CBSE Class 10 Mathematics Polynomials MCQs Set 17

Answer: (a) 2

Explanation:
1. If the two zeroes are reciprocal of each other, their product is 1.
2. Product of zeroes = \( \frac{k+2}{k^2} = 1 \), so \( k^2 - k - 2 = 0 \), i.e. \( (k-2)(k+1) = 0 \).
3. Since \( k > 0 \), \( k = 2 \).

Teacher's Note:
1. Reciprocal zeroes always have product 1.
2. Reject the root that does not satisfy the given condition on k.

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

Explanation:
1. Sum of the zeroes = \( -\frac{-8}{2} = 4 \).
2. Other zero = \( 4 - \frac{5}{2} = \frac{3}{2} \).

Teacher's Note:
1. Use sum of zeroes = −b/a; the value of m is not needed.
2. The sum of the two zeroes is fixed by the coefficients of x² and x.

Answer: (b) 14

Explanation:
1. \( f(a) = a \) gives \( a^2 - 27a + 196 = a \), i.e. \( a^2 - 28a + 196 = 0 \).
2. This is \( (a - 14)^2 = 0 \), so \( a = 14 \).

Teacher's Note:
1. Substituting x = a and equating to a gives a quadratic equation in a.
2. A perfect square has one repeated root.

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 the given value of x in each polynomial separately.
2. Take care of the signs while adding the two values.

Answer: (c) 15, −5

Explanation:
1. Split the middle term: \( x^2 - 15x + 5x - 75 = x(x - 15) + 5(x - 15) = (x - 15)(x + 5) \).
2. So \( p(x) = 0 \) when \( x = 15 \) or \( x = -5 \).

Teacher's Note:
1. Look for two numbers with product −75 and sum −10.
2. Zeroes are the values of x that make the polynomial 0.

Answer: (c) 3

Explanation:
1. Sum \( a + b = -\frac{-4}{2} = 2 \) and product \( ab = \frac{3}{2} \).
2. \( a^2b + ab^2 = ab(a + b) = \frac{3}{2} \times 2 = 3 \).

Teacher's Note:
1. Factorise the expression as ab(a + b) before using sum and product.
2. The zeroes need not be real for this relation to hold.

Answer: (b) \( -m, m + 3 \)

Explanation:
1. \( x^2 - 3x - m(m+3) = x^2 - (m+3)x + mx - m(m+3) \).
2. \( = x[x - (m+3)] + m[x - (m+3)] = (x + m)(x - (m+3)) \).
3. So the zeroes are \( -m \) and \( m + 3 \).

Teacher's Note:
1. Write −3 as −(m + 3) + m and group the terms.
2. Check with sum of zeroes: −m + m + 3 = 3.

Answer: (d) 1

Explanation:
1. \( x^2 - 1 = (x - 1)(x + 1) \) vanishes at \( x = 1 \) and \( x = -1 \).
2. \( x^2 - 2x + 1 = (x - 1)^2 \) vanishes only at \( x = 1 \).
3. The common value is \( x = 1 \).

Teacher's Note:
1. Find the zeroes of each polynomial and pick the common one.
2. A perfect square has a single repeated zero.

Answer: (b) −2

Explanation:
1. \( x^2 + 4x + 2 = (x + 2)^2 - 2 \).
2. \( (x + 2)^2 \ge 0 \), so the least value is \( -2 \), at \( x = -2 \).

Teacher's Note:
1. Completing the square shows the minimum of an upward-opening parabola.
2. The square term is never negative.

Answer: (a) 18

Explanation:
1. A quadratic with the single zero \( -1 \) has it twice: \( p(x) = k(x + 1)^2 \).
2. \( p(-2) = k(-1)^2 = k = 2 \), so \( p(x) = 2(x + 1)^2 \).
3. \( p(2) = 2 \times 9 = 18 \).

Teacher's Note:
1. Only one zero of a quadratic means the zero is repeated.
2. Use the given value to find the constant multiplier k.

Answer: (b) −10

Explanation:
1. Since 2 is a zero, \( 2^2 + 3(2) + k = 0 \).
2. \( 10 + k = 0 \), so \( k = -10 \).

Teacher's Note:
1. A zero of a polynomial satisfies p(x) = 0.
2. Substitute the zero and solve for the unknown.

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

Explanation:
1. Put \( x = -3 \): \( 9(k - 1) - 3k + 1 = 0 \).
2. \( 9k - 9 - 3k + 1 = 0 \), so \( 6k = 8 \) and \( k = \frac{4}{3} \).

Teacher's Note:
1. Substitute the zero into the polynomial and equate to 0.
2. Expand carefully: (−3)² = 9.

Answer: (b) both negative

Explanation:
1. Sum of zeroes = −99 < 0 and product = 127 > 0.
2. A positive product means the zeroes have the same sign, and a negative sum means that sign is negative.
3. The discriminant \( 99^2 - 4(127) > 0 \), so the zeroes are real and distinct. Both are negative.

Teacher's Note:
1. Use the sign of the sum and the product to decide the signs of the zeroes.
2. Check that the discriminant is positive so that real zeroes exist.

Answer: (d) \( \sqrt{5}x^2 + \frac{1}{2}x + \frac{3}{7} \)

Explanation:
1. A polynomial has only whole-number powers of the variable.
2. \( \frac{2}{x} = 2x^{-1} \), \( \sqrt{x} = x^{1/2} \) and \( \frac{2}{3x-4} \) all fail this test.
3. In (d) the powers are 2, 1, 0, so it is a polynomial.

Teacher's Note:
1. Irrational or fractional coefficients are allowed; irrational or negative powers of x are not.
2. A variable in the denominator makes the expression a non-polynomial.

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

Explanation:
1. The zero polynomial has no non-zero term, so its degree is not defined; (a) is false.
2. A non-zero constant polynomial has degree 0, so (c) is false.
3. A polynomial of degree n has at most n zeroes, so (d) is false.
4. Hence only (b) is correct.

Teacher's Note:
1. The degree of the zero polynomial is undefined, not 0.
2. Degree n means at most n zeroes, not exactly n.

Answer: (c) \( x^3 - 3x + 1 \)

Explanation:
1. \( \frac{1}{x} = x^{-1} \), \( \sqrt{x} = x^{1/2} \) and \( x^{3/2} \) have powers that are not whole numbers.
2. Only \( x^3 - 3x + 1 \) has all powers as whole numbers.

Teacher's Note:
1. All powers of x must be 0, 1, 2, 3, and so on.
2. Rewrite roots and reciprocals as powers to test them.

Answer: (c) \( a \) is a non-zero real number and \( b, c \) are any real numbers

Explanation:
1. The coefficients of a quadratic polynomial are real numbers.
2. The coefficient \( a \) of \( x^2 \) must be non-zero, otherwise the degree is not 2.
3. So (c) is the correct statement.

Teacher's Note:
1. a ≠ 0 is what makes the polynomial quadratic.
2. b and c may be zero.

Answer: (d) \( D = 0 \)

Explanation:
1. The roots are \( \frac{-b \pm \sqrt{D}}{2a} \).
2. They are equal only when \( \sqrt{D} = 0 \), i.e. \( D = 0 \).

Teacher's Note:
1. D > 0 gives two distinct real roots; D < 0 gives none.
2. D = 0 means the parabola touches the x-axis.

Answer: (b) \( \frac{1}{4} \)

Explanation:
1. Product of zeroes \( = \alpha \cdot \frac{1}{\alpha} = 1 \).
2. Also product \( = \frac{8k}{2} = 4k \), so \( 4k = 1 \) and \( k = \frac{1}{4} \).

Teacher's Note:
1. Zeroes α and 1/α always have product 1.
2. Product of zeroes = c/a.

Answer: (c) neither touches nor intersects the x-axis

Explanation:
1. Zeroes of the polynomial are the x-coordinates where the graph meets the x-axis.
2. \( x^2 + 1 = 0 \) has no real solution since \( x^2 + 1 \ge 1 \).
3. So the parabola lies wholly above the x-axis.

Teacher's Note:
1. No real zeroes means the graph does not meet the x-axis.
2. Here the discriminant is 0 − 4 = −4 < 0.

Answer: (c) \( k\left(x^2 + px - \frac{1}{p}\right) \)

Explanation:
1. A quadratic polynomial with sum S and product P of zeroes is \( k[x^2 - Sx + P] \).
2. Here \( S = -p \) and \( P = -\frac{1}{p} \), so it is \( k\left[x^2 + px - \frac{1}{p}\right] \).

Teacher's Note:
1. Remember the form k[x² − (sum)x + product].
2. Watch the sign when the sum is negative.

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

Explanation:
1. Reciprocal zeroes have product 1: \( \frac{4k}{k^2 + 4} = 1 \), so \( k^2 - 4k + 4 = 0 \), i.e. \( (k - 2)^2 = 0 \) and \( k = 2 \). The Assertion is true.
2. The Reason is also true (a factor gives a zero), but it is not what is used to find k, so it does not explain the Assertion: option (b).

Teacher's Note:
1. Here the key idea is product of zeroes = 1.
2. A true Reason that is not the method used is not the explanation.

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

Explanation:
1. \( x^3 + x = x(x^2 + 1) \) and \( x^2 + 1 \neq 0 \) for real x, so the only real zero is 0. The Assertion is true.
2. A polynomial of degree n has at most n real zeroes, not exactly n, so the Reason is false: option (c).

Teacher's Note:
1. At most n zeroes, not always n.
2. Factorise to count the real zeroes.

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

Explanation:
1. The zero polynomial has no non-zero term, so its degree is not defined: the Assertion is true.
2. A non-zero constant such as 5 = 5x⁰ has degree 0: the Reason is true, but it says nothing about the zero polynomial, so it does not explain the Assertion: option (b).

Teacher's Note:
1. Constant 0 is the special case; other constants have degree 0.
2. Both statements are true but separate.

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: the Assertion is true.
2. For the Reason, sum \( = 4 \) and product \( = 4 - 3 = 1 \), so the polynomial is \( x^2 - 4x + 1 \), not \( x^2 - 4x + 7 \): the Reason is false: option (c).

Teacher's Note:
1. Check the product: (2 + √3)(2 − √3) = 4 − 3 = 1.
2. Use x² − (sum)x + product.

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

Explanation:
1. \( x^2 + 4x - 5 = (x + 5)(x - 1) \), so it has the two zeroes −5 and 1: the Assertion is true.
2. The Reason is also true, and it states the general fact (at most two zeroes) of which the Assertion is a case: option (a).

Teacher's Note:
1. A quadratic has 0, 1 or 2 real zeroes; this one has 2.
2. The Reason gives the upper limit, the Assertion shows it is reached.