JEE Mathematics Polynomials MCQs Set 01

Answer: (b) \( 3(a^2-b^2) \)

Explanation:
1. Use \( p^2-q^2=(p+q)(p-q) \) with \( p=2a+b,\ q=2b+a \).
2. \( p+q=3a+3b \) and \( p-q=a-b \).
3. The product is \( 3(a+b)(a-b)=3(a^2-b^2) \).

Teacher's Note:
1. Differences of squares are fastest through the factor form.
2. Take out the common factor 3 before multiplying.

Answer: (a) \( 9-4\sqrt5 \)

Explanation:
1. Compute the product \( (9+4\sqrt5)(9-4\sqrt5)=81-80=1 \).
2. A product equal to 1 means the two numbers are reciprocals.
3. So the reciprocal is \( 9-4\sqrt5 \).

Teacher's Note:
1. Conjugate surds multiply to a rational number, which is why rationalising works.
2. Always check the product equals 1 to confirm a reciprocal.

Answer: (d) 12

Explanation:
1. By the remainder theorem put \( x=\tfrac12 \).
2. \( 4\cdot\tfrac14-8\cdot\tfrac12+15=1-4+15=12 \).
3. The remainder is 12, so the polynomial is not exactly divisible.

Teacher's Note:
1. Substitute the zero of the divisor into the dividend to get the remainder.
2. The zero of \( 2x-1 \) is \( \tfrac12 \), not 1.

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

Explanation:
1. The numerator is the cube expansion \( (x-y)^3 \).
2. So \( \dfrac{(x-y)^3}{x-y}=(x-y)^2 \).

Teacher's Note:
1. Spot the pattern 1, 3, 3, 1 in the coefficients, which signals a perfect cube.
2. Cancel a common factor only after factorising.

Answer: (a) \( (m-n)^2 \)

Explanation:
1. The expression is \( m^2-n^2-(2mn-2n^2) \).
2. Remove the brackets: \( m^2-n^2-2mn+2n^2=m^2-2mn+n^2 \).
3. This equals \( (m-n)^2 \).

Teacher's Note:
1. A minus sign before a bracket changes every sign inside it.
2. Collect like terms before trying to factorise.

Answer: (b) \( 2-5x \)

Explanation:
1. Innermost bracket: \( 3x+4x-5+3=7x-2 \).
2. Then \( 4x-(7x-2)-5x=-8x+2 \).
3. Finally \( 3x-8x+2=2-5x \), which depends on \( x \).

Teacher's Note:
1. Remove brackets from the innermost pair outwards.
2. Be careful with the minus sign before each curly bracket.

Answer: (b) 5

Explanation:
1. The zero of \( 2x-1 \) is \( x=\tfrac12 \).
2. \( 14\cdot\tfrac18-3\cdot\tfrac14+4\cdot\tfrac12+2=\tfrac74-\tfrac34+2+2=5 \).
3. The remainder is 5.

Teacher's Note:
1. The remainder theorem avoids long division.
2. Evaluate fractions carefully with a common denominator.

Answer: (d) 16

Explanation:
1. Write \( (x^2+1)^4=\left[(x^2+1)^2\right]^2=(x^4+2x^2+1)^2 \).
2. Then \( (x^4+2x^2+1)^2-(x^4-6x^2+1)^2=(2x^4-4x^2+2)(8x^2) \).
3. This is \( 16x^2(x^2-1)^2=16(x^3-x)^2 \), so \( k=16 \).

Teacher's Note:
1. Use the difference of squares on the two large squares.
2. Note that \( x^2(x^2-1)^2=(x^3-x)^2 \).

Answer: (a) 8

Explanation:
1. \( (a+b)^3=a^3+b^3+3ab(a+b) \).
2. With \( a+b=2 \) this gives \( 8=a^3+b^3+6ab \).
3. So the value is 8.

Teacher's Note:
1. Learn the cube identity in the form with \( 3ab(a+b) \).
2. Substituting the given sum directly turns the expression into a cube.

Answer: (a) \( -x+y+3z \)

Explanation:
1. Collect only the terms that contain \( a \).
2. The coefficient is \( (x+y+z)-(x+y-z)+(y+z-x) \).
3. Simplifying gives \( -x+y+3z \).

Teacher's Note:
1. To find a coefficient, look at each bracket's own coefficient of that letter.
2. Watch the sign of the term \( -a \) in the second bracket.

Answer: (b) \( 2xy \)

Explanation:
1. The numerator is \( 2(x^2+y^2) \).
2. The bracket is \( \dfrac{x^2+y^2}{xy} \).
3. Dividing gives \( 2(x^2+y^2)\cdot\dfrac{xy}{x^2+y^2}=2xy \).

Teacher's Note:
1. Combine the fractions in the bracket first.
2. Dividing by a fraction means multiplying by its reciprocal.

Answer: (a) \( -4 \)

Explanation:
1. \( x+y-z=1-2-3=-4 \), so the first square is 16.
2. \( y-z-x=-2-3-1=-6 \), so the second square is 36.
3. \( z-x-y=3-1+2=4 \), so the third is 16. The value is \( 16-36+16=-4 \).

Teacher's Note:
1. Simplify each bracket numerically before squaring.
2. Keep the minus sign in front of the second square.

Answer: (b) \( -1 \)

Explanation:
1. Compare the coefficient of \( x^2y \) on both sides.
2. On the left it is 1. On the right the term in \( x^2y \) is \( N\cdot x\cdot y\cdot(-x)=-Nx^2y \).
3. So \( 1=-N \) and \( N=-1 \).

Teacher's Note:
1. In an identity you may compare the coefficient of one convenient term.
2. A quick check with \( x=0,y=1,z=2 \) also gives \( N=-1 \).

Answer: (a) 2

Explanation:
1. \( (a+b)^2-(a^2+b^2)=2ab \).
2. Dividing by \( ab \) gives \( \dfrac{2ab}{ab}=2 \).

Teacher's Note:
1. The cross term of \( (a+b)^2 \) is exactly \( 2ab \).
2. This idea is often used in simplification problems.

Answer: (d) 4

Explanation:
1. \( ax=x+6 \) gives \( x(a-1)=6 \), so \( x=\dfrac{6}{a-1} \).
2. \( x \) is natural when \( a-1 \) is a positive divisor of 6, i.e. 1, 2, 3 or 6.
3. This gives \( a=2,3,4,7 \) with \( x=6,3,2,1 \), so there are 4 values of \( a \).

Teacher's Note:
1. Rearrange to isolate the product \( x(a-1) \) and use divisors of 6.
2. Count the divisors carefully, there are four positive ones.

Answer: (d) \( x^3+4 \)

Explanation:
1. Dividend = divisor times quotient plus remainder.
2. \( (x+1)(x^2-x+1)+3=x^3+1+3 \).
3. So the expression is \( x^3+4 \).

Teacher's Note:
1. Remember the division algorithm in the form dividend = divisor x quotient + remainder.
2. \( (x+1)(x^2-x+1) \) is the sum of cubes identity.

Answer: (c) \( -\dfrac1{27} \)

Explanation:
1. Here \( a=(x+1)^3,\ b=(x+1)^2,\ c=x+1 \).
2. So \( a-3b+3c-1=(x+1)^3-3(x+1)^2+3(x+1)-1=(x+1-1)^3=x^3 \).
3. At \( x=-\tfrac13 \) the value is \( -\dfrac1{27} \).

Teacher's Note:
1. Recognise the pattern \( t^3-3t^2+3t-1=(t-1)^3 \) with \( t=x+1 \).
2. This avoids heavy arithmetic with fractions.

Answer: (a) \( xy=a \)

Explanation:
1. Rearrange: \( x-y=\dfrac ay-\dfrac ax=\dfrac{a(x-y)}{xy} \).
2. Since \( x\ne y \), divide by \( x-y \) to get \( 1=\dfrac{a}{xy} \).
3. Hence \( xy=a \).

Teacher's Note:
1. Dividing by \( x-y \) is allowed only because it is non-zero.
2. Bring like terms of each variable together first.

Answer: (a) 2

Explanation:
1. Since \( c=-a-b \), \( a^2-bc=a^2+b(a+b)=a^2+ab+b^2 \), and similarly \( b^2-ca \) and \( c^2-ab \) equal \( a^2+ab+b^2 \).
2. So all three denominators are equal and the sum is \( \dfrac{a^2+b^2+c^2}{a^2+ab+b^2} \).
3. Because \( c^2=(a+b)^2 \), the numerator is \( 2(a^2+ab+b^2) \) and the value is 2.

Teacher's Note:
1. When \( a+b+c=0 \), the three expressions \( a^2-bc \), \( b^2-ca \), \( c^2-ab \) are all equal.
2. Always substitute the third variable using the condition.

Answer: (b) 1

Explanation:
1. Expand the denominator: \( 1+2mn+m^2n^2-m^2-2mn-n^2=1-m^2-n^2+m^2n^2 \).
2. This factorises as \( (1-m^2)(1-n^2) \).
3. So \( p(1-n^2)=\dfrac{(1-m^2)(1-n^2)}{(1-m^2)(1-n^2)}=1 \).

Teacher's Note:
1. Expand the squares and group to spot the factorisation.
2. The denominator factorises into the two quantities in the question.