CBSE Class 10 Mathematics Polynomials MCQs Set 15

Answer: (a) \(-\frac{5}{2}\)

Explanation:
1. Let \(\alpha, \beta\) be the zeroes of \(4x^2 - 5x - 6\): \(\alpha + \beta = \frac{5}{4}\).
2. The zeroes of \(x^2 + px + q\) are \(2\alpha\) and \(2\beta\), so their sum is \(2(\alpha + \beta) = \frac{5}{2}\).
3. The sum of zeroes of \(x^2 + px + q\) is \(-p\), so \(p = -\frac{5}{2}\).

Teacher's Note:
1. Find the sum of the new zeroes first.
2. The coefficient of x gives minus the sum of the zeroes.

Answer: (b) \(-\frac{5}{4}\)

Explanation:
1. \(\alpha + \beta = -\frac{-5}{6} = \frac{5}{6}\) and \(\alpha\beta = \frac{-4}{6} = -\frac{2}{3}\).
2. \(\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{5/6}{-2/3} = -\frac{5}{4}\).

Teacher's Note:
1. Use \(\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta}\).
2. Check the sign of the product carefully.

Answer: (d) more than 3

Explanation:
1. The general form is \(k\left[x^2 - (\alpha + \beta)x + \alpha\beta\right] = k(x^2 + 8x + 15)\), where \(k\) is any real number.
2. Since \(k\) can take infinitely many values, there are more than 3 such polynomials.

Teacher's Note:
1. A non-zero constant multiple keeps the same zeroes.
2. So the number of such polynomials is never a small finite count.

Answer: (d) 0

Explanation:
1. Comparing \(x^2 - 1\) with \(ax^2 + bx + c\) gives \(b = 0\).
2. So \(\alpha + \beta = -\frac{b}{a} = 0\).

Teacher's Note:
1. Read off b from the coefficient of x.
2. A quadratic with no x term has zeroes that add to zero.

Answer: (d) \(a = 0,\ b = -6\)

Explanation:
1. Sum of zeroes: \(2 + (-3) = -1 = -(a + 1)\), so \(a = 0\).
2. Product of zeroes: \(2 \times (-3) = -6 = b\).

Teacher's Note:
1. Use the sum and the product of the zeroes.
2. Substitute the values back to check.

Answer: (d) \(-\frac{2}{3}\)

Explanation:
1. Sum \(= -\frac{2}{k}\) and product \(= \frac{3k}{k} = 3\).
2. \(-\frac{2}{k} = 3\), so \(k = -\frac{2}{3}\).

Teacher's Note:
1. Write both the sum and the product in terms of k.
2. Set them equal and solve.

Answer: (a) \(-2\)

Explanation:
1. Sum of zeroes \(= -2\), product of zeroes \(= 1\).
2. \(\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{-2}{1} = -2\).

Teacher's Note:
1. Use \(\frac{\alpha + \beta}{\alpha\beta}\) directly.
2. Here the product is 1, so the answer equals the sum.

Answer: (c) \(\frac{13}{2}\)

Explanation:
1. Here \(a = 2\), \(b = -13\), \(c = 6\).
2. \(\alpha + \beta = -\frac{b}{a} = \frac{13}{2}\).

Teacher's Note:
1. The sum of zeroes is \(-\frac{b}{a}\).
2. Be careful with the sign of b.

Answer: (d) \(-\frac{6}{5}\)

Explanation:
1. Substitute \(x = 2\): \(4k + 6 + k = 0\), so \(5k = -6\).
2. \(k = -\frac{6}{5}\).

Teacher's Note:
1. A zero makes the polynomial equal to zero.
2. Substitute the value and solve for k.

Answer: (d) 3

Explanation:
1. A cubic polynomial has degree 3.
2. A polynomial of degree n has at most n zeroes, so at most 3.

Teacher's Note:
1. Use the degree of the polynomial.
2. The degree is the highest power of x.

Answer: (b) \(-10\)

Explanation:
1. Substitute \(x = 2\): \(4 + 6 + k = 0\), so \(k = -10\).

Teacher's Note:
1. Substitute the zero into the polynomial.
2. Check: 4 + 6 − 10 = 0.

Answer: (a) \(x^2 + 5x + 6\)

Explanation:
1. \(p(x) = x^2 - (\alpha + \beta)x + \alpha\beta = x^2 + 5x + 6\).

Teacher's Note:
1. Use \(x^2 - (\text{sum})x + \text{product}\).
2. The sum here is negative, so the middle term is positive.

Answer: (a) 3

Explanation:
1. Let the zeroes be \(\alpha\) and \(\frac{1}{\alpha}\). Their product is \(\alpha \cdot \frac{1}{\alpha} = 1\).
2. Product of zeroes \(= \frac{k}{3} = 1\), so \(k = 3\).

Teacher's Note:
1. Two reciprocals always multiply to 1.
2. Product of zeroes is \(\frac{c}{a}\).

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

Explanation:
1. Put \(x = -m\): \((-m)^2 - 3(-m) - m(m + 3) = m^2 + 3m - m^2 - 3m = 0\). So \(-m\) is a zero.
2. Put \(x = m + 3\): \((m + 3)^2 - 3(m + 3) - m(m + 3) = (m + 3)[m + 3 - 3 - m] = 0\). So \(m + 3\) is a zero.

Teacher's Note:
1. Test each option by substituting it into the polynomial.
2. Two zeroes are enough to identify the answer.

Answer: (d) 1

Explanation:
1. \(x^2 - 1 = (x - 1)(x + 1)\) and \(x^2 - 2x + 1 = (x - 1)^2\).
2. Both vanish when \(x = 1\).

Teacher's Note:
1. Factorise both polynomials.
2. A common zero is a value that makes both factors zero.

Answer: (a) \(\frac{15}{4}\)

Explanation:
1. \(\alpha + \beta = \frac{1}{1} = 1\) and \(\alpha\beta = \frac{-4}{1} = -4\).
2. \(\frac{1}{\alpha} + \frac{1}{\beta} - \alpha\beta = \frac{\alpha + \beta}{\alpha\beta} - \alpha\beta = \frac{1}{-4} + 4 = \frac{15}{4}\).

Teacher's Note:
1. Find \(\alpha + \beta\) and \(\alpha\beta\) first.
2. Remember that \(-\alpha\beta\) becomes \(+4\).

Answer: (a) positive for all the real numbers

Explanation:
1. For \(x = 0\) or \(x = 1\), the value is \(1 > 0\).
2. For \(x < 0\), every term is positive, so the value is greater than 0. Hence the value is positive for all real \(x\).

Teacher's Note:
1. Test a few values to see the pattern.
2. Look at the sign of each term when x is negative.

Answer: (b) −2

Explanation:
1. \(x^2 + 4x + 2 = (x + 2)^2 - 2\).
2. \((x + 2)^2\) is never negative, and it is 0 when \(x = -2\). So the lowest value is \(-2\).

Teacher's Note:
1. Complete the square.
2. A square is never negative, so its smallest value is 0.

Answer: (c) 4

Explanation:
1. Sum of zeroes \(= \frac{3k}{2}\). So \(6 = \frac{3k}{2}\).
2. \(k = \frac{12}{3} = 4\).

Teacher's Note:
1. For a cubic \(ax^3 + bx^2 + \ldots\), the sum of zeroes is \(-\frac{b}{a}\).
2. Here b = −3k and a = 2.

Answer: (d) \(\pm 18\)

Explanation:
1. Sum of zeroes \(= -p\) and product \(= 45\).
2. \((\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta\), so \(p^2 - 180 = 144\).
3. \(p^2 = 324\), so \(p = \pm 18\).

Teacher's Note:
1. Use \((\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta\).
2. Remember that the answer has both signs.

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

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

Teacher's Note:
1. Substitute the zero into the polynomial.
2. Collect the terms in k at the end.

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

Explanation:
1. Sum of zeroes \(= 1\) and product \(= -12\).
2. \(p(x) = x^2 - x - 12\). Dividing by 2 gives \(\frac{x^2}{2} - \frac{x}{2} - 6\), which has the same zeroes.

Teacher's Note:
1. Use \(x^2 - (\text{sum})x + \text{product}\).
2. Any non-zero multiple of this polynomial also has the same zeroes.

Answer: (d) \(a = 0,\ b = -6\)

Explanation:
1. This is the same as question 5 in this set. Put \(x = 2\) and \(x = -3\) into the polynomial.
2. From \(x = 2\): \(2a + b = -6\). From \(x = -3\): \(3a - b = 6\). Adding gives \(a = 0\), so \(b = -6\).

Teacher's Note:
1. Use two equations and solve them together.
2. Check that a = 0, b = −6 satisfies both.

Answer: (b) both negative

Explanation:
1. Product of zeroes \(= 127 > 0\), so both zeroes have the same sign.
2. Sum of zeroes \(= -99 < 0\), so both zeroes are negative.

Teacher's Note:
1. Product positive means the signs match.
2. Sum negative means both are negative.

Answer: (a) cannot both be positive

Explanation:
1. Sum of zeroes \(= -k\) and product \(= k\).
2. If \(k < 0\), the product is negative, so the zeroes have opposite signs. If \(k > 0\), the sum is negative and the product is positive, so both zeroes are negative. In either case they cannot both be positive.

Teacher's Note:
1. Split into the cases k < 0 and k > 0.
2. Compare the sign of the sum and the product in each case.

Answer: (c) c and a have the same sign

Explanation:
1. Zeroes are equal, so they have the same sign. Then \(\alpha\beta > 0\), i.e. \(\frac{c}{a} > 0\).
2. So \(c\) and \(a\) have the same sign.

Teacher's Note:
1. Equal zeroes multiply to a positive number.
2. Use \(\alpha\beta = \frac{c}{a}\).

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

Explanation:
1. Let the zeroes be \(\alpha\) and \(-\alpha\). Sum \(= 0\), so there is no linear term.
2. Product \(= -\alpha^2 < 0\), so the constant term is negative.

Teacher's Note:
1. A zero and its negative add to zero.
2. The product of a number and its negative is negative.

Answer: (d) Graph (d)

Explanation:
1. A quadratic polynomial has at most two zeroes, so its graph meets the x-axis at most twice.
2. Graph (d) cuts the x-axis at three points, so it cannot be the graph of a quadratic polynomial.

Teacher's Note:
1. Count the points where the curve meets the x-axis.
2. Graphs (a), (b) and (c) meet the axis at most twice.

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

Explanation:
1. A quadratic can touch the x-axis at one point when it has equal zeroes (discriminant 0). So (A) is false.
2. Reason (R) is a true statement, so (R) is true.

Teacher's Note:
1. A graph touching the axis at one point can still be quadratic.
2. Check the discriminant when a question mentions one touching point.

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

Explanation:
1. The highest power of x in \(p(x)\) is 4, so the degree is 4. Assertion (A) is false.
2. Reason (R) is correct.

Teacher's Note:
1. Find the largest power of x before the statement about degree.
2. The degree is the highest power, here 4 from the term 8x⁴.

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

Explanation:
1. Irrational zeros of a polynomial with rational coefficients occur in conjugate pairs, so the other zero is \(2 + \sqrt{3}\).
2. Reason (R) is the correct explanation of Assertion (A).

Teacher's Note:
1. Conjugate pairs explain why irrational zeros come in pairs.
2. Both statements are true here.

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

Explanation:
1. Product of zeroes \(= \frac{4k}{k^2 + 4} = 1\), so \(k^2 - 4k + 4 = 0\), giving \(k = 2\). Assertion (A) is true.
2. Reason (R) is true (the factor theorem), but it does not explain why k = 2.

Teacher's Note:
1. Use the product of zeroes to find k.
2. A true reason need not explain the assertion.

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

Explanation:
1. \(x^3 + x = x(x^2 + 1)\). Since \(x^2 + 1 \neq 0\) for all real \(x\), there is only one real zero (\(x = 0\)). Assertion (A) is true.
2. Reason (R) is false: a polynomial of degree n has at most n real zeroes, not exactly n.

Teacher's Note:
1. Factorise to find the real zeroes.
2. Degree n gives at most n zeroes, not always n.

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

Explanation:
1. Zeroes are equal in magnitude and opposite in sign, so their sum is 0. Sum \(= 2k = 0\), so \(k = 0\), not \(\frac{1}{2}\). Assertion (A) is false.
2. Reason (R) is true.

Teacher's Note:
1. Opposite zeroes add to zero.
2. Check the value of k against the sum of zeroes.