CBSE Class 10 Mathematics Polynomials MCQs Set 20

Answer: (b) 8

Explanation:
1. The zeroes are α and \( \frac{1}{\alpha} \), so their product is α × \( \frac{1}{\alpha} \) = 1.
2. For 4x² − 2x + (k − 4), the product of the zeroes = \( \frac{k-4}{4} \).
3. So \( \frac{k-4}{4} = 1 \), giving k − 4 = 4 and k = 8.

Teacher's Note:
1. Reciprocal zeroes always have product 1.
2. Substitute k = 8 to check: 4x² − 2x + 4 has product 1.

Answer: (d) x² + 3x − 40

Explanation:
1. Sum of zeroes = 5 + (−8) = −3 and product = 5 × (−8) = −40.
2. The polynomial is x² − (sum)x + product = x² + 3x − 40.

Teacher's Note:
1. Use x² − (sum of zeroes)x + (product of zeroes).
2. Check: (x − 5)(x + 8) = x² + 3x − 40.

Answer: (d) infinitely many

Explanation:
1. A polynomial with zeroes 1 and −2 has the factor (x − 1)(x + 2).
2. Any non-zero constant multiple k(x − 1)(x + 2) also has exactly these zeroes, and k can be any non-zero number.
3. So there are infinitely many such polynomials.

Teacher's Note:
1. The zeroes do not decide the leading coefficient.
2. For example x² + x − 2 and 3x² + 3x − 6 have the same zeroes.

Answer: (a) cannot both be positive

Explanation:
1. For x² + kx + k, the sum of the zeroes = −k and the product = k.
2. If both zeroes were positive, the sum would be positive (−k > 0, so k < 0) but the product would then be negative (k < 0). That is a contradiction.
3. So the zeroes cannot both be positive. Both negative is possible, for example k = 4 gives (x + 2)².

Teacher's Note:
1. Use the sum and the product together.
2. Try k = 4 to see that equal negative zeroes do occur.

Answer: (a) both positive

Explanation:
1. Sum of zeroes = 34 and product = 288, both positive.
2. So both zeroes have the same sign and, as their sum is positive, both are positive.
3. Check: x² − 34x + 288 = (x − 16)(x − 18), zeroes 16 and 18.

Teacher's Note:
1. Positive product means the same signs; the sign of the sum then decides.
2. Factorise to confirm.

Answer: (b) −36

Explanation:
1. Since −4 is a zero, put x = −4: (−4)² − 5(−4) + k = 0.
2. 16 + 20 + k = 0, so k = −36.

Teacher's Note:
1. A zero satisfies the polynomial.
2. Check: x² − 5x − 36 = (x − 9)(x + 4) has zero −4.

Answer: (c) If the zeroes of a quadratic polynomial ax² + bx + c are both negative, then a, b and c all have the same sign.

Explanation:
1. If both zeroes are negative, their sum \( -\frac{b}{a} \) is negative and their product \( \frac{c}{a} \) is positive. So \( \frac{b}{a} > 0 \) and \( \frac{c}{a} > 0 \), which means a, b and c have the same sign. (c) is true.
2. A quadratic polynomial has one variable and degree 2, so (a) is false.
3. Equal zeroes need k² − 4k = 0, so k = 4, which is even. (b) is false.
4. A non-zero constant has degree 0, so (d) is false.

Teacher's Note:
1. Translate every statement into sum and product of the zeroes.
2. The degree of a non-zero constant polynomial is 0.

Answer: (d) The degree of a quadratic polynomial is less than or equal to 2.

Explanation:
1. A quadratic polynomial has degree exactly 2, so saying that it is 'less than or equal to 2' is wrong in general. (d) is false.
2. A cubic graph always cuts the x-axis at least once, so (a) is true.
3. A polynomial of degree n has at most n zeroes, so (b) is true.
4. A cubic can touch the axis at one point and cut it at another, giving exactly two points. (c) is true.

Teacher's Note:
1. Degree is exact for a quadratic: a ≠ 0 forces degree 2.
2. A cubic can have 1, 2 or 3 real zeroes.

Answer: (b) The degree of the zero polynomial is not defined.

Explanation:
1. The zero polynomial 0 can be written as 0·x^n for every n, so no single degree can be assigned.
2. Hence the degree of the zero polynomial is not defined.

Teacher's Note:
1. Do not confuse the zero polynomial with a non-zero constant, whose degree is 0.
2. Only the zero polynomial has an undefined degree.