CBSE Class 10 Mathematics Polynomials MCQs Set 27

Answer: (c) -1

Explanation:
1. p(-1) = 1 + 2 - 3 = 0.
2. p(0) = -3, p(1) = -4 and p(-3) = 12, none of which is zero.

Teacher's Note:
1. A zero makes the polynomial equal to 0.
2. Factorising gives (x - 3)(x + 1), so the zeros are 3 and -1.

Answer: (b) 5

Explanation:
1. Product of zeros = k/5.
2. Reciprocal zeros have product 1, so k/5 = 1 and k = 5.

Teacher's Note:
1. Product of zeros of ax² + bx + c is c/a.
2. Reciprocal roots always give a product of 1.

Answer: (b) 1 - c

Explanation:
1. f(x) = x² - px - (p + c), so α + β = p and αβ = -(p + c).
2. (α + 1)(β + 1) = αβ + α + β + 1 = -(p + c) + p + 1 = 1 - c.

Teacher's Note:
1. Rewrite the polynomial in standard form first.
2. Expand (α + 1)(β + 1) in terms of the sum and product.

Answer: (c) x² + x - 2

Explanation:
1. Sum of zeros = 1 + (-2) = -1 and product = -2.
2. Polynomial = x² - (sum)x + product = x² + x - 2.

Teacher's Note:
1. The standard form is x² - (sum)x + product.
2. Check by substituting x = 1: 1 + 1 - 2 = 0.

Answer: (b) c is a factor of b

Explanation:
1. c divides ab and has no common factor with a.
2. So all prime factors of c must divide b, hence c divides b.

Teacher's Note:
1. This is Euclid's lemma for co-prime numbers.
2. Try a = 3, b = 10, c = 5 to see c | b.

Answer: (a) max{m, n}

Explanation:
1. The term of higher degree cannot be cancelled by the other polynomial.
2. So the difference has degree max{m, n}.

Teacher's Note:
1. If m = n the leading terms may cancel and the degree can drop.
2. Subtraction never increases the degree beyond the larger one.

Answer: (c) \( p(x)q(x) = \ell h \)

Explanation:
1. For polynomials, as for numbers, product = LCM × HCF.
2. So p(x)q(x) = ℓh (up to a constant factor).

Teacher's Note:
1. The relation HCF × LCM = product holds for two polynomials.
2. Check with p = x(x+1) and q = x(x+2).

Answer: (c) \( x + a \)

Explanation:
1. x³ + a³ = (x + a)(x² - ax + a²).
2. (x + a)² = (x + a)(x + a).
3. The common factor is (x + a).

Teacher's Note:
1. Use the identity a³ + b³ = (a + b)(a² - ab + b²).
2. x² - ax + a² has no factor x + a.

Answer: (c) 6

Explanation:
1. α + β = 5 and α - β = 1, so α = 3 and β = 2.
2. k = αβ = 6.

Teacher's Note:
1. Solve the two simultaneous equations first.
2. Product of zeros = constant term.

Answer: (a) \( 2p^3 = pq - r \)

Explanation:
1. Let the zeros be m - d, m, m + d. Their sum is 3m = 3p, so m = p.
2. Since p is a zero: p³ - 3p·p² + qp - r = 0, i.e. -2p³ + pq - r = 0.
3. So 2p³ = pq - r.

Teacher's Note:
1. When zeros are in A.P., the middle zero is the sum divided by 3.
2. Substitute the middle zero in the polynomial.

Answer: (b) pq = r

Explanation:
1. Let the zeros be a, -a and c. Then p = c, q = -a² and r = -a²c.
2. So r = q × c = qp, i.e. pq = r.

Teacher's Note:
1. Use the sum, sum of products and product of zeros.
2. Alternatively, x = p must be a zero when two zeros cancel.

Answer: (b) (-1, -4)

Explanation:
1. 4u² + 8u = 4(u + 1)² - 4.
2. The vertex is at u = -1 with value -4, i.e. (-1, -4).

Teacher's Note:
1. Complete the square to find the vertex.
2. The vertex u-coordinate is -b/2a = -8/8 = -1.

Answer: (c) \( -\frac{c}{a} \)

Explanation:
1. Factorise: (ax + b)(bx - c). The zeros are -b/a and c/b.
2. Product = (-b/a)(c/b) = -c/a.

Teacher's Note:
1. Factorise before expanding.
2. The product of zeros can be read off directly from the factors.

Answer: (d) b = 0

Explanation:
1. Zeros are k and -k, so their sum is 0.
2. Sum of zeros = -b/a = 0, hence b = 0.

Teacher's Note:
1. Equal and opposite zeros means the sum is zero.
2. Example: x² - 4 has zeros 2 and -2 and no x term.

Answer: (b) \( \gamma = \alpha \)

Explanation:
1. Product of zeros = γ/α.
2. Reciprocal zeros have product 1, so γ = α.

Teacher's Note:
1. Reciprocal zeros multiply to 1.
2. Check with 2x² + 5x + 2, whose zeros -2 and -1/2 have product 1.

Answer: (a) \( \frac{n - q}{m - p} \)

Explanation:
1. x = -a is a zero of both: a² - pa + q = 0 and a² - ma + n = 0.
2. Subtract: (m - p)a + (q - n) = 0, so a = (n - q)/(m - p).

Teacher's Note:
1. Substitute x = -a in both polynomials.
2. Subtracting removes the a² term.

Answer: (a) a + c + e = 0 and b + d = 0

Explanation:
1. x = 1 gives a + b + c + d + e = 0 and x = -1 gives a - b + c - d + e = 0.
2. Adding gives a + c + e = 0 and subtracting gives b + d = 0.

Teacher's Note:
1. x² - 1 = (x - 1)(x + 1), so both x = 1 and x = -1 are zeros.
2. Adding and subtracting the two equations separates the odd and even powers.

Answer: (a) 0 or 1

Explanation:
1. Put x = 1: 3 + 9 - 7 - 5 - 3a + 3a² = 0.
2. This gives 3a² - 3a = 0, so a(a - 1) = 0.
3. a = 0 or a = 1.

Teacher's Note:
1. Divisible by x - 1 means p(1) = 0 (Factor Theorem).
2. The constants 3 + 9 - 7 - 5 cancel to zero.

Answer: (b) 0

Explanation:
1. Put x = 2y: 3(8y³) - 2(4y²)y - 13(2y)y² + 10y³ = 24y³ - 8y³ - 26y³ + 10y³ = 0.

Teacher's Note:
1. Use the Remainder Theorem with x = 2y.
2. A zero remainder means x - 2y is a factor.

Answer: (c) 9 or -9

Explanation:
1. (x - y)² = x² + y² - 2xy gives 1 = 41 - 2xy, so xy = 20.
2. (x + y)² = 41 + 40 = 81, so x + y = 9 or -9.
3. Both occur: (5, 4) and (-4, -5).

Teacher's Note:
1. Use (x + y)² = (x - y)² + 4xy.
2. Do not forget the negative root.

Answer: (a) \( \frac{q - m}{p - \ell} \)

Explanation:
1. x = -a is a zero of both: a² - pa + q = 0 and a² - ℓa + m = 0.
2. Subtracting: (ℓ - p)a + (q - m) = 0, so a = (q - m)/(p - ℓ).

Teacher's Note:
1. Substitute x = -a in both and subtract.
2. Watch the sign when rearranging.