JEE Mathematics Polynomials MCQs Set 02

Answer: (d) binomial

Explanation:
1. It has two terms (8 and 2b), so it is a binomial.
2. Its degree is 1, not 8.

Teacher's Note:
1. Count the terms separated by + or - signs.
2. Degree is the highest power of the variable.

Answer: (b) binomial

Explanation:
1. The two terms \( abc^2 \) and \( 5abc^3 \) are unlike terms and cannot be combined.
2. So the expression is a binomial.

Teacher's Note:
1. Like terms must have exactly the same letters with the same powers.
2. \( abc^2 \) and \( abc^3 \) differ in the power of c.

Answer: (b) \( 2x^4y^2 \)

Explanation:
1. Multiply the coefficients: \( \left(-\frac{2}{3}\right)(-3) = 2 \).
2. Multiply the letters by adding powers: \( x^2 \cdot x^2 = x^4 \) and \( y \cdot y = y^2 \).

Teacher's Note:
1. A negative times a negative is positive.
2. Add exponents of the same letter when multiplying.

Answer: (d) \( 4x^2 + 2xy - 4xz \)

Explanation:
1. The second factor is \( -2x - y + 2z \).
2. \( (-2x)(-2x) = 4x^2,\ (-2x)(-y) = 2xy,\ (-2x)(2z) = -4xz \).

Teacher's Note:
1. Remove the bracket in the second factor first.
2. Multiply each term by \( -2x \) and watch the signs.

Answer: (a) \( 3x^2 + 2xy - 3xz - y^2 + yz \)

Explanation:
1. Second factor: \( x + 2y - y - z = x + y - z \).
2. \( (3x - y)(x + y - z) = 3x^2 + 3xy - 3xz - xy - y^2 + yz \).
3. \( = 3x^2 + 2xy - 3xz - y^2 + yz \).

Teacher's Note:
1. Simplify the bracket first.
2. Collect the xy terms: \( 3xy - xy = 2xy \).

Answer: (c) \( -2x^4 + 5x^3 - 4x^2 + x \)

Explanation:
1. The first factor simplifies to \( x - 2x^2 \).
2. \( (x - 2x^2)(x^2 - 2x + 1) = x^3 - 2x^2 + x - 2x^4 + 4x^3 - 2x^2 \).
3. \( = -2x^4 + 5x^3 - 4x^2 + x \).

Teacher's Note:
1. Cancel \( x^3 \) and \( -x^3 \) before multiplying.
2. Arrange the answer in descending powers.

Answer: (c) \( 4mn \)

Explanation:
1. \( (m + n)^2 = m^2 + 2mn + n^2 \) and \( (m - n)^2 = m^2 - 2mn + n^2 \).
2. Subtracting: \( 4mn \).

Teacher's Note:
1. Use \( A^2 - B^2 = (A - B)(A + B) \): \( (2n)(2m) = 4mn \).
2. This identity gives a quick way to find products.

Answer: (a) \( 2a(a - b) \)

Explanation:
1. Take \( (b - a) \) as common factor: \( (b - a)[(b - a) - (a + b)] \).
2. \( = (b - a)(-2a) = 2a(a - b) \).

Teacher's Note:
1. Factorising is quicker than expanding here.
2. \( -2a(b - a) = 2a(a - b) \).

Answer: (d) \( 27x^3 + 8 \)

Explanation:
1. \( 9x^2 - 6x + 4 = (3x)^2 - (3x)(2) + 2^2 \), so the product is \( (3x)^3 + 2^3 \).
2. \( = 27x^3 + 8 \).

Teacher's Note:
1. Recognise the identity \( (a + b)(a^2 - ab + b^2) = a^3 + b^3 \).
2. Direct multiplication gives the same result.

Answer: (b) \( 3b^4c - 2b^2c^2 - c^3 \)

Explanation:
1. \( b^2 \cdot 3b^2c = 3b^4c,\ b^2 \cdot c^2 = b^2c^2 \).
2. \( -c \cdot 3b^2c = -3b^2c^2,\ -c \cdot c^2 = -c^3 \).
3. Adding: \( 3b^4c + b^2c^2 - 3b^2c^2 - c^3 = 3b^4c - 2b^2c^2 - c^3 \).

Teacher's Note:
1. Combine the like terms in \( b^2c^2 \).
2. Four products give three distinct terms.

Answer: (b) \( 9x^4 - 16x^2 + 40x - 25 \)

Explanation:
1. Write the factors as \( 3x^2 - (4x - 5) \) and \( 3x^2 + (4x - 5) \).
2. Product \( = 9x^4 - (4x - 5)^2 = 9x^4 - (16x^2 - 40x + 25) \).
3. \( = 9x^4 - 16x^2 + 40x - 25 \).

Teacher's Note:
1. Spot the difference of squares pattern.
2. Squaring \( 4x - 5 \) gives the middle term 40x.

Answer: (d) \( m^5 - 41m - 120 \)

Explanation:
1. Multiply term by term: \( m^5 + 4m^4 + 5m^3 - 4m^4 - 16m^3 - 20m^2 + 11m^3 + 44m^2 + 55m - 24m^2 - 96m - 120 \).
2. Collect: \( m^5 + (4 - 4)m^4 + (5 - 16 + 11)m^3 + (-20 + 44 - 24)m^2 + (55 - 96)m - 120 \).
3. \( = m^5 - 41m - 120 \).

Teacher's Note:
1. Several terms cancel, so the answer is short.
2. Check with m = 1: \( (-16)(10) = -160 = 1 - 41 - 120 \).

Answer: (a) \( x + y \)

Explanation:
1. \( x^3 + y^3 = (x + y)(x^2 - xy + y^2) \).
2. So the quotient is \( x + y \) and the remainder is 0.

Teacher's Note:
1. Learn the factorisation of a sum of cubes.
2. Check with x = y = 1: 2 / 1 = 2.

Answer: (c) \( a - b \)

Explanation:
1. \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \).
2. So the quotient is \( a - b \).

Teacher's Note:
1. The difference of cubes factorises with a plus sign in the quadratic factor.
2. Check with a = 2, b = 1: 7 / 7 = 1 = a - b.

Answer: (c) \( 1 + x + x^2 + x^3 - 4x^4 \)

Explanation:
1. Check by multiplying: \( (1 - x)(1 + x + x^2 + x^3 - 4x^4) = 1 - x^4 - 4x^4 + 4x^5 \).
2. \( = 1 - 5x^4 + 4x^5 \), which matches the dividend.

Teacher's Note:
1. Verify any division by multiplying the quotient by the divisor.
2. The remainder is 0 because x = 1 makes the dividend zero.

Answer: (a) \( 2a - 3b + c \)

Explanation:
1. The dividend is \( (2a)^2 + (-3b)^2 + c^2 + 2(2a)(-3b) + 2(-3b)(c) + 2(2a)(c) \).
2. This is \( (2a - 3b + c)^2 \), so dividing by \( 2a - 3b + c \) gives \( 2a - 3b + c \).

Teacher's Note:
1. Recognise \( (p + q + r)^2 \) in the dividend.
2. The cross terms -12ab, -6bc, +4ac confirm it.

Answer: (d) \( \frac{1}{4}x^2 - \frac{1}{3}x + \frac{1}{9} \)

Explanation:
1. The dividend is \( \left(\frac{x}{2}\right)^3 - 3\left(\frac{x}{2}\right)^2\left(\frac{1}{3}\right) + 3\left(\frac{x}{2}\right)\left(\frac{1}{3}\right)^2 - \left(\frac{1}{3}\right)^3 = \left(\frac{x}{2} - \frac{1}{3}\right)^3 \).
2. Dividing by one factor leaves \( \left(\frac{x}{2} - \frac{1}{3}\right)^2 = \frac{x^2}{4} - \frac{x}{3} + \frac{1}{9} \).

Teacher's Note:
1. Spot the cube of a binomial.
2. Check at x = 2: the dividend is \( 1 - 1 + \frac{1}{3} - \frac{1}{27} = \frac{8}{27} \), the divisor is \( \frac{2}{3} \) and the quotient is \( \frac{4}{9} \), and \( \frac{2}{3} \cdot \frac{4}{9} = \frac{8}{27} \).

Answer: (b) 2

Explanation:
1. By the remainder theorem, the remainder on dividing by \( 3x - 2 \) is the value at \( x = \frac{2}{3} \).
2. \( 15\left(\frac{4}{9}\right) + 2\left(\frac{2}{3}\right) - 10 = \frac{20}{3} + \frac{4}{3} - 10 = -2 \).
3. The remainder is -2, so we must add 2. Then \( 15x^2 + 2x - 8 = (3x - 2)(5x + 4) \).

Teacher's Note:
1. Add the negative of the remainder.
2. Check by factorising the new quadratic.

Answer: (b) \( (a^2 - b^2)(c^2 - d^2) \)

Explanation:
1. Expand: \( a^2c^2 + 2abcd + b^2d^2 - a^2d^2 - 2abcd - b^2c^2 \).
2. \( = a^2c^2 - a^2d^2 - b^2c^2 + b^2d^2 = a^2(c^2 - d^2) - b^2(c^2 - d^2) \).
3. \( = (a^2 - b^2)(c^2 - d^2) \).

Teacher's Note:
1. The cross terms 2abcd cancel.
2. Factor by grouping.

Answer: (d) A = 5, B = 3

Explanation:
1. Compare coefficients: \( A + B = 8 \) and \( -3A + 2B = -9 \).
2. From the first, \( B = 8 - A \). Then \( -3A + 16 - 2A = -9 \Rightarrow A = 5 \) and \( B = 3 \).
3. Check at x = 3: \( 15 = 5B \), B = 3; at x = -2: \( -25 = -5A \), A = 5.

Teacher's Note:
1. Putting x = 3 and x = -2 also gives A and B quickly.
2. Both methods agree.