c) 6/5
(d) −6/5
Answer
(d) -6/5
Since 2 is a zero of kx2 + 3x + k, we have:
k × (2)2 + 3(2) + k = 0
⇒ 4k + k + 6 = 0
⇒ 5k = -6
⇒ k = -6/5
15. If one zero of the quadratic polynomial (k – 1)x2 – kx + 1 is - 4, then the value of k is
(a) −5/4
(b) 5/4
(c) −4/3
(d) 4/3
Solution
(b) 5/4
Since –4 is a zero of (k – 1) x2 + kx + 1, we have:
(k – 1) × (-4)2 + k × (-4) + 1 = 0
⇒ 16k – 16 – 4k + 1 = 0
⇒ 12k – 15 = 0
⇒ k = 15/12 = 5/4
⇒ k = 5/4
16. If -2 and 3 are the zeroes of the quadratic polynomial x2 + (a + 1)x + b, then
(a) a = -2, b = 6
(b) a = 2, b = -6
(c) a = -2, b = -6
(d) a = 2, b = 6
Solution
(c) a = –2, b = –6
Given: –2 and 3 are the zeroes of x2 + (a + 1) x + b.
Now, (–2)2 + (a + 1) × (–2) + b = 0 ⇒ 4 – 2a – 2 + b = 0
⇒ b – 2a = –2 ….(1)
Also, 32 + (a + 1) × 3 + b = 0 ⇒ 9 + 3a + 3 + b = 0
⇒ b + 3a = –12 ….(2)
On subtracting (1) from (2), we get a = –2
∴ b = –2 – 4 = –6 [From (1)]
17. If one zero of 3x2 – 8x + k be the reciprocal of the other, then k = ?
(a) 3
(b) -3
(c) 1/3
(d) −1/3
Solution
(a) k = 3
Let α and 1/α be the zeroes of 3x2 – 8x + k.
Then the product of zeroes = k/3
⇒ α × 1/α = k/3
⇒ 1 = k/3
⇒ k = 3
18. If the sum of the zeroes of the quadratic polynomial kx2 + 2x + 3k is equal to the product of its zeroes, then k = ?
(a) 1/3
(b) −1/3
(c) 2/3
(d) −2/3
Solution
(d) -2/3
Let α and β be the zeroes of kx2 + 2x + 3k.
Then α + β = -2/k and αβ = 3
⇒ α + β = αβ
⇒ -2/k = 3
⇒ k = -2/3
19. If α, β are the zeroes of the polynomial x2 + 6x + 2, then (1/α + 1/β) = ?
(a) 3
(b) -3
(c) 12
(d) -12
Solution
(b) –3
Since α and β be the zeroes of x2 + 6x + 2, we have:
α + β = –6 and αβ = 2
∴ (1/α + 1/β) = (α + β)/αβ = -6/2 = -3
20. If α, β, γ be the zeroes of the polynomial x3 – 6x2 – x + 30, then (αβ + βγ + γα) = ?
(a) -1
(b) 1
(c) -5
(d) 30
Solution
(a) -1
It is given that α, β and γ are the zeroes of x3 – 6x2 – x + 30.
∴ (αβ + βγ + γα) = (co-efficient of x)/(co-efficient of x3) = -1/1 = -1
21. If α, β, γ be the zeroes of the polynomial 2x3 + x2 – 13x + 6, then αβγ = ?
(a) -3
(b) 3
(c) -1/2
(d) -13/2
Solution
(a) –3
Since, α, β and γ are the zeroes of 2x3 + x2 – 13x + 6, we have:
αβγ = (-constant term)/(coefficient of x3) = -6/2 = -3
22. If α, β, γ be the zeroes of the polynomial p(x) such that (α + β + γ) = 3, (αβ + βγ + γα) = –10 and αβγ = –24, then p(x) = ?
(a) x3 + 3x2 – 10x + 24
(b) x3 + 3x2 + 10x – 24
(c) x3 – 3x2 – 10x + 24
(d) none of these
Solution
(c) x3 – 3x2 – 10x + 24
Given: α, β and γ are the zeroes of polynomial p(x).
Also, (α + β + γ) = 3, (αβ + βγ + γα) ) = –10 and αβγ = –24
∴ p(x) = x3– (α + β + γ)x2 + (αβ + βγ + γα)x – αβγ
= x3 – 3x2 – 10x + 24
23. If two of the zeroes of the cubic polynomial ax3 + bx2 + cx + d are 0, then the third zero is
(a) -b/a
(b) b/a
(c) c/a
(d) -d/a
Solution
(a) -b/a
Let α, 0 and 0 be the zeroes of ax3 + bx2 + cx + d = 0
Then the sum of zeroes = -b/a
⇒ α + 0 + 0 = -b/a
⇒ α = -b/a
Hence, the third zero is -b/a.
24. If one of the zeroes of the cubic polynomial ax3 + bx2 + cx + d is 0, then the product of the other two zeroes is
(a) -c/a
(b) c/a
(c) 0
(d) -b/a
Solution
(b) c/a
Let α, β and 0 be the zeroes of ax3 + bx2 + cx + d.
Then, sum of the products of zeroes taking two at a time is given by
(αβ + β × 0 + α × 0) = c/α
⇒ αβ = c/α
∴ The product of the other two zeroes is c/α.
25. If one of the zeroes of the cubic polynomial x3 + ax2 + bx + c is -1, then the product of the other two zeroes is
(a) a – b – 1
(b) b – a – 1
(c) 1 – a + b
(d) 1 + a – b
Solution
(c) 1 – a + b
Since –1 is a zero of x3 + ax2 + bx + c, we have:
(–1)3 + a × (–1)2 + b × (–1) + c = 0
⇒ a – b + c + 1 = 0
⇒ c = 1 – a + b
Also, product of all zeroes is given by
αβ × (–1) = – c
⇒ αβ = c
⇒ αβ = 1 – a + b
26. If α, β be the zeroes of the polynomial 2x2 + 5x + k such that (α + β)2 – αβ = 21/4, then k = ?
(a) 3
(b) -3
(c) -2
(d) 2
Solution
(d) 2
Since α and β are the zeroes of 2x2 + 5x + k, we have:
α + β = -5/2 and αβ = k/2
Also, it is given that α2 + β2 + αβ = 21/4.
⇒ (α + β)2 – αβ = 21/4
⇒ (−5/2)2 - k/2 = 21/4
⇒ 25/4 - k/2 = 21/4
⇒ k/2 = 25/4 - 21/4 = 4/4 = 1
⇒ k = 2
27. On dividing a polynomial p(x) by a non-zero polynomial q(x), let g(x) be the quotient and r(x) be the remainder, then p(x) = q(x). g(x) + r(x), where
(a) r(x) = 0 always
(b) deg r(x) ˂ deg g(x) always
(c) either r(x) = 0 or deg r(x) ˂ deg g(x)
(d) r(x) = g(x)
Solution
(c) either r(x) = 0 or deg r(x) ˂ deg g(x)
By division algorithm on polynomials, either r(x) = 0 or deg r(x) ˂ deg g(x).
28. Which of the following is a true statement?
(a) x2 + 5x – 3 is a linear polynomial.
(b) x2 + 4x – 1 is a binomial
(c) x + 1 is a monomial
(d) 5x2is a monomial
Solution
(d) 5x2 is a monomial.
5x2 consists of one term only. So, it is a monomial.
Exercise – Assesment
1. The zeroes of the polynomial P(x) = x2 – 2x – 3 are
(a) -3, 1
(b) -3, -1
(c) 3, -1
(d) 3, 1
Solution
(c) 3, -1
Here, p(x) = x2 – 2x – 3
Let x2 – 2x – 3 = 0
⇒ x2– (3 – 1)x – 3 = 0
⇒ x2– 3x + x – 3 = 0
⇒ x(x – 3) + 1(x – 3) = 0
⇒ (x – 3) (x + 1) = 0
⇒ x = 3, –1
2. If α, β, γ be the zeroes of the polynomial x3 – 6x2 – x + 3, then the values of (αβ + βγ + γα) = ?
(a) -1
(b) 1
(c) -5
(d) 3
Solution
(a) -1
Here, p(x) = x3 – 6x2 – x + 3
Comparing the given polynomial with x3 – (α + β + γ)x2 + ((αβ + βγ + γα)x – αβγ, we get: (αβ + βγ + γα) = -1
3. If α, β are the zeros of kx2 – 2x + 3k is equal α + β = αβ then k = ?
(a) 1/3
(b) -1/3
(c) 2/3
(d) -2/3
Solution
(c) 2/3
Here, p(x) = x2 – 2x + 3k
Comparing the given polynomial with ax2 + bx + c, we get:
a = 1, b = – 2 and c = 3k
It is given that α and β are the roots of the polynomial.
∴ α + β = -b/a
⇒ α + β = – (-2)/1
⇒ α + β = 2 ….(i)
Also, αβ = c/α
⇒ αβ = 3k/1
⇒ αβ = 3k ….(ii)
Now, α + β = αβ
⇒ 2 = 3k [Using (i) and (ii)]
⇒ k = 2/3
4. It is given that the difference between the zeroes of 4x2 – 8kx + 9 is 4 and k > 0. Then, k = ?
(a) 1/2
(b) 3/2
(c) 5/2
(d) 7/2
Solution
(c) 5/2
Let the zeroes of the polynomial be α and α + 4
Here, p(x) = 4x2 – 8kx + 9
Comparing the given polynomial with ax2 + bx + c, we get:
a = 4, b = -8k and c = 9
Now, sum of the roots = −b/a
⇒ α + α + 4 = -(-8)/4
⇒ 2α + 4 = 2k
⇒ α + 2 = k
⇒ α = (k – 2) ….(i)
Also, product of the roots, αβ = c/α
⇒ α(α + 4) = 9/4
⇒ (k – 2) (k – 2 + 4) = 9/4
⇒ (k – 2) (k + 2) = 9/4
⇒ k2 – 4 = 9/4
⇒ 4k2 – 16 = 9
⇒ 4k2 = 25
⇒ k2 = 25/4
⇒ k = 5/2 (∵ k > 0)
5. Find the zeroes of the polynomial x2 + 2x – 195.
Solution
Here, p(x) = x2 + 2x – 195
Let p(x) = 0
⇒ x2 + (15 – 13)x – 195 = 0
⇒ x2 + 15x – 13x – 195 = 0
⇒ x (x + 15) – 13(x + 15) = 0
⇒ (x + 15) (x – 13) = 0
⇒ x = –15, 13
Hence, the zeroes are –15 and 13.
6. If one zero of the polynomial (a2 + 9)x2 – 13x + 6a is the reciprocal of the other, find the value of a.
Solution
(a + 9)x2 – 13x + 6a = 0
Here, A = (a2 + 9), B = 13 and C = 6a
Let α and 1/α be the two zeroes.
Then, product of the zeroes = C/A
⇒ α.1/α = 6α/(α2 + 9)
⇒ 1 = 6α/(α2 + 9)
⇒ a2 + 9 = 6a
⇒ a2 – 6a + 9 = 0
⇒ a2 – 2 × a × 3 + 32 = 0
⇒ (a – 3)2 = 0
⇒ a – 3 = 0
⇒ a = 3
7. Find a quadratic polynomial whose zeroes are 2 and -5.
Solution
It is given that the two roots of the polynomial are 2 and -5.
Let α = 2 and β = -5
Now, the sum of the zeroes, α + β = 2 + (-5) = -3
Product of the zeroes, αβ = 2 × (-5) = -15
∴ Required polynomial = x2 – (α + β)x + αβ
= x2 – (-3)x + 10
= x2 + 3x – 10
8. If the zeroes of the polynomial x3– 3x2 + x + 1 are (a – b), a and (a + b), find the values of a and b.
Solution
The given polynomial = x3 – 3x2 + x + 1 and its roots are (a – b), a and (a + b).
Comparing the given polynomial with Ax3 + Bx2 + Cx + D, we have:
A = 1, B = -3, C = 1 and D = 1
Now, (a – b) + a + (a + b) = -B/A
⇒ 3 a = -3/1
⇒ a = 1
Also, (a – b) × a × (a + b) = -D/A
⇒ a (a2 – b2) = -1/1
⇒ 1 (12 – b2) = -1
⇒ 1 – b2 = -1
⇒ b2 = 2
⇒ b = ±√2
∴ a = 1 and b = ±√2
9. Verify that 2 is a zero of the polynomial x3 + 4x2 – 3x – 18.
Solution
Let p(x) = x3 + 4x2 – 3x – 18
Now, p(2) = 23 + 4 × 22 – 3 × 2 – 18 = 0
∴ 2 is a zero of p(x).
10. Find the quadratic polynomial, the sum of whose zeroes is -5 and their product is 6.
Solution
Given:
Sum of the zeroes = -5
Product of the zeroes = 6
∴ Required polynomial = x2 – (sum of the zeroes) x + product of the zeroes
= x2 – (-5) x + 6
= x2 + 5x + 6
11. Find a cubic polynomial whose zeroes are 3, 5 and -2.
Solution
Let α, β and γ are the zeroes of the required polynomial.
Then we have:
α + β + γ = 3 + 5 + (-2) = 6
αβ + βγ + γα = 3×5 + 5×(-2) + (-2)×3 = -1
and αβγ = 3 × 5 × -2 = -30
Now, p(x) = x3– x2 (α + β + γ) + x (αβ + βγ + γα) – αβγ
= x3 – x2 × 6 + x × (-1) – (-30)
= x3 – 6x2 – x + 30
So, the required polynomial is p(x) = x3 – 6x2 – x + 30.
12. Using remainder theorem, find the remainder when p(x) = x3 + 3x2 – 5x + 4 is divided by (x – 2).
Solution
Given:
p(x) = x3 + 3x2 – 5x + 4
Now, p(2) = 23 + 3(22) – 5(2) + 4
= 8 + 12 – 10 + 4
= 14
13. Show that (x + 2) is a factor of f(x) = x3 + 4x2 + x – 6.
Solution
Given: f(x) = x3 + 4x2 + x – 6
Now, f(-2) = (-2)3 + 4(-2)2 + (-2) - 6
= - 8 + 16 – 2 – 6
= 0
∴ (x + 2) is a factor of f(x) = x3 + 4x2 + x – 6.
14. If α, β, γ are the zeroes of the polynomial p(x) = 6x3 + 3x2 – 5x + 1, find the value of (1/α + 1/β + 1/γ).
Solution
Given: p(x) = 6x3 + 3x2 – 5x + 1
= 6x3 – (–3) x2 + (–5) x – 1
Comparing the polynomial with x3 – x2 (α + β + γ) + x(αβ + βγ + γα) – αβγ, we get: αβ + βγ + γα = –5
and αβγ = – 1
∴ (1/α + 1/β + 1/γ)
= (βγ + αγ + αβ)/αβγ
= (-5/-1)
= 5
15. If α, β are the zeroes of the polynomial f(x) = x2 – 5x + k such that α – β = 1, find the value of k.
Solution
Given: x2 – 5x + k
The co-efficients are a = 1, b = -5 and c = k.
∴ α + β = -b/a
⇒ α + β = (-5)/1
⇒ α + β = 5 ...(1)
Also, α – β = 1 ...(2)
From (1) and (2), we get:
2α = 6
⇒ α = 3
Putting the value of α in (1), we get β = 2.
Now, αβ = c/α
⇒ 3 × 2 = k/1
∴ k = 6
16. Show that the polynomial f(x) = x2 + 4x + 6 has no zero.
Solution
Let t = x2
So, f(t) = t2 + 4t + 6
Now, to find the zeroes, we will equate f(t) = 0
⇒ t2 + 4t + 6 = 0
The zeroes of a polynomial should be real numbers.
∴ The given f(x) has no zeroes.
17. If one zero of the polynomial p(x) = x3 – 6x2 + 11x – 6 is 3, find the other two zeroes.
Solution
p(x) = x3 – 6x2 + 11x – 6 and its factor, x + 3
Let us divide p(x) by (x – 3).
Here, x3 – 6x2 + 11x – 6 = (x – 3) (x2 – 3x + 2)
= (x – 3) [(x2 – (2 + 1)x + 2]
= (x – 3) (x2 – 2x – x + 2)
= (x – 3) [x (x – 2) – 1(x – 2)]
= (x – 3) (x – 1) (x – 2)
∴ The other two zeroes are 1 and 2.
18. If two zeroes of the polynomial p(x) = 2x4 – 3x3 – 3x2 + 6x – 2 are √2 and – √2, find its other two zeroes.
Solution
Given: p(x) = 2x4 – 3x3 – 3x2 + 6x – 2 and the two zeroes, √2 and – √2
So, the polynomial is (x + √2) (x – √2) = x2 – 2.
Let us divide p(x) by (x2 – 2)
Here, 2x4 – 3x3 – 3x2 + 6x – 2 = (x2 – 2) (2x2 – 3x + 1)
= (x2 – 2) [(2x2 – (2 + 1) x + 1]
= (x2 – 2) (2x2 – 2x – x + 1)
= (x2 – 2) [(2x (x – 1) –1(x – 1)]
= (x2 – 2) (2x – 1) (x – 1)
The other two zeroes are 1/2 and 1.
19. Find the quotient when p(x) = 3x4 + 5x3– 7x2 + 2x + 2 is divided by (x2 + 3x + 1)
Solution
Given: p(x) = 3x4 + 5x3 – 7x2 + 2x + 2
Dividing p(x) by (x2 + 3x + 1), we have:
∴ The quotient is 3x2 – 4x + 2
20. Use remainder theorem to find the value of k, it being given that when x3 + 2x2 + kx + 3 is divided by (x – 3), then the remainder is 21.
Solution
Let p(x) = x3 + 2x2 + kx + 3
Now, p(3) = (3)3 + 2(3)2 + 3k + 3
= 27 + 18 + 3k + 3
= 48 + 3k
It is given that the reminder is 21
∴ 3k + 48 = 21
⇒3k = –27
⇒ k = –9