CBSE Class 10 Mathematics Polynomials MCQs Set 18

Answer: (a) 3x² − 8x + 2 = 0

Explanation:
1. The roots of ax² + bx + c = 0 are \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
2. Comparing, −b = 8 so b = −8, 2a = 6 so a = 3, and 4ac = 4 × 3 × 2 so c = 2.
3. The equation is 3x² − 8x + 2 = 0: (a).

Teacher's Note:
1. Match the given expression with the quadratic formula.
2. The sign inside the formula is −b, so b = −8.

Answer: (c) 2

Explanation:
1. Since x⁰ = 1, the term with x⁰ is the constant term.
2. The constant term is 2.

Teacher's Note:
1. x⁰ = 1 for x ≠ 0.
2. The coefficient of x⁰ is the constant term.

Answer: (a) a ≠ 0

Explanation:
1. A quadratic polynomial has degree 2, so the coefficient of x² must not be zero.
2. Hence a ≠ 0.

Teacher's Note:
1. If a = 0 the expression becomes linear.
2. b and c can be any real numbers.

Answer: (c) both p = 1 and p = −1

Explanation:
1. It is not quadratic when the coefficient of x² is zero.
2. p² − 1 = 0 gives p = ±1.
3. So both p = 1 and p = −1 work.

Teacher's Note:
1. Set the x² coefficient equal to 0.
2. Do not forget the negative root.

Answer: (d) 3, −2

Explanation:
1. x² − x − 6 = x² − 3x + 2x − 6 = (x − 3)(x + 2).
2. Setting each factor to 0 gives x = 3 and x = −2: (d).

Teacher's Note:
1. Split the middle term using numbers with product −6 and sum −1.
2. Check: 3² − 3 − 6 = 0.

Answer: (c) 9

Explanation:
1. f(−4) = 0 gives 16 + 4 − (2k + 2) = 0.
2. 18 − 2k = 0, so k = 9.

Teacher's Note:
1. Substitute x = −4 into the polynomial.
2. Solve the linear equation in k.

Answer: (b) x² + 6x − 4

Explanation:
1. A quadratic polynomial is x² − (α + β)x + αβ.
2. Here it is x² − (−6)x + (−4) = x² + 6x − 4.

Teacher's Note:
1. Use x² − (sum)x + product.
2. Be careful with the sign of the sum.

Answer: (d) −1

Explanation:
1. α + β = −7/2 and αβ = 5/2.
2. α + β + αβ = −7/2 + 5/2 = −1.

Teacher's Note:
1. Use sum = −b/a and product = c/a.
2. Add the sum and the product.

Answer: (a) 3

Explanation:
1. Put y = 1 in the first equation: a + a + 3 = 0, so a = −3/2.
2. Put y = 1 in the second: 1 + 1 + b = 0, so b = −2.
3. ab = (−3/2)(−2) = 3.

Teacher's Note:
1. Substitute y = 1 in each equation separately.
2. Multiply the two values.

Answer: (c) p, −(p + 1)

Explanation:
1. x² + x − p(p + 1) = x² + (p + 1)x − px − p(p + 1) = (x + p + 1)(x − p).
2. The roots are x = p and x = −(p + 1).

Teacher's Note:
1. Split x into (p + 1)x − px.
2. Check: sum of roots = −1 and product = −p(p + 1).

Answer: (b) −m, m + 3

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

Teacher's Note:
1. Write −3x as −(m + 3)x + mx.
2. Check: sum = 3 and product = −m(m + 3).

Answer: (b) −10

Explanation:
1. Substituting x = 2: 4 + 6 + k = 0.
2. So k = −10.

Teacher's Note:
1. A zero makes the polynomial 0.
2. Substitute and solve.

Answer: (a) x² + 5x + 6

Explanation:
1. The polynomial is x² − (sum)x + product = x² − (−5)x + 6.
2. It is x² + 5x + 6.

Teacher's Note:
1. Take care with the sign of the sum.
2. Check: x² + 5x + 6 = (x + 2)(x + 3) has zeroes −2 and −3.

Answer: (a) 2

Explanation:
1. Substitute x = 1/2: 1/4 + k/2 − 5/4 = 0.
2. So k/2 = 1, giving k = 2.

Teacher's Note:
1. A root satisfies the equation.
2. Combine the fractions with denominator 4.

Answer: (d) more than 3

Explanation:
1. One such polynomial is (x + 2)(x − 5) = x² − 3x − 10.
2. Multiplying it by any non-zero constant gives another polynomial with the same zeroes.
3. So there are more than 3 (in fact infinitely many).

Teacher's Note:
1. Multiplying by a constant does not change the zeroes.
2. Cubic and higher polynomials with extra factors also work.

Answer: (d) none of these

Explanation:
1. In a polynomial every power of x must be a non-negative integer.
2. In (a) the power of x is 1/2, in (b) it is −2 and in (c) the powers are 1/2 and −1/2.
3. So none of them is a polynomial: (d).

Teacher's Note:
1. Check every exponent of x.
2. Fractional and negative exponents are not allowed.

Answer: (b) 2

Explanation:
1. The number of zeroes equals the number of points where the graph meets the x-axis.
2. The curve meets the x-axis at two points, so f(x) has 2 zeroes.

Teacher's Note:
1. Count the points where the graph touches or crosses the x-axis.

Answer: (a) \( \frac{7}{2} \)

Explanation:
1. α + β = 7/5 and αβ = 2/5.
2. \( \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{7/5}{2/5} = \frac{7}{2} \).

Teacher's Note:
1. Combine the reciprocals into one fraction.
2. Use sum = −b/a and product = c/a.

Answer: (b) 3

Explanation:
1. A polynomial of degree n has at most n zeroes.
2. For degree 3 the graph meets the x-axis at most 3 times.

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

Answer: (b) ±3

Explanation:
1. Sum = 27/4 and product = 3k²/4.
2. 27/4 = 3k²/4 gives k² = 9, so k = ±3.

Teacher's Note:
1. Equate −b/a and c/a.
2. Both +3 and −3 satisfy k² = 9.

Answer: (a) 8

Explanation:
1. Sum = 2/4 = 1/2 and product = (k − 4)/4.
2. Sum = ½ × product gives 1/2 = (k − 4)/8.
3. So k − 4 = 4 and k = 8.

Teacher's Note:
1. Write sum and product in terms of k.
2. Half of the product is (k − 4)/8.

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

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

Teacher's Note:
1. Substitute the zero into the polynomial.
2. Expand 9(k − 1) carefully.

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

Explanation:
1. x² − (sum)x + product = x² − x − 12 for zeroes −3 and 4.
2. Any constant multiple has the same zeroes; ½(x² − x − 12) = x²/2 − x/2 − 6 is option (c).
3. Option (d) has zeroes 3 and −4.

Teacher's Note:
1. Multiplying by a constant keeps the zeroes.
2. Check each option's sum and product of zeroes.

Answer: (d) a = 0, b = −6

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

Teacher's Note:
1. Use sum = −(coefficient of x) and product = constant term.
2. Check with the zero 2: 4 + 2 − 6 = 0.

Answer: (a) quadratic

Explanation:
1. The highest power of x is 2.
2. So the polynomial is quadratic.

Teacher's Note:
1. The degree gives the type.

Answer: (a) 5

Explanation:
1. x² − 7x + 10 = (x − 5)(x − 2), so the zeroes are 5 and 2.
2. Since a > b, a = 5.

Teacher's Note:
1. Factorise to find the zeroes.
2. a is the larger zero.

Answer: (b) 2

Explanation:
1. The zeroes are 5 and 2 and b < a, so b = 2.

Teacher's Note:
1. b is the smaller zero.

Answer: (c) x² + 7x + 12

Explanation:
1. Sum = −7 and product = 12.
2. Polynomial = x² − (−7)x + 12 = x² + 7x + 12.

Teacher's Note:
1. Use x² − (sum)x + product.

Answer: (c) 3

Explanation:
1. Sum of zeroes = 5, so the other zero = 5 − 2 = 3.

Teacher's Note:
1. Sum of zeroes = −b/a = 5.

Answer: (c) 2

Explanation:
1. The graph cuts the x-axis at two points.
2. So it has 2 zeroes.

Teacher's Note:
1. Count the x-axis crossings.

Answer: (a) 2, −3

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

Teacher's Note:
1. Read the x-values where y = 0.

Answer: (b) x² + x − 6

Explanation:
1. Sum of zeroes = 2 − 3 = −1 and product = −6.
2. Polynomial = x² − (−1)x − 6 = x² + x − 6.

Teacher's Note:
1. Two zeroes means a quadratic.

Answer: (c) quadratic

Explanation:
1. The degree is 2, so it is quadratic.

Teacher's Note:
1. Degree 2 means quadratic.

Answer: (c) both x = 3 and x = −4

Explanation:
1. x² + x − 6 = 6 gives x² + x − 12 = 0, i.e. (x − 3)(x + 4) = 0.
2. So x = 3 or x = −4: (c).

Teacher's Note:
1. Set the polynomial equal to 6 and solve.

Answer: (c) 3

Explanation:
1. The curve cuts the x-axis at three points, so there are 3 zeroes.

Teacher's Note:
1. Count the x-axis crossings.

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

Explanation:
1. Expanding, (x + 4)(x − 2)(x − 3) = x³ − x² − 14x + 24, which has degree 3, so the Assertion is false.
2. The Reason is a true standard result.
3. Assertion false, Reason true: option (d).

Teacher's Note:
1. The degree is the number of linear factors multiplied here, which is 3.
2. Do not confuse the degree with the number of terms.

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

Explanation:
1. The zero polynomial 0 has no highest power of x with a non-zero coefficient, so its degree is not defined. The Assertion is true.
2. A non-zero constant such as 7 = 7x⁰ has degree 0, so the Reason is true.
3. The Reason does not explain the Assertion: option (b).

Teacher's Note:
1. The zero polynomial and non-zero constants are different cases.
2. The Reason talks about non-zero constants only.

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

Explanation:
1. The highest power of x in p(x) is 4 (from 5x⁴), so the degree is 4 and the Assertion is false.
2. The Reason is the definition of degree and is true.
3. Option (d).

Teacher's Note:
1. Look at all terms, not only the first one written.
2. Terms need not be written in descending order.

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

Explanation:
1. Let the zeroes be a and 1/a. Product of zeroes = 4k/(k² + 4) = 1.
2. So k² − 4k + 4 = 0, i.e. (k − 2)² = 0 and k = 2. The Assertion is true.
3. The Reason (factor theorem) is true, but the Assertion is solved with the product of zeroes, not with the Reason: option (b).

Teacher's Note:
1. Reciprocal zeroes have product 1.
2. The factor theorem is not what gives the value of k here.

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

Explanation:
1. x² + 7x + 12 = (x + 3)(x + 4), so the zeroes are −3 and −4, which are real. The Assertion is false.
2. The Reason is true.
3. Option (d).

Teacher's Note:
1. Factorise before judging.
2. Real zeroes are x-coordinates where the graph meets the x-axis.