Table of contents | |
Previous Year Questions 2024 | |
Previous Year Questions 2023 | |
Previous Year Questions 2022 | |
Previous Year Questions 2021 | |
Previous Year Questions 2020 | |
Previous Year Questions 2019 |
Q1: What should be added from the polynomial x2 – 5x + 4, so that 3 is the zero of the resulting polynomial? (2024)
(a) 1
(b) 2
(c) 4
(d) 5
Ans: (b)
Let, f(x) = x2 – 5x + 4
Let p should be added to f(x) then 3 becomes zero of polynomial.
So, f(3) + p = 0
⇒ 32 – 5 × 3 + 4 + p = 0
⇒ 9 + 4 – 15 + p = 0
⇒ – 2 + p = 0
⇒ p = 2 So, 2 should be added.
Q2: Find the zeroes of the quadratic polynomial x2 – 15 and verify the relationship between the zeroes and the coefficients of the polynomial. (2024)
Ans:
x2 – 15 = 0
x2 = 15
x = ± √15
Zeroes will be α = √15 , β = – √15
Verification: Givenpolynomial x2 – 15
On comparing above polynomial with
ax2 + bx + c, we have
a = 1, b = 0, c = –15
sum of zeros = α + β
Productofzeros = αβ
(15 ×− 15 ) = −115 = ac Hence,verified.
Hence, verified.
(a) 3
(b) 1
(c) 2
(d) 0
Ans: (b)
Sol: Here, y = p(x) touches the x-axis at one point
So, number of zeros is one.
Q4: If α, β are the zeroes of a polynomial p(x) = x2 + x - 1, then 1/α + 1/β equals to (2023)
(a) 1
(b) 2
(c) -1
(d) -1/2
Ans: (a)
Sol: We have, p(x) = x2 + x - 1, α + β = -1 and α . β = -1
Now, 1/α + 1/β = α + β/αβ = -1/-1 = 1
Q5: If α, β are the zeroes of a polynomial p(x) = x2 - 1, then the value of (α + β) is (2023)
(a) 1
(b) 2
(c) -1
(d) 0
Ans: (d)
Since, α, β are the zeroes of polynomial x2 - 1
∴ x2 + 0x - 1 = 0
∴ Sum of zeroes, (α + β) = 0
Q6: If α, β are the zeroes of a polynomial p(x) = 4x2 - 3x - 7, then (1/α + 1/β) is equal to (2023)
(a) 7/3
(b) -7/3
(c) 3/7
(d) -3/7
Ans: (d)
Since, α, β are the zeroes of polynomial p(x) = 4x2 - 3x - 7
∴ Sum of zeroes, (α + β) =3/4
and product of zeroes (αβ) = -7/4
Now,
= -3/7
Q7: If one of the zeroes of a quadratic polynomial ( k - 1 )x2 + kx + 1 is - 3 , then the value of k is (2022)
(a) 4/3
(b) -4/3
(c) 2/3
(d) -2/3
Ans: (a)
Sol: Given. -3 is a zero of quadratic polynomial (k - 1)2+ kx + 1.
∴ (k - 1) (-3)2 + k(-3) +1 = 0
⇒ 9k - 9 - 3k + 1 = 0 ⇒ 6k - 8 = 0
⇒ k = 8/6
⇒ k = 4/3
Q8: If the path traced by the car has zeroes at -1 and 2, then it is given by (2022)
(a) x2 + x + 2
(b) x2 - x + 2
(c) x2 - x - 2
(d) x2 + x - 2
Ans: (c)
Sol: The polynomial having zeroes α,β is k[x2 - (α + β)x + αβ], where k is real.
Here α = - 1 and β= 2
∴ Required polynomial = k[x2 - (-1 + 2)x + (-1) x (2)]
= [x2 - x - 2] (for k = 1)
Q9: The number of zeroes of the polynomial representing the whole curve, is (2022)
(a) 4
(b) 3
(c) 2
(d) 1
Ans: (a)
Sol: Given curve cuts the x-axis at four distinct points.
So, number of zeroes will be 4 .
Q10: The distance between C and G is (2022)
(a) 4 units
(b) 6 units
(c) 8 units
(d) 7 units
Ans: (b)
Sol: The distance between point C and G is 6 units.
Q11: The quadratic polynomial, the sum of whose zeroes is -5 and their product is 6. (2022)
(a) x2 + 5x + 6
(b) x2 - 5x + 6
(c) x2 - 5 x - 6
(d) - x2 + 5x + 6
Ans: (a)
Sol: Let α, β be the zeroes of required polynomial p(x).
Given, α + β=-5 and α.β=6
∴ p(x)=k[x2 - (-5)x + 6] = k[x2 + 5x + 6]
Thus, one of the polynomial which satisfy the given condition is x2+ 5x + 6
Q12: If one zero of the quadratic polynomial x2 + 3x + k is 2 then find the value of k. (2021)
View AnswerAns: Given, polynomial is f(x) =x2 + 3x + k
Since, 2 is zero of the polynomial f(x).
∴ f(2) = 0
⇒ f(2) =(2)2 + 3 x 2 + k
⇒ 4 + 6 + k = 0
⇒ k = -10
Ans: (a)
Sol: Since, the polynomial has two zeroes only. So. the degree of the polynomial is 2.
Q14: If one of the zeroes of the quadratic polynomial x2 + 3x + k is 2. then the value of k is (2020)
(a) 10
(b) - 10
(c) -7
(d) -2
Ans: (b)
Sol: Given, 2 is a zero of the polynomial
p(x) = x2 + 3x + k
∴ p (2) = 0
⇒ (2)2 + 3(2) + k = 0
⇒ 4 + 6 + k = 0 =
⇒ 10 + k = 0
⇒ k= -10
Q15: The quadratic polynomial, the sum of whose zeroes is -5 and their product is 6________ is (2020)
(a) x2 + 5x + 6
(b) x2 - 5x + 6
(c) x2- 5x - 6
(d) -x2 + 5x + 6
Ans: (a)
Sol: Let α, β be the zeroes of required polynomial p(x)
Given, α+ β = -5 and αβ = 6
p(x) = k[x2 - (- 5)x + 6]
= k[x2 + 5x + 6]
Thus, one of the polynomial which satisfy the given condition is x2 + 5x + 6.
Q16: Form a quadratic polynomial, the sum and product of whose zeroes are -3 and 2 respectively. (2020)
Ans: Let α, β be the zeroes of required polynomial Given, α + β = -3 and αβ = 2
∴ p(x) = k[x2= - (-3)x + 2] = k(x2 + 3x + 2)
For k = 1 , p (x) = x2 + 3x + 2
Hence, one of the polynomial which satisfy the given condition is x2 + 3x + 2.
Ans: 7
The given polynomial is x2 -(k + 6)x + 2(2k - 1)
According to the question
Sum of zeroes = 1/2(Product of Zeroes ):
⇒ k + 6 = 1/2 x 2 (2k - 1)
⇒ k + 6 = 2k - 1
⇒ k = 7
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1. What are polynomials, and what are their main components? |
2. How do you add or subtract polynomials? |
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