Class 10 Maths Previous Year Questions - Polynomials

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.

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.

Ans: (b)
Sol: Here, y = p(x) touches the x-axis at one point
So, number of zeros is one.

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

Ans: (d)
Since, α, β are the zeroes of polynomial x² - 1
∴ x² + 0x - 1 = 0
∴ Sum of zeroes, (α + β) = 0

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

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

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)

Ans: (a)
Sol: Given curve cuts the x-axis at four distinct points.
So, number of zeroes will be 4 .

Ans: (b)
Sol: The distance between point C and G is 6 units.

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

Ans: 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.

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

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.

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