CBSE Class 10 Mathematics Polynomials MCQs Set 22

Answer: (a) 9

Explanation:
1. Sum of zeroes \( = \frac{k}{3} = 3 \).
2. So \( k = 9 \).

Teacher's Note:
1. Sum = -b/a with b = -k.
2. Product (6/3 = 2) does not depend on k.

Answer: (b) at most 3

Explanation:
1. The graph of a cubic meets the x-axis at most 3 times.
2. It can have 1, 2 or 3 real zeroes, so the correct statement is 'at most 3'.

Teacher's Note:
1. A polynomial of degree n has at most n real zeroes.
2. x³ + 1 has only one real zero.

Answer: (c) 756

Explanation:
1. \( A + B = 12 \) and \( AB = 27 \).
2. \( A^3 + B^3 = (A + B)^3 - 3AB(A + B) = 1728 - 3 \times 27 \times 12 = 1728 - 972 = 756 \).

Teacher's Note:
1. Use the identity for the sum of cubes in terms of sum and product.
2. The zeroes are 3 and 9: 27 + 729 = 756.

Answer: (d) \( \frac{20}{3} \)

Explanation:
1. \( \alpha + \beta = 12 \) and \( \alpha\beta = 32 \).
2. \( \alpha^2 + \beta^2 = 144 - 64 = 80 \).
3. \( \frac{80}{12} = \frac{20}{3} \).

Teacher's Note:
1. The roots are 4 and 8, which also gives 80/12.
2. Simplify the fraction at the end.

Answer: (a) -1, -1

Explanation:
1. The factor for these zeroes is \( x^2 - \frac{5}{3} \), i.e. \( 3x^2 - 5 \).
2. Dividing: \( 3x^4 + 6x^3 - 2x^2 - 10x - 5 = (3x^2 - 5)(x^2 + 2x + 1) \).
3. \( x^2 + 2x + 1 = (x + 1)^2 \), so the other zeroes are -1 and -1.

Teacher's Note:
1. Divide by the quadratic factor formed from the known pair of zeroes.
2. A repeated zero is counted twice.

Answer: (b) ax² - bx + c = 0

Explanation:
1. New sum \( = -(\alpha + \beta) = \frac{b}{a} \) and new product \( = \alpha\beta = \frac{c}{a} \).
2. Equation: \( x^2 - \frac{b}{a}x + \frac{c}{a} = 0 \), i.e. \( ax^2 - bx + c = 0 \).

Teacher's Note:
1. Replacing x by -x gives the same result: a(-x)² + b(-x) + c.
2. Only the sign of the middle term changes.

Answer: (d) 8

Explanation:
1. Let the roots be r and 2r. Sum: \( 3r = 2 \), so \( r = \frac{2}{3} \).
2. Product: \( 2r^2 = \frac{8}{9} = \frac{k}{9} \), so \( k = 8 \).

Teacher's Note:
1. Use both the sum and the product conditions.
2. Check: 9x² - 18x + 8 = (3x - 2)(3x - 4) with roots 2/3 and 4/3.

Answer: (c) 3

Explanation:
1. Product of zeroes = 1, so \( \frac{6a}{a^2 + 9} = 1 \).
2. \( a^2 - 6a + 9 = 0 \), so \( (a - 3)^2 = 0 \) and \( a = 3 \).

Teacher's Note:
1. Reciprocal zeroes multiply to 1.
2. Product of zeroes = c/a.

Answer: (b) 16x² - 9x + 1

Explanation:
1. \( \alpha + \beta = 3 \), \( \alpha\beta = -2 \). Then \( (2\alpha + \beta) + (2\beta + \alpha) = 3(\alpha + \beta) = 9 \).
2. \( (2\alpha + \beta)(2\beta + \alpha) = 2\alpha^2 + 2\beta^2 + 5\alpha\beta = 2(9 + 4) - 10 = 16 \).
3. New sum \( = \frac{9}{16} \), new product \( = \frac{1}{16} \), so the polynomial is \( 16x^2 - 9x + 1 \).

Teacher's Note:
1. Find the sum and product of the new zeroes by combining the fractions.
2. Express α² + β² through (α + β)² - 2αβ.

Answer: (a) \( a = \frac{1}{2},\ c = 5 \)

Explanation:
1. \( m + n = \frac{5}{a} = 10 \), so \( a = \frac{1}{2} \).
2. \( mn = \frac{c}{a} = 10 \), so \( c = 5 \).

Teacher's Note:
1. Use sum = -b/a with b = -5.
2. Then product gives c.

Answer: (c) \( -\frac{1}{3},\ \frac{1}{2} \)

Explanation:
1. \( 6x^2 - x - 1 = 6x^2 - 3x + 2x - 1 = (2x - 1)(3x + 1) \).
2. Zeroes: \( x = \frac{1}{2} \) and \( x = -\frac{1}{3} \).
3. Check: sum \( = \frac{1}{6} = -\frac{b}{a} \); product \( = -\frac{1}{6} = \frac{c}{a} \).

Teacher's Note:
1. Split the middle term using two numbers whose product is -6 and sum is -1.
2. Verify with sum and product of zeroes.

Answer: (a) \( \pm\frac{9}{7} \)

Explanation:
1. \( 49x^2 - 81 = (7x)^2 - 9^2 = (7x - 9)(7x + 9) \).
2. Zeroes: \( x = \pm\frac{9}{7} \).

Teacher's Note:
1. Use a² - b² = (a + b)(a - b).
2. Sum of zeroes is 0 because there is no x term.

Answer: (d) x² - 6x + 2

Explanation:
1. Sum \( = 6 \) and product \( = 9 - 7 = 2 \).
2. Polynomial: \( x^2 - 6x + 2 \).

Teacher's Note:
1. The product is a difference of squares: (3 + √7)(3 - √7) = 9 - 7.
2. Irrational parts cancel in the sum.

Answer: (b) \( 3x^2 + 10\sqrt{3}x + 21 \)

Explanation:
1. \( k\left(x^2 + \frac{10}{\sqrt{3}}x + 7\right) \).
2. Taking \( k = 3 \): \( 3x^2 + \frac{30}{\sqrt{3}}x + 21 = 3x^2 + 10\sqrt{3}x + 21 \).

Teacher's Note:
1. Rationalise: 30/√3 = 10√3.
2. Choose k to remove the denominator.

Answer: (b) 2

Explanation:
1. Reciprocal roots have product 1.
2. Product of roots \( = \frac{b}{2} = 1 \), so \( b = 2 \).

Teacher's Note:
1. Product of roots is c/a, here b/2.
2. The value of a does not matter.

Answer: (d) \( 2,\ -\frac{1}{2} \)

Explanation:
1. Divide by \( x + 3 \): \( 2x^3 + 3x^2 - 11x - 6 = (x + 3)(2x^2 - 3x - 2) \).
2. \( 2x^2 - 3x - 2 = (2x + 1)(x - 2) \).
3. The other zeroes are 2 and \( -\frac{1}{2} \).

Teacher's Note:
1. Use the known zero to get a factor, then divide.
2. Product of all zeroes = 6/2 = 3; (-3)(2)(-1/2) = 3.

Answer: (a) a = 5, b = 1

Explanation:
1. \( x^4 - 3x^2 + 5x + 3 = (x^2 - 2)(x^2 - 1) + 5x + 1 \).
2. So the remainder is \( 5x + 1 \), i.e. a = 5 and b = 1.

Teacher's Note:
1. Long division: x⁴ ÷ x² gives x², then -x² ÷ x² gives -1.
2. Check by multiplying back.

Answer: (c) 1 and -2

Explanation:
1. \( \alpha + \beta = -\frac{6}{R} \) and \( \alpha\beta = \frac{9}{R} \).
2. \( \alpha^2 + \beta^2 = \frac{36}{R^2} - \frac{18}{R} = 18 \) gives \( R^2 + R - 2 = 0 \).
3. \( (R + 2)(R - 1) = 0 \), so \( R = 1 \) or \( R = -2 \).

Teacher's Note:
1. Multiply by R² to clear fractions.
2. Check R = -2: sum = 3, product = -9/2, so squares add to 9 + 9 = 18.

Answer: (d) 12

Explanation:
1. \( \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = 64 - 2k = 40 \).
2. So \( k = 12 \).

Teacher's Note:
1. Substitute the sum 8 first.
2. Check: x² - 8x + 12 has zeroes 2 and 6; 4 + 36 = 40.

Answer: (b) -16

Explanation:
1. \( \alpha + \beta = 6 \), so \( 2\alpha + 2\beta = 12 \).
2. Subtracting from \( 3\alpha + 2\beta = 20 \) gives \( \alpha = 8 \) and \( \beta = -2 \).
3. \( k = \alpha\beta = -16 \).

Teacher's Note:
1. Use the sum to eliminate one zero.
2. Product of the zeroes equals k.

Answer: (a) 25x² + 4x + 4

Explanation:
1. \( \alpha + \beta = 2 \), \( \alpha\beta = \frac{5}{2} \), so \( \alpha^2 + \beta^2 = 4 - 5 = -1 \).
2. New sum \( = \frac{\alpha^2 + \beta^2}{(\alpha\beta)^2} = -\frac{4}{25} \) and new product \( = \frac{4}{25} \).
3. Polynomial: \( x^2 + \frac{4}{25}x + \frac{4}{25} \), i.e. \( 25x^2 + 4x + 4 \).

Teacher's Note:
1. Write 1/α² + 1/β² = (α² + β²)/(αβ)².
2. Here α² + β² is negative because the zeroes are not real.

Answer: (c) \( -\frac{1}{5} \)

Explanation:
1. \( \alpha + \beta = -\frac{3}{5} \) and \( \alpha\beta = \frac{k}{5} \).
2. \( \frac{9}{25} - \frac{2k}{5} = \frac{11}{25} \), so \( -\frac{2k}{5} = \frac{2}{25} \) and \( k = -\frac{1}{5} \).

Teacher's Note:
1. Subtract carefully: 11/25 - 9/25 = 2/25.
2. Check: with k = -1/5 the product of zeroes is -1/25.

Answer: (a) 2

Explanation:
1. Sum \( = -\frac{2a + 3}{a + 1} = -1 \), so \( 2a + 3 = a + 1 \) and \( a = -2 \).
2. The polynomial becomes \( -x^2 - x - 2 \); product \( = \frac{-2}{-1} = 2 \).

Teacher's Note:
1. Find a from the sum first.
2. Then use product = c/a.

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

Explanation:
1. Put \( x = -\sqrt{2} \): \( 2k + 2 + 1 = 0 \).
2. So \( k = -\frac{3}{2} \).

Teacher's Note:
1. By the Factor Theorem the value at x = -√2 must be 0.
2. (-√2)² = 2 and -√2 × (-√2) = +2.

Answer: (c) -5

Explanation:
1. The polynomial is \( x^2 - px - (p + k) \), so \( \alpha + \beta = p \) and \( \alpha\beta = -(p + k) \).
2. \( (\alpha + 1)(\beta + 1) = \alpha\beta + \alpha + \beta + 1 = -p - k + p + 1 = 1 - k \).
3. \( 1 - k = 6 \), so \( k = -5 \).

Teacher's Note:
1. Expand (α + 1)(β + 1) before substituting.
2. The p terms cancel.