CBSE Class 10 Mathematics Polynomials MCQs Set 23

Answer: (b) 0

Explanation:
1. For x² − 2x + 3: D = b² − 4ac = 4 − 12 = −8.
2. D < 0, so the graph does not touch or cross the x-axis.
3. The number of real zeros is 0.

Teacher's Note:
1. If the discriminant is negative the graph of the quadratic does not meet the x-axis.
2. The vertex (1, 2) lies above the x-axis and the parabola opens upwards.

Answer: (a) downwards

Explanation:
1. The coefficient of x² is −2, which is negative, so the parabola opens downwards.

Teacher's Note:
1. The sign of a decides the direction: a > 0 opens up, a < 0 opens down.
2. Here a = −2.

Answer: (d) \( \frac{1}{2},\ \frac{1}{2} \)

Explanation:
1. 4s² − 4s + 1 = (2s − 1)².
2. 2s − 1 = 0 gives s = 1/2 twice.
3. Check: sum = 1 = −b/a and product = 1/4 = c/a.

Teacher's Note:
1. Recognise the perfect square (2s − 1)².
2. Verify with sum = −b/a and product = c/a.

Answer: (c) \( \sqrt{15},\ -\sqrt{15} \)

Explanation:
1. t² − 15 = (t − √15)(t + √15).
2. The zeros are √15 and −√15.
3. Sum = 0 = −0/1 and product = −15 = c/a.

Teacher's Note:
1. Use a² − b² = (a − b)(a + b) with b = √15.
2. Check with sum and product of zeros.

Answer: (b) 0 and −2

Explanation:
1. 4u² + 8u = 4u(u + 2).
2. u = 0 or u + 2 = 0, so u = 0 or u = −2.
3. Check: sum = −2 = −b/a and product = 0 = c/a.

Teacher's Note:
1. Take out the common factor 4u.
2. Each factor equals zero for a zero of the polynomial.

Answer: (a) \( \frac{1}{\sqrt{2}},\ -\frac{1}{\sqrt{2}} \)

Explanation:
1. 6x² − 3 = 0 gives x² = 1/2, so x = ±1/√2. Sum = 0 = −b/a and product = −1/2 = −3/6.

Teacher's Note:
1. Set the polynomial to zero and solve for x².
2. Take both the positive and the negative root.

Answer: (d) \( x^2 - 8x + 12 \)

Explanation:
1. Sum of zeros = 8 and product = 12, so the polynomial is x² − (sum)x + product = x² − 8x + 12.

Teacher's Note:
1. The monic quadratic is x² − (α + β)x + αβ.
2. Sum = 8 and product = 12.

Answer: (c) \( 9x^2 + (9a + 1)x + a \)

Explanation:
1. Sum of zeros = −a − 1/9, product = (−a)(−1/9) = a/9.
2. x² − (sum)x + product = x² + (a + 1/9)x + a/9.
3. Multiply by 9: 9x² + (9a + 1)x + a.

Teacher's Note:
1. Form x² − (sum)x + product.
2. Multiply by 9 to clear the fraction.

Answer: (b) \( x^3 - 6x^2 + 11x - 6 \)

Explanation:
1. Sum = 6, sum of products in pairs = 6 + 2 + 3 = 11 and product = 6, so the polynomial is x³ − 6x² + 11x − 6.

Teacher's Note:
1. A monic cubic is x³ − (sum)x² + (pair sum)x − product.
2. Check by expanding (x − 1)(x − 2)(x − 3).

Answer: (a) 6, 11 and 6

Explanation:
1. α + β + γ = 3 + 2 + 1 = 6.
2. αβ + βγ + γα = 6 + 2 + 3 = 11.
3. αβγ = 3 × 2 × 1 = 6.

Teacher's Note:
1. Substitute the values directly.
2. Pair products are 3×2, 2×1 and 1×3.

Answer: (d) 0

Explanation:
1. z⁴ + 4z² + 5 = (z² + 2)² + 1 ≥ 1 for every real z, so it is never zero.

Teacher's Note:
1. Complete the square in z².
2. A sum of a square and a positive number is always positive.

Answer: (c) 2 and 2

Explanation:
1. x³ − 3x² + 4 ÷ (x + 1) = x² − 4x + 4.
2. x² − 4x + 4 = (x − 2)².
3. The other zeros are 2 and 2.

Teacher's Note:
1. Divide the cubic by x + 1.
2. The quotient is a perfect square.

Answer: (b) 2 and −5

Explanation:
1. x² + 3x − 10 = x² + 5x − 2x − 10 = (x + 5)(x − 2).
2. Zeros: x = −5 and x = 2.
3. Sum = −3 = −b/a, product = −10 = c/a.

Teacher's Note:
1. Split the middle term: 5x − 2x.
2. Verify with the sum and product of the zeros.

Answer: (a) \( x^3 - x^2 - 10x - 8 \)

Explanation:
1. x³ − (sum)x² + (pair sum)x − product = x³ − x² − 10x − 8.

Teacher's Note:
1. Use the standard form x³ − (α + β + γ)x² + (αβ + βγ + γα)x − αβγ.
2. Watch the signs of the last two terms.

Answer: (d) 0

Explanation:
1. Product of zeros = (k + 4)/4.
2. α · (1/α) = 1, so (k + 4)/4 = 1.
3. k = 0.

Teacher's Note:
1. The product of the zeros equals c/a.
2. Here the product is 1.

Answer: (c) \( x^3 - 3x^2 - 6x + 8 \)

Explanation:
1. (x + 2)(x − 1)(x − 4) = (x² + x − 2)(x − 4) = x³ − 3x² − 6x + 8.

Teacher's Note:
1. Multiply the three factors step by step.
2. Sum of zeros 3, pair sum −6, product −8.

Answer: (b) 1, 4 and 7

Explanation:
1. Let the zeros be a − d, a, a + d: sum = 3a = 12, so a = 4.
2. Divide f(x) by (x − 4): x² − 8x + 7 = (x − 1)(x − 7).
3. Zeros: 1, 4, 7 (common difference 3).

Teacher's Note:
1. Let the zeros be a − d, a and a + d.
2. The sum of the zeros is 12.

Answer: (a) 2, 3, −1/2 and 1/2

Explanation:
1. (x − 2)(x − 3) = x² − 5x + 6.
2. 4x⁴ − 20x³ + 23x² + 5x − 6 = (x² − 5x + 6)(4x² − 1).
3. 4x² − 1 = 0 gives x = ±1/2.

Teacher's Note:
1. Use the factor x² − 5x + 6.
2. The quotient is 4x² − 1.

Answer: (d) \( 14x - 10 \)

Explanation:
1. (4x² + 3x − 2)(2x² + 2x − 1) = 8x⁴ + 14x³ − 2x² − 7x + 2.
2. Subtract from the given polynomial: remainder = 14x − 10.
3. So 14x − 10 must be subtracted.

Teacher's Note:
1. Perform long division of the quartic by the quadratic.
2. The remainder is the amount to subtract.

Answer: (c) 1 and 2

Explanation:
1. Divide by (x + 2): quotient x² − 3x + 2.
2. x² − 3x + 2 = (x − 1)(x − 2).
3. Other zeros: 1 and 2.

Teacher's Note:
1. Divide by x + 2 and factorise the quotient.
2. Check: sum of zeros = −2 + 1 + 2 = 1.

Answer: (b) \( \frac{b^4 - 4ab^2c + 2a^2c^2}{a^4} \)

Explanation:
1. α + β = −b/a and αβ = c/a.
2. α² + β² = (α + β)² − 2αβ = (b² − 2ac)/a².
3. α⁴ + β⁴ = (α² + β²)² − 2(αβ)² = (b² − 2ac)²/a⁴ − 2c²/a².
4. = (b⁴ − 4ab²c + 2a²c²)/a⁴.

Teacher's Note:
1. First find α² + β², then square it.
2. α²β² = c²/a².

Answer: (a) \( \frac{b^4 - 4ab^2c + 2a^2c^2}{a^2c^2} \)

Explanation:
1. α²/β² + β²/α² = (α⁴ + β⁴)/(α²β²).
2. α⁴ + β⁴ = (b⁴ − 4ab²c + 2a²c²)/a⁴ and α²β² = c²/a².
3. Dividing gives (b⁴ − 4ab²c + 2a²c²)/(a²c²).

Teacher's Note:
1. Combine over the common denominator α²β².
2. Reuse the value of α⁴ + β⁴.

Answer: (d) 108

Explanation:
1. α⁴β³ + α³β⁴ = (αβ)³(α + β).
2. α + β = 4 and αβ = 3.
3. Value = 3³ × 4 = 108.

Teacher's Note:
1. Factorise α³β³ out of both terms.
2. Use sum = 4 and product = 3.

Answer: (c) \( x^2 - 24x + 128 \)

Explanation:
1. Adding: 2α = 32, so α = 16; then β = 8.
2. Sum = 24 and product = 16 × 8 = 128.
3. Polynomial: x² − 24x + 128.

Teacher's Note:
1. Solve the two equations for α and β.
2. Then use sum and product.

Answer: (b) 0

Explanation:
1. Zeros α and −α have sum zero, so 8k/4 = 0 and k = 0.

Teacher's Note:
1. If the zeros are negatives, their sum is 0.
2. Sum of zeros = −b/a = 8k/4.

Answer: (a) \( 2p^3 - 9pq + 27r = 0 \)

Explanation:
1. Let the zeros be a − d, a, a + d: 3a = p, so a = p/3.
2. a is a zero: (p/3)³ − p(p/3)² + q(p/3) − r = 0.
3. −2p³/27 + pq/3 − r = 0; multiply by −27: 2p³ − 9pq + 27r = 0.

Teacher's Note:
1. The middle zero of an A.P. is the average, p/3.
2. Substitute it into f(x).

Answer: (d) −28

Explanation:
1. Sum of zeros = 12, so the middle zero is 4.
2. f(4) = 64 − 192 + 156 + k = 28 + k = 0.
3. k = −28.

Teacher's Note:
1. The middle zero of three zeros in A.P. is one third of their sum.
2. Substitute x = 4.

Answer: (c) \( 2 - \sqrt{3},\ 7,\ -5 \)

Explanation:
1. The zero 2 − √3 follows from 2 + √3 (conjugate).
2. (x − (2 + √3))(x − (2 − √3)) = x² − 4x + 1.
3. f(x) ÷ (x² − 4x + 1) = x² − 2x − 35 = (x − 7)(x + 5).
4. Other zeros: 2 − √3, 7 and −5.

Teacher's Note:
1. Irrational zeros of this kind occur in conjugate pairs.
2. Divide f(x) by x² − 4x + 1.