CBSE Class 10 Mathematics Polynomials MCQs Set 16

Answer: (b) 1

Explanation:
1. Sum of zeroes = −(coefficient of x) ÷ (coefficient of x²) = −m.
2. −m = −1, so m = 1.

Teacher's Note:
1. Use the sum of zeroes = −b/a.
2. Check: x² + x − 2 = (x + 2)(x − 1) has zeroes −2 and 1 with sum −1.

Answer: (a) \( 12x^2 + 5x - 3 \)

Explanation:
1. Sum of zeroes = \( \dfrac{1}{3} - \dfrac{3}{4} = -\dfrac{5}{12} \) and product = \( -\dfrac{1}{4} \).
2. Polynomial: \( x^2 + \dfrac{5}{12}x - \dfrac{1}{4} \), multiply by 12 to get \( 12x^2 + 5x - 3 \).

Teacher's Note:
1. Form x² − (sum)x + product, then clear fractions.
2. Check with the factors (3x − 1)(4x + 3) = 12x² + 5x − 3.

Answer: (c) 5

Explanation:
1. Zeroes α and 1/α have product 1.
2. Product = m/5 = 1, so m = 5.

Teacher's Note:
1. Reciprocal zeroes give product = 1, so c/a = 1.
2. Here a = 5 and c = m.

Answer: (c) 2

Explanation:
1. Product of zeroes = 1, so \( \dfrac{4\lambda}{\lambda^2 + 4} = 1 \).
2. \( \lambda^2 - 4\lambda + 4 = 0 \), that is \( (\lambda - 2)^2 = 0 \), so λ = 2.

Teacher's Note:
1. Set the product c/a equal to 1.
2. The quadratic in λ is a perfect square.

Answer: (d) a = 0, b = −6

Explanation:
1. Sum of zeroes = 2 + (−3) = −1 = −(a + 1), so a = 0.
2. Product of zeroes = 2 × (−3) = −6 = b. So a = 0, b = −6.

Teacher's Note:
1. Use sum = −b/a and product = c/a for the polynomial.
2. Check: x² + x − 6 = (x + 3)(x − 2).

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. The graph meets the x-axis at four points, so f(x) has 4 zeroes. A is true.
2. The number of zeroes equals the number of points where the graph cuts or touches the x-axis. R is true and explains A.

Teacher's Note:
1. Zeroes are the x-coordinates of the points where the graph meets the x-axis.
2. Count the points of intersection carefully.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. \( x^2 + 7x + 12 = (x + 3)(x + 4) \), so it has two real zeroes −3 and −4. A is true.
2. A quadratic polynomial has at most two zeroes. R is true.
3. R is a general statement that does not explain why this particular polynomial has two zeroes, so R is not the correct explanation of A.

Teacher's Note:
1. Factorise to find the actual zeroes.
2. The reason gives only an upper limit, not the actual number.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. α + β = 8 and αβ = 1 (by R, which is a true statement).
2. \( \dfrac{1}{\alpha} + \dfrac{1}{\beta} - \alpha\beta = \dfrac{\alpha + \beta}{\alpha\beta} - \alpha\beta = 8 - 1 = 7 \). So A is true and R is the formula used.

Teacher's Note:
1. Write 1/α + 1/β as (α + β)/αβ.
2. Substitute the sum and the product.

Answer: (c) A is true, but R is false.

Explanation:
1. \( x^2 - 4x - 5 = (x - 5)(x + 1) \), so the zeroes are 5 and −1. A is true.
2. For zeroes \( 2 \pm \sqrt{3} \): sum = 4 and product = 4 − 3 = 1, so the polynomial is \( x^2 - 4x + 1 \), not \( x^2 - 4x + 7 \). R is false.

Teacher's Note:
1. Compute the sum and product of the given zeroes.
2. (a + b)(a − b) = a² − b² gives the product 4 − 3.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. For a quadratic with rational coefficients, the sum and the product of the zeroes are rational. If one zero is 2 − √3, the other must be 2 + √3 (sum 4, product 1).
2. R states exactly this property, so both A and R are true and R explains A.

Teacher's Note:
1. Both statements need the condition that the coefficients are rational.
2. Without that condition, the other zero could be any number.

Answer: (a) a parabola

Explanation:
1. The graph of every quadratic polynomial \( ax^2 + bx + c \) (a ≠ 0) is a curve called a parabola.
2. So the shape is a parabola.

Teacher's Note:
1. Linear polynomial: straight line. Quadratic polynomial: parabola.
2. The parabola is U-shaped (a > 0) or inverted U (a < 0).

Answer: (c) 2

Explanation:
1. The zeroes of a polynomial are the x-coordinates of the points where its graph meets the x-axis.
2. The graph meets the x-axis at 2 points, so the number of zeroes is 2.

Teacher's Note:
1. Count the points where the curve meets the x-axis.
2. The zeroes here are −2 and 4.

Answer: (a) \( -\dfrac{2}{\sqrt{3}} \) and \( \dfrac{\sqrt{3}}{4} \)

Explanation:
1. Split the middle term: \( 4\sqrt{3}x^2 + 8x - 3x - 2\sqrt{3} = 4x(\sqrt{3}x + 2) - \sqrt{3}(\sqrt{3}x + 2) = (\sqrt{3}x + 2)(4x - \sqrt{3}) \).
2. Zeroes: \( x = -\dfrac{2}{\sqrt{3}} \) and \( x = \dfrac{\sqrt{3}}{4} \).

Teacher's Note:
1. Split 5x as 8x − 3x because 8 × (−3) = −24 = 4√3 × (−2√3).
2. Check: sum = −5/(4√3) = −b/a.

Answer: (a) a parabola

Explanation:
1. p(t) is a quadratic polynomial (degree 2).
2. The graph of a quadratic polynomial is a parabola.

Teacher's Note:
1. Degree 2 means a parabola.
2. Here a = 24 > 0, so it opens upwards.

Answer: (c) 2

Explanation:
1. Discriminant: \( b^2 - 4ac = 41^2 - 4 \times 24 \times 12 = 1681 - 1152 = 529 = 23^2 \), which is positive.
2. So the graph meets the t-axis at 2 points and p(t) has 2 zeroes.

Teacher's Note:
1. A positive discriminant means two distinct real zeroes.
2. The graph of the polynomial cuts the t-axis at those two points.

Answer: (a) \( \dfrac{3}{8} \) and \( \dfrac{4}{3} \)

Explanation:
1. Split the middle term: \( 24t^2 - 32t - 9t + 12 = 8t(3t - 4) - 3(3t - 4) = (8t - 3)(3t - 4) \).
2. Zeroes: \( t = \dfrac{3}{8} \) and \( t = \dfrac{4}{3} \).

Teacher's Note:
1. Find two numbers with product 24 × 12 = 288 and sum −41: −32 and −9.
2. Check the sum: 3/8 + 4/3 = 41/24.

Answer: (c) \( \dfrac{x^2}{2} - \dfrac{x}{2} - 6 \)

Explanation:
1. Sum of zeroes = 1 and product = −12, so one polynomial is \( x^2 - x - 12 \), and every constant multiple of it also has these zeroes.
2. \( \dfrac{x^2}{2} - \dfrac{x}{2} - 6 = \dfrac{1}{2}(x^2 - x - 12) \), so the answer is (c). The others have the wrong sign of the constant term or of the sum of zeroes.

Teacher's Note:
1. Write the polynomial as k[x² − (sum)x + product].
2. Check each option's sum and product.

Answer: (a) has no linear term and the constant term is negative

Explanation:
1. Zeroes are k and −k. Sum = 0 = −a, so a = 0 and there is no linear term.
2. Product = −k² = b, which is negative (for k ≠ 0). So the answer is (a).

Teacher's Note:
1. Sum of zeroes zero means no x term.
2. Product of k and −k is −k².

Answer: (d) graph (d)

Explanation:
1. The graph of a quadratic polynomial is a parabola: a U-shaped or inverted U-shaped curve with a single turning point.
2. Graphs (a), (b) and (c) are parabolas. Graph (d) has two turning points (it is wave-like) and is the graph of a cubic polynomial, so it is not a quadratic polynomial.

Teacher's Note:
1. A parabola turns only once.
2. A curve with two turning points is the graph of a polynomial of degree 3.

Answer: (a) \( \dfrac{4}{3} \)

Explanation:
1. Put x = −3: \( 9(k - 1) - 3k + 1 = 0 \).
2. \( 6k - 8 = 0 \), so k = 4/3.

Teacher's Note:
1. A zero of a polynomial satisfies p(x) = 0.
2. Expand carefully: 9k − 9 − 3k + 1.

Answer: (c) graph (c)

Explanation:
1. The zeroes are x = 1 (twice) and x = −2. So the graph touches the x-axis at x = 1 and crosses it at x = −2.
2. Also at x = −1 the value is \( (-2)^2 \times 1 = 4 \), which is positive and for large x the graph goes up. Only graph (c) has these features.

Teacher's Note:
1. A repeated zero means the graph touches the axis there.
2. Test a point like x = −1 to separate the graphs.

Answer: (b) −m, m + 3

Explanation:
1. For \( x^2 - 3x - m(m + 3) \), the sum of the zeroes is 3 and the product is −m(m + 3).
2. Take the zeroes −m and (m + 3): their sum is −m + m + 3 = 3 and their product is −m(m + 3). So the polynomial factorises as \( (x + m)(x - m - 3) \) and the zeroes are −m and m + 3.

Teacher's Note:
1. Test the options with sum = 3 and product = −m(m + 3).
2. Only −m and m + 3 give both.

Answer: (a) \( k[x^2 - 8x + 5] \)

Explanation:
1. A quadratic polynomial is \( k[x^2 - (\text{sum})x + \text{product}] \).
2. Here it is \( k[x^2 - 8x + 5] \).

Teacher's Note:
1. Do not swap the sum and the product.
2. k is any non-zero constant.

Answer: (b) 2

Explanation:
1. Value at x = 3: \( 9 - 15 + 4 = -2 \).
2. To make the value 0, add 2.

Teacher's Note:
1. Find the value at the given point first.
2. Add the number that makes it zero.

Answer: (d) \( \dfrac{-3}{7} \)

Explanation:
1. α + β = 3/4 and αβ = −7/4.
2. \( \dfrac{1}{\alpha} + \dfrac{1}{\beta} = \dfrac{\alpha + \beta}{\alpha\beta} = \dfrac{3/4}{-7/4} = -\dfrac{3}{7} \).

Teacher's Note:
1. Combine the fractions over αβ.
2. Watch the sign: the product is negative.

Answer: (b) 7

Explanation:
1. α + β = k + 6 and αβ = 2(2k − 1).
2. k + 6 = ½ × 2(2k − 1) = 2k − 1, so k = 7.

Teacher's Note:
1. Use the sum and product from the polynomial.
2. Solve the linear equation in k.

Answer: (d) 0

Explanation:
1. The polynomial has no x term, so the sum of zeroes = −0/1 = 0.
2. (The zeroes are 1 and −1.) So α + β = 0.

Teacher's Note:
1. Sum = −b/a, and b = 0 here.
2. Check: 1 + (−1) = 0.

Answer: (d) parabola

Explanation:
1. The curve has one turning point and opens upwards, like the graph of a quadratic polynomial.
2. So the wire is bent in the shape of a parabola.

Teacher's Note:
1. A quadratic polynomial always gives a parabola.
2. A linear polynomial gives a straight line.

Answer: (a) 2

Explanation:
1. The zeroes of a polynomial are the x-coordinates of the points where its graph meets the x-axis.
2. There are two such points, so the number of zeroes is 2.

Teacher's Note:
1. Count the points of intersection with the x-axis.
2. A quadratic polynomial can have at most 2 zeroes.

Answer: (b) −1, 3

Explanation:
1. The graph meets the x-axis at x = −1 and x = 3.
2. So the zeroes are −1 and 3.

Teacher's Note:
1. Read the x-coordinates where the curve touches or crosses the x-axis.
2. Check with the next question: (x + 1)(x − 3) has these zeroes.

Answer: (c) \( x^2 - 2x - 3 \)

Explanation:
1. Sum of zeroes = −1 + 3 = 2 and product = −3, so the polynomial is \( x^2 - 2x - 3 \).
2. Check with the graph: at x = 0 the value is −3, which matches the y-intercept.

Teacher's Note:
1. Use x² − (sum)x + product.
2. Verify with the point where the graph cuts the y-axis.

Answer: (d) 0

Explanation:
1. \( p(-1) = (-1)^2 - 2(-1) - 3 = 1 + 2 - 3 = 0 \).
2. So the value is 0, because −1 is a zero of the polynomial.

Teacher's Note:
1. Substitute x = −1 carefully.
2. The value at a zero is always 0.

Answer: (b) \( \alpha = \beta + \dfrac{v - t}{s - u} \)

Explanation:
1. Let γ = α − β. Then \( \gamma^2 + s\gamma + t = 0 \) and \( \gamma^2 + u\gamma + v = 0 \).
2. Subtracting: \( (s - u)\gamma + (t - v) = 0 \), so \( \gamma = \dfrac{v - t}{s - u} \).
3. Hence \( \alpha - \beta = \dfrac{v - t}{s - u} \) and \( \alpha = \beta + \dfrac{v - t}{s - u} \).

Teacher's Note:
1. A common zero satisfies both equations, so subtract them to remove x².
2. Then move β to the right side.

Answer: (a) The graph opens upwards if a > 0.

Explanation:
1. For a > 0 the parabola opens upwards (U-shape) and for a < 0 it opens downwards.
2. If a = 0 the polynomial is no longer quadratic. So only the first statement is true.

Teacher's Note:
1. Remember the sign of a decides the opening direction.
2. a = 0 gives a straight line.

Answer: (d) 7

Explanation:
1. Zeroes k and −k have sum 0.
2. Sum of zeroes = −(p − 7) = 0, so p = 7. (Then the equation is α² − 25 = 0 with zeroes 5 and −5.)

Teacher's Note:
1. Additive inverse means the zeroes are k and −k.
2. The coefficient of the middle term must be 0.

Answer: (b) 4

Explanation:
1. p + q = −5/2 and pq = k/2.
2. \( p^2 + q^2 + pq = (p + q)^2 - pq = \dfrac{25}{4} - \dfrac{k}{2} = \dfrac{17}{4} \), so \( \dfrac{k}{2} = 2 \) and k = 4.

Teacher's Note:
1. Write p² + q² + pq as (p + q)² − pq.
2. Then substitute the sum and product.

Answer: (c) A is true, but R is false.

Explanation:
1. Product = −5/2 < 0, so the zeroes have opposite signs: one positive and one negative. A is true.
2. The sum of the zeroes is −b/a, not b/a. R is false.

Teacher's Note:
1. A negative product means opposite signs.
2. Sum of zeroes = −b/a.

Answer: (c) A is true, but R is false.

Explanation:
1. Sum = 8/9 and product = 0, so the polynomial is \( x^2 - \dfrac{8}{9}x \), that is \( 9x^2 - 8x \). A is true.
2. If the zeroes are p and q, the factors are (x − p) and (x − q), so \( p(x) = (x - p)(x - q) \). R is false because of the + sign.

Teacher's Note:
1. Sum is 8/9 and product is 0, so x² − (8/9)x.
2. The factor for the zero q is (x − q), not (x + q).

Answer: (a) 1

Explanation:
1. Product of zeroes = \( \alpha \times \dfrac{2}{\alpha} = 2 \) = 2a.
2. So a = 1.

Teacher's Note:
1. The product of the zeroes is c/a.
2. The α cancels in the product.

Answer: (a) \( -\dfrac{2}{3} \)

Explanation:
1. α + β = 2/4 = 1/2 and αβ = −3/4.
2. \( \dfrac{1}{\alpha} + \dfrac{1}{\beta} = \dfrac{\alpha + \beta}{\alpha\beta} = \dfrac{1/2}{-3/4} = -\dfrac{2}{3} \).

Teacher's Note:
1. Combine as (α + β)/αβ.
2. Dividing by a negative number gives a negative result.