CBSE Class 10 Mathematics Polynomials MCQs Set 28

Answer: (b) 3t² + 11

Explanation:
1. \( t^{6} = (t^{3} + 1)(t^{3} - 1) + 1 \), so \( t^{6} + 3t^{2} + 10 = (t^{3} + 1)(t^{3} - 1) + 3t^{2} + 11 \).
2. The remainder \( 3t^{2} + 11 \) has degree 2, which is less than 3, so the division stops.

Teacher's Note:
1. Rewrite the highest power in terms of the divisor to find the remainder quickly.
2. The remainder must have lower degree than the divisor.

Answer: (b) \( \frac{n-q}{m-p} \)

Explanation:
1. \( x = -a \) is a zero of both: \( a^{2} \) − pa \( + q = 0 \) and \( a^{2} \) − ma \( + n = 0 \).
2. Subtract: \( a(m - p) = n - q \), so \( a = (n - q)/(m - p) \).

Teacher's Note:
1. Substitute the zero of the common factor into both polynomials.
2. Subtracting removes the \( a^{2} \) term.

Answer: (c) −3, −8

Explanation:
1. \( x = 2 \): \( 16 + 4a + 2b + 12 = 0 \), so \( 2a + b = -14 \).
2. \( x = -2 \): \( -16 + 4a - 2b + 12 = 0 \), so \( 2a - b = 2 \).
3. Adding: \( 4a = -12 \), so \( a = -3 \) and \( b = -8 \).

Teacher's Note:
1. A factor \( x^{2} - 4 = (x - 2)(x + 2) \) gives two conditions.
2. Solve the pair of linear equations in a and \( b \).

Answer: (d) 50

Explanation:
1. Put \( x = -1 \): \( (-1)^{51} + 51 = -1 + 51 = 50 \).

Teacher's Note:
1. An odd power of −1 is −1.
2. Use the remainder theorem: the remainder on dividing by \( (x - k) \) is the value at \( x = k \).

Answer: (b) \( \frac{18}{127} \)

Explanation:
1. \( R_{1} = 64a + 48 - 3 = 64a + 45 \) and \( R_{2} = 128 - 20 + a = a + 108 \).
2. \( 2R_{1} - R_{2} = 128a + 90 - a - 108 = 127a - 18 = 0 \), so \( a = 18/127 \).

Teacher's Note:
1. Find each remainder by putting \( x = 4 \).
2. Collect like terms carefully.

Answer: (b) 2x² + 6x + 4

Explanation:
1. It has the form \( k(x + 1)(x + 2) \). At \( x = -3 \): \( k(-2)(-1) = 2k = 4 \), so \( k = 2 \).
2. The polynomial is \( 2(x^{2} + 3x + 2) = 2x^{2} + 6x + 4 \).

Teacher's Note:
1. Write the zeros as factors with an unknown constant \( k \).
2. Use the remainder condition to find \( k \).

Answer: (a) −34

Explanation:
1. \( x^{2} - 2x - 3 = (x - 3)(x + 1) \), so \( p(3) = 0 \) and \( p(-1) = 0 \).
2. \( p(3) = 72 + 3a + b = 0 \) and \( p(-1) = -4 - a + b = 0 \), so \( b = a + 4 \).
3. \( 72 + 3a + a + 4 = 0 \) gives \( a = -19, b = -15 \), so \( a + b = -34 \).

Teacher's Note:
1. Factorise the divisor and apply the factor theorem to each factor.
2. Two equations in a and \( b \) give the answer.

Answer: (d) 6x − 5

Explanation:
1. Let the remainder be ax \( + b \). \( p(1) = 1 - 2 + 1 - 2 + 1 + 1 + 1 = 1 \), so \( a + b = 1 \).
2. \( p(2) = 2^{200} - 2^{200} + 2^{50} - 2^{50} + 4 + 2 + 1 = 7 \), so \( 2a + b = 7 \).
3. Hence \( a = 6, b = -5 \), and the remainder is \( 6x - 5 \).

Teacher's Note:
1. When the divisor is quadratic, the remainder is linear: ax \( + b \).
2. Use the two zeros of the divisor to find a and \( b \).

Answer: (a) 16

Explanation:
1. \( p(x) - x \) vanishes at 1, 2, 3, so \( p(x) - x = k(x - 1)(x - 2)(x - 3) \).
2. At \( x = 4 \): \( 5 - 4 = 6k \), so \( k = 1/6 \).
3. \( p(6) = 6 + (5 \cdot 4 \cdot 3)/6 = 6 + 10 = 16 \).

Teacher's Note:
1. Subtract \( x \) so that three values become zero.
2. Use the fourth value to find the constant \( k \).

Answer: (d) I and III

Explanation:
1. Adding: \( a^{2} = 25 \); subtracting: \( b^{2} = 4 \). So \( a = \pm 5 \) and \( b = \pm 2 \).
2. ab = ±10, so −10 and 10 are both possible, while 5√2 is not.

Teacher's Note:
1. Solve for \( a^{2} \) and \( b^{2} \) first.
2. Do not forget the negative values of a and \( b \).

Answer: (d) 8a³

Explanation:
1. \( (a + b)^{3} + (a - b)^{3} = 2a^{3} \) + 6ab².
2. \( 6a(a^{2} - b^{2}) = 6a^{3} \) − 6ab².
3. The sum is \( 8a^{3} \).

Teacher's Note:
1. Expand the cubes and cancel the ab² terms.

Answer: (a) 1

Explanation:
1. \( a^{3} + b^{3} + c^{3} \) − 3abc = (\( a + b + c)(a^{2} + b^{2} + c^{2} \) − ab − bc − ca).
2. The denominator is −(a² + b² + c² − ab − bc − ca), so the ratio is \( -(a + b + c) \).
3. \( a + b + c = -5 - 6 + 10 = -1 \), so the value is 1.

Teacher's Note:
1. Recognise the identity for \( a^{3} + b^{3} + c^{3} \) − 3abc.
2. The denominator is the negative of the second factor.

Answer: (b) 1

Explanation:
1. \( x^{2} + y^{2} + z^{2} = 1 - 2(-1) = 3 \).
2. \( x^{3} + y^{3} + z^{3} \) − 3xyz = (\( x + y + z)(x^{2} + y^{2} + z^{2} \) − xy − yz − zx) = \( 1 \times (3 + 1) = 4 \).
3. So \( x^{3} + y^{3} + z^{3} = 4 + 3(-1) = 1 \).

Teacher's Note:
1. Find \( x^{2} + y^{2} + z^{2} \) from the square of the sum.
2. Then use the sum-of-cubes identity.

Answer: (b) \( x+y+z=3x^{1/3}y^{1/3}z^{1/3} \)

Explanation:
1. Let \( p = x^{1/3}, q = y^{1/3}, r = z^{1/3} \). Since \( p + q + r = 0, p^{3} + q^{3} + r^{3} \) = 3pqr.
2. That is \( x + y + z = 3x^{1/3}y^{1/3}z^{1/3} \).

Teacher's Note:
1. If \( a + b + c = 0 \) then \( a^{3} + b^{3} + c^{3} \) = 3abc.
2. Apply it to the cube roots.

Answer: (b) 36

Explanation:
1. \( (a^{2} + 1/a^{2})^{2} = 119 + 2 = 121 \), so \( a^{2} + 1/a^{2} = 11 \).
2. \( (a - 1/a)^{2} = 11 - 2 = 9 \). Since \( a \gt 1, a - 1/a = 3 \).
3. \( a^{3} - 1/a^{3} = (a - 1/a)(a^{2} + 1 + 1/a^{2}) = 3 \times 12 = 36 \).

Teacher's Note:
1. Work downwards: fourth power, square, then the plain expression.
2. The condition \( a \gt 1 \) makes \( a - 1/a \) positive.

Answer: (d) 3

Explanation:
1. Put the three terms over the common denominator \( (a - b)(b - c)(c - a) \): the numerator is \( (a - b)^{3} + (b - c)^{3} + (c - a)^{3} \).
2. The three bases add to zero, so the numerator is \( 3(a - b)(b - c)(c - a) \).
3. The value is 3.

Teacher's Note:
1. Use \( p^{3} + q^{3} + r^{3} \) = 3pqr when \( p + q + r = 0 \).
2. Here \( p = a - b, q = b - c, r = c - a \).

Answer: (b) 1

Explanation:
1. With \( a = 0.76 \) and \( b = 0.24 \) the expression is (\( a^{3} + b^{3})/(a^{2} \) − ab + b²) = a + b.
2. \( a + b = 1 \).

Teacher's Note:
1. Spot the identity a³ + b³ = (a + b)(a² − ab \( + b^{2} \)).
2. No long multiplication is needed.

Answer: (b) 7

Explanation:
1. \( x \) and \( y \) are the roots of \( t^{2} - 3t + 2 = 0 \), so \( x = 2 \) and \( y = 1 \).
2. \( x^{3} - y^{3} = 8 - 1 = 7 \).

Teacher's Note:
1. Find the numbers directly from the sum and product.
2. The condition \( x \gt y \) fixes the order.

Answer: (a) \( \frac{5}{4} \)

Explanation:
1. \( 1 + 1/x = 3 \), so \( 1 + 1/3 = 4/3 \) and \( 1/(4/3) = 3/4 \).
2. \( x + 3/4 = 1/2 + 3/4 = 5/4 \).

Teacher's Note:
1. Evaluate a continued fraction from the bottom upwards.

Answer: (d) m² + m + 1

Explanation:
1. \( (m^{3} + 1)(m^{2} + m + 1) = m^{5} + m^{4} + m^{3} + m^{2} + m + 1 \).
2. So the missing factor is \( m^{2} + m + 1 \).

Teacher's Note:
1. Group the terms as \( m^{3}(m^{2} + m + 1) + (m^{2} + m + 1) \).

Answer: (d) 12

Explanation:
1. \( x^{2} = 2 + \sqrt{2} \), so \( x^{4} = 6 + 4\sqrt{2} \).
2. \( 4/x^{4} = 4/(6 + 4\sqrt{2}) = 4(6 - 4\sqrt{2})/(36 - 32) = 6 - 4\sqrt{2} \).
3. The sum is 12.

Teacher's Note:
1. Square twice and rationalise the reciprocal.

Answer: (d) 15616

Explanation:
1. Use \( a^{3} - b^{3} = (a - b)^{3} \) + 3ab(a \( - b \)) with \( a = 4x \) and \( b = 5z \).
2. (16)³ + 3(20xz)(16) = \( 4096 + 3 \times 240 \times 16 = 4096 + 11520 = 15616 \).

Teacher's Note:
1. \( 64x^{3} = (4x)^{3} \) and \( 125z^{3} = (5z)^{3} \).
2. Use the cube of a difference.

Answer: (b) y − x

Explanation:
1. Put \( p = 1/x \) and \( q = 1/y \). The fraction is (\( p^{3} - q^{3})/(p^{3}q + p^{2}q^{2} \) + pq³) = (\( p - q)(p^{2} \) + pq + q²)/[pq(p² + pq + q²)].
2. = (\( p \) − q)/(pq) = \( 1/q - 1/p = y - x \).

Teacher's Note:
1. Substitute reciprocals to avoid negative exponents.
2. Factor \( p^{3} - q^{3} \) and cancel the common quadratic.

Answer: (d) 16 : 1

Explanation:
1. The numerator is \( a + b \) + 2√(ab) = \( (\sqrt{a} + \sqrt{b})^{2} \).
2. So \( (\sqrt{a} + \sqrt{b})/(\sqrt{a} - \sqrt{b}) = 5/3 \), giving \( 3\sqrt{a} + 3\sqrt{b} = 5\sqrt{a} - 5\sqrt{b} \), i.e. \( \sqrt{a} = 4\sqrt{b} \).
3. Therefore a : \( b = 16 \) : 1.

Teacher's Note:
1. The numerator simplifies to a perfect square.
2. Then cancel one factor with \( a - b = (\sqrt{a} - \sqrt{b})(\sqrt{a} + \sqrt{b}) \).

Answer: (b) 2

Explanation:
1. \( 1 + x + x^{2} = (1 - x^{3})/(1 - x) \).
2. Adding \( x^{3}/(1 - x) \) gives \( 1/(1 - x) = 1/0.5 = 2 \).

Teacher's Note:
1. Use the formula for \( 1 + x + x^{2} \).

Answer: (c) p³ − 6p − 6 = 0

Explanation:
1. Let \( a = 2^{1/3} \), so \( p = a^{2} + a \). \( p^{3} = a^{6} + a^{3} + 3a^{3}(a^{2} + a) = 4 + 2 + 3 \times 2 \times p \).
2. So \( p^{3} = 6 + 6p \), i.e. \( p^{3} - 6p - 6 = 0 \).

Teacher's Note:
1. Cube a sum using \( (u + v)^{3} = u^{3} + v^{3} \) + 3uv(u \( + v \)).
2. Note that uv = 2 here.

Answer: (a) 2

Explanation:
1. The product of zeros is \( k/2 \) and it equals 1 for reciprocal zeros.
2. So \( k = 2 \).

Teacher's Note:
1. Reciprocal zeros have product 1.

Answer: (a) 3

Explanation:
1. \( \alpha + \beta = 4 \) and \( \alpha\beta = 1 \).
2. \( 1/\alpha + 1/\beta = (\alpha + \beta)/(\alpha\beta) = 4 \), so the value is \( 4 - 1 = 3 \).

Teacher's Note:
1. Write the sum of reciprocals over a common denominator.

Answer: (a) 5

Explanation:
1. \( \alpha + \beta = 5 \) and \( \alpha^{2} + \beta^{2} = 15 \), so \( 2\alpha\beta = 25 - 15 = 10 \).
2. \( C = \alpha\beta = 5 \).

Teacher's Note:
1. Use \( \alpha^{2} + \beta^{2} = (\alpha + \beta)^{2} - 2\alpha\beta \).

Answer: (c) a perfect cube but not a perfect square

Explanation:
1. Expand: \( 3^{18} + 3^{13}\cdot 2^{5} + 2^{15} + 2^{10}\cdot 3^{7} \). Try \( (3^{6} + 2^{5})^{3} \) ... check using \( u = 3^{6} = 729 \) and \( v = 2^{5} = 32 \) with \( u^{3} = 3^{18} \) and \( v^{3} = 2^{15} \).
2. \( 3u^{2}v = 3 \cdot 3^{12} \cdot 2^{5} = 3^{13} \cdot 2^{5} \) and 3uv² = \( 3 \cdot 3^{6} \cdot 2^{10} = 3^{7} \cdot 2^{10} \), which match the other two terms.
3. So the number is \( (3^{6} + 2^{5})^{3} = 761^{3}, a \) perfect cube. Since 761 is prime, it is not a perfect square.

Teacher's Note:
1. Recognise the pattern \( u^{3} + 3u^{2}v \) + 3uv² \( + v^{3} \).
2. A cube of a prime is never a perfect square.