Polynomials
Previous Year Solved Questions 2024
1. If one of the zeroes of the quadratic polynomial (a-1)x² + ax + 1 is -3, then the value of a is: [CBSE 2024]
(a) -2/3 (b) 2/3 (c) 4/3 (d) 3/4
(a) -2/3 (b) 2/3 (c) 4/3 (d) 3/4
View Answer
Ans: (c) 4/3
Since -3 is a zero:
(a-1)(-3)² + a(-3) + 1 = 0
(a-1)(9) – 3a + 1 = 0
9a – 9 – 3a + 1 = 0
6a – 8 = 0
6a = 8
a = 8/6 = 4/3
Since -3 is a zero:
(a-1)(-3)² + a(-3) + 1 = 0
(a-1)(9) – 3a + 1 = 0
9a – 9 – 3a + 1 = 0
6a – 8 = 0
6a = 8
a = 8/6 = 4/3
2. For what value of k, the product of zeroes of the polynomial kx² – 4x – 7 is 2? [CBSE 2024]
(a) -1/14 (b) -7/2 (c) 7/2 (d) -2/7
(a) -1/14 (b) -7/2 (c) 7/2 (d) -2/7
View Answer
Ans: (b) -7/2
Product of zeroes = constant term/coefficient of x²
= (-7)/k
Given: (-7)/k = 2
-7 = 2k
k = -7/2
Product of zeroes = constant term/coefficient of x²
= (-7)/k
Given: (-7)/k = 2
-7 = 2k
k = -7/2
3. Assertion (A): Zeroes of a polynomial p(x) = x² – 2x – 3 are -1 and 3.
Reason (R): The graph of polynomial p(x) = x² – 2x – 3 intersects x-axis at (-1, 0) and (3, 0).
Select the correct option: [CBSE 2024]
(a) Both A and R are true. R explains A completely.
(b) Both A and R are true. R does not explain A.
(c) A is true but R is false.
(d) A is false but R is true.
Reason (R): The graph of polynomial p(x) = x² – 2x – 3 intersects x-axis at (-1, 0) and (3, 0).
Select the correct option: [CBSE 2024]
(a) Both A and R are true. R explains A completely.
(b) Both A and R are true. R does not explain A.
(c) A is true but R is false.
(d) A is false but R is true.
View Answer
Ans: (a) Both A and R are true. R explains A completely.
x² – 2x – 3 = 0
(x+1)(x-3) = 0
Zeroes: x = -1, 3 ✓
Graph intersects x-axis at zeroes: (-1,0) and (3,0) ✓
R correctly explains that zeroes are x-intercepts.
x² – 2x – 3 = 0
(x+1)(x-3) = 0
Zeroes: x = -1, 3 ✓
Graph intersects x-axis at zeroes: (-1,0) and (3,0) ✓
R correctly explains that zeroes are x-intercepts.
4. The zeroes of a polynomial x² + px + q are twice the zeroes of the polynomial 4x² – 5x – 6. The value of p is: [CBSE 2024]
(a) -5/2 (b) 5/2 (c) -5 (d) 10
(a) -5/2 (b) 5/2 (c) -5 (d) 10
View Answer
Ans: (a) -5/2
Let zeroes of 4x² – 5x – 6 be α, β
Sum = α+β = -(-5)/4 = 5/4
Zeroes of x² + px + q are 2α, 2β
Sum = 2α+2β = 2(α+β) = 2×(5/4) = 5/2
But sum for x² + px + q = -p
So -p = 5/2 ⇒ p = -5/2
Let zeroes of 4x² – 5x – 6 be α, β
Sum = α+β = -(-5)/4 = 5/4
Zeroes of x² + px + q are 2α, 2β
Sum = 2α+2β = 2(α+β) = 2×(5/4) = 5/2
But sum for x² + px + q = -p
So -p = 5/2 ⇒ p = -5/2
5. If the sum of zeroes of the polynomial p(x) = 2x² – k√2x + 1 is √2, then value of k is: [CBSE 2024]
(a) √2 (b) 2 (c) 2√2 (d) 1/2
(a) √2 (b) 2 (c) 2√2 (d) 1/2
View Answer
Ans: (b) 2
Sum of zeroes = -b/a = -(-k√2)/2 = k√2/2
Given: k√2/2 = √2
k/2 = 1
k = 2
Sum of zeroes = -b/a = -(-k√2)/2 = k√2/2
Given: k√2/2 = √2
k/2 = 1
k = 2