CBSE Class 10 Mathematics Polynomials MCQs Set 19

Answer: (b) 3

Explanation:
1. Degree of quotient = degree of dividend − degree of divisor = 7 − 4 = 3.

Teacher's Note:
1. The quotient degree is a simple subtraction.
2. This holds when the remainder has a smaller degree than the divisor.

Answer: (a) 1

Explanation:
1. p(1) = 0 gives a − 3(a − 1) − 1 = 0.
2. −2a + 2 = 0, so a = 1.

Teacher's Note:
1. A zero makes the polynomial value 0.
2. Substitute x = 1 and solve for a.

Answer: (b) −2

Explanation:
1. f(1) = 2 − 6 + 4 − 5 = −5.
2. g(−2) = 3(4) − 9 = 3.
3. f(1) + g(−2) = −5 + 3 = −2.

Teacher's Note:
1. Substitute each value separately.
2. Mind the sign when squaring a negative number.

Answer: (b) −1, 0

Explanation:
1. p(−1) = 0: 2α + β = −2. p(−2) = 0: 4α + β = −4.
2. Subtracting: 2α = −2, so α = −1 and β = 0.

Teacher's Note:
1. A factor x + c gives p(−c) = 0.
2. Solve the two linear equations.

Answer: (c) \( \frac{3}{2} \)

Explanation:
1. Sum of zeroes = 8/2 = 4.
2. Other zero = 4 − 5/2 = 3/2.

Teacher's Note:
1. Sum of zeroes = −b/a.
2. Subtract the known zero.

Answer: (a) 2, 0

Explanation:
1. f(0) = b = 0.
2. f(2) = 16 − 20 + 2a = 0, so a = 2.

Teacher's Note:
1. Substitute each zero in turn.
2. Start with x = 0 because it gives b at once.

Answer: (c) 3

Explanation:
1. The highest power of y is 3, so the degree is 3.

Teacher's Note:
1. Degree is the highest power of the variable.
2. Constants and coefficients do not change it.

Answer: (a) 2

Explanation:
1. Product of zeroes = \( \frac{k+2}{k^2} \) = 1.
2. So k² − k − 2 = 0, that is (k − 2)(k + 1) = 0. Since k > 0, k = 2.

Teacher's Note:
1. Reciprocal zeroes have product 1.
2. Reject the negative root.

Answer: (a) 18

Explanation:
1. p(x) has only one zero, so p(x) = a(x + 1)².
2. p(−2) = a = 2, so p(2) = 2 × 9 = 18.

Teacher's Note:
1. A quadratic with one zero is a perfect square.
2. Use p(−2) to find the constant.

Answer: (b) −m, m + 3

Explanation:
1. x² − 3x − m(m + 3) = (x − (m + 3))(x + m).
2. The zeroes are m + 3 and −m.

Teacher's Note:
1. Factorise by splitting the middle term.
2. Check the sum = 3 and product = −m(m + 3).

Answer: (c) 3

Explanation:
1. a + b = 2 and ab = 3/2.
2. a²b + ab² = ab(a + b) = 3/2 × 2 = 3.

Teacher's Note:
1. Factor out ab.
2. Use sum and product of zeroes.

Answer: (b) 6

Explanation:
1. α + β = 5 and α − β = 1 give α = 3 and β = 2.
2. k = αβ = 6.

Teacher's Note:
1. Solve for α and β.
2. The product of zeroes is k.

Answer: (b) 1

Explanation:
1. x³ + y³ = (x + y)³ − 3xy(x + y) = 1 − 3xy.
2. So x³ + y³ + 3xy = 1.

Teacher's Note:
1. Use the cube of a sum.
2. The xy terms cancel.

Answer: (c) 15, −5

Explanation:
1. x² − 10x − 75 = (x − 15)(x + 5).
2. The zeroes are 15 and −5.

Teacher's Note:
1. Find two numbers with product −75 and sum −10.
2. Check: 15 − 5 = 10.

Answer: (d) parabola

Explanation:
1. The graph of a quadratic polynomial is a parabola, and this curve opens upward with one turning point.

Teacher's Note:
1. Quadratic polynomial graphs are parabolas.
2. This one opens upward.

Answer: (a) 2

Explanation:
1. The curve cuts the x-axis at two points, so it has 2 zeroes.

Teacher's Note:
1. Zeroes are the x-intercepts.
2. Count the intersections with the x-axis.

Answer: (b) −1, 3

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

Teacher's Note:
1. Read the x-intercepts off the graph.
2. Check: (−1)² − 2(−1) − 3 = 0.

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

Explanation:
1. Zeroes −1 and 3 give sum 2 and product −3, so the polynomial is x² − 2x − 3.
2. Check with (4, 5): 16 − 8 − 3 = 5.

Teacher's Note:
1. Use x² − (sum)x + product.
2. Check with a point on the curve.

Answer: (d) 0

Explanation:
1. (−1)² − 2(−1) − 3 = 1 + 2 − 3 = 0, because −1 is a zero.

Teacher's Note:
1. Substitute x = −1.
2. A zero gives the value 0.

Answer: (b) −2

Explanation:
1. x² + 4x + 2 = (x + 2)² − 2.
2. The square is at least 0, so the least value is −2 (at x = −2).

Teacher's Note:
1. Complete the square.
2. A square is never negative.