JEE Mathematics Algebraic Expressions MCQs Set 01

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

Explanation:
1. A polynomial has only non-negative integral powers of the variable.
2. The expressions with powers \( \frac{3}{2} \), \( \frac{1}{3} \), \( -1 \) or \( \frac{1}{2} \) are not polynomials.
3. So the first expression, with integral powers 3, 2, 1, 0, is a polynomial.

Teacher's Note:
1. Fractional or negative exponents rule out a polynomial.
2. An expression containing \( x^{3/2} \) and \( x^{1/3} \) is not a polynomial.

Answer: (d) x = 0 and y = 0

Explanation:
1. The square of a real number is never negative, so \( x^2 \ge 0 \) and \( y^2 \ge 0 \).
2. A sum of two non-negative numbers is zero only when both are zero.
3. So x = 0 and y = 0.

Teacher's Note:
1. This idea is used in many 'sum of squares equals zero' problems.
2. If complex numbers were allowed, the result would fail.

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

Explanation:
1. \( \left(x + \frac{1}{x}\right)^2 = x^2 + 2 + \frac{1}{x^2} \).
2. So \( x^2 + \frac{1}{x^2} = m^2 - 2 \).

Teacher's Note:
1. Square the given relation and subtract the middle term 2.
2. Check with x = 1: m = 2 and \( x^2 + \frac{1}{x^2} = 2 = 4 - 2 \).

Answer: (c) a rational function

Explanation:
1. It is a ratio of two polynomials, so it is a rational function.
2. It is not a polynomial because the denominator is not a constant.

Teacher's Note:
1. Rational function means polynomial divided by polynomial.
2. It is defined wherever the denominator is not zero.

Answer: (c) not identical, because \( f(x) \ne g(x) \) for most values of x

Explanation:
1. Two functions are identical only if they have the same value at every x.
2. At x = 2: \( f(2) = 2 \) but \( g(2) = 4 \). They agree only at x = 0 and x = 1.
3. So f and g are not identical.

Teacher's Note:
1. One counter-example is enough to show two functions are different.
2. Having a common point does not make functions identical.

Answer: (a) 4, which is independent of x

Explanation:
1. Use \( (A + B)^2 - (A - B)^2 = 4AB \) with \( A = x \) and \( B = \frac{1}{x} \).
2. \( 4AB = 4 \cdot x \cdot \frac{1}{x} = 4 \).

Teacher's Note:
1. The identity \( (A+B)^2 - (A-B)^2 = 4AB \) saves long expansion.
2. The answer does not contain x.

Answer: (d) \( 2pq \)

Explanation:
1. Numerator \( = pqx + pxy + pqx + qxy - pxy - qxy = 2pqx \).
2. Dividing by x: \( 2pq \), which is independent of x and y.

Teacher's Note:
1. Expand each product and cancel the terms pxy and qxy.
2. What is left is twice pqx.

Answer: (b) odd

Explanation:
1. Let the numbers be n and \( n + 1 \). The sum of squares is \( n^2 + (n+1)^2 = 2n^2 + 2n + 1 \).
2. \( 2n^2 + 2n \) is even, so adding 1 gives an odd number for every n.

Teacher's Note:
1. Test with 1 and 2: 1 + 4 = 5; with 2 and 3: 4 + 9 = 13.
2. One of the two numbers is even and the other is odd, so the squares have different parity.

Answer: (b) 5

Explanation:
1. Let the number be x. Then \( \frac{x + 45}{5} = 2x \).
2. \( x + 45 = 10x \Rightarrow 9x = 45 \Rightarrow x = 5 \).

Teacher's Note:
1. Translate the words into an equation first.
2. Check: \( \frac{5 + 45}{5} = 10 = 2 \times 5 \).

Answer: (d) 14

Explanation:
1. The identity holds for all x, so put x = 0: \( e = (2)(7) = 14 \).
2. Alternatively e is the product of the constant terms.

Teacher's Note:
1. Putting x = 0 isolates the constant term.
2. The constant term of a product is the product of the constant terms.

Answer: (a) -6

Explanation:
1. Collect the \( x^4 \) terms of the product: \( x^3 \cdot (-3x) + (-3x^2)(x^2) = -3x^4 - 3x^4 = -6x^4 \).
2. So a = -6.

Teacher's Note:
1. Only two pairs of terms produce \( x^4 \).
2. Expanding fully: \( x^5 - 6x^4 + 20x^3 - 31x^2 + 22x + 14 \).

Answer: (c) 5

Explanation:
1. The degree is the highest power of x with a non-zero coefficient.
2. The highest power is \( x^5 \), so the degree is 5.

Teacher's Note:
1. Ignore the order in which the terms are written.
2. Degree is about exponents, not coefficients.

Answer: (c) \( 10x^2 + xy - 11y^2 \)

Explanation:
1. \( x^2 \) terms: \( 1 + 1 + 8 = 10 \).
2. xy terms: \( -2 + 3 + 0 = 1 \).
3. \( y^2 \) terms: \( 1 - 2 - 10 = -11 \).

Teacher's Note:
1. Add like terms column by column.
2. A missing term counts as zero.

Answer: (a) \( 2x^2 + 7x + 12 \)

Explanation:
1. Excess of A over B means \( A - B \).
2. \( (4x^2 - 3x + 5) - (2x^2 - 10x - 7) = 2x^2 + 7x + 12 \).

Teacher's Note:
1. Change all the signs of the second polynomial when subtracting.
2. \( 5 - (-7) = 12 \).

Answer: (d) \( 11x^4 + 2x^3 + 12x^2 + 31 \)

Explanation:
1. The shortfall is the larger polynomial minus the smaller: \( (27x^4 - 9x^3 + 7x^2 + 19) - (16x^4 - 11x^3 - 5x^2 - 12) \).
2. \( = 11x^4 + 2x^3 + 12x^2 + 31 \).

Teacher's Note:
1. Falls short of: the amount that must be added to the first to get the second.
2. Check by adding it back to the first polynomial.

Answer: (b) \( (a^4 - b^4)(a^8 - b^8) \)

Explanation:
1. \( (a^2 - b^2)(a^2 + b^2) = a^4 - b^4 \).
2. \( (a^4 + b^4)(a^4 - b^4) = a^8 - b^8 \).
3. The product is \( (a^4 - b^4)(a^8 - b^8) \).

Teacher's Note:
1. Group each pair so that it forms a difference of squares.
2. Do not multiply all four factors blindly.

Answer: (b) \( x = \frac{55}{2} \)

Explanation:
1. Multiply every term by 45: \( 5(2x - 1) + 9(11 - x) = 225 - 3(7 + x) \).
2. \( 10x - 5 + 99 - 9x = 225 - 21 - 3x \Rightarrow x + 94 = 204 - 3x \).
3. \( 4x = 110 \Rightarrow x = \frac{55}{2} \).

Teacher's Note:
1. The LCM of 9, 5 and 15 is 45.
2. Check: at x = 27.5 the left side is 6 - 3.3 = 2.7 and the right side is 5 - 2.3 = 2.7.

Answer: (d) 0

Explanation:
1. \( A^2 - B^2 = (A - B)(A + B) \) with \( A = 2x + 3 \) and \( B = 2x - 3 \).
2. \( A - B = 6 \) and \( A + B = 4x \), so \( 24x = 0 \Rightarrow x = 0 \).

Teacher's Note:
1. Check: \( 3^2 - (-3)^2 = 0 \).
2. Expanding gives \( 24x = 0 \) directly.

Answer: (a) -103

Explanation:
1. \( x^6 \) arises from \( 7x^4 \cdot (-7x^2) = -49x^6 \) and \( -6x^3 \cdot 9x^3 = -54x^6 \).
2. Sum \( = -49 - 54 = -103 \).

Teacher's Note:
1. List only the pairs whose exponents add up to 6.
2. The first polynomial has degree 4 and the second has terms of degree 0, 2, 3.

Answer: (c) \( \frac{4}{a^2}(2a^2 - 1) \)

Explanation:
1. \( x^2 + 2xy - 3y^2 = (x - y)(x + 3y) \).
2. \( x - y = \frac{2}{a} \) and \( x + 3y = 4a - \frac{2}{a} \).
3. Product \( = \frac{2}{a}\left(4a - \frac{2}{a}\right) = 8 - \frac{4}{a^2} = \frac{4}{a^2}(2a^2 - 1) \).

Teacher's Note:
1. Factorise the quadratic expression in x and y first.
2. Check with a = 1: x = 2, y = 0 and the value is 4 = (4/1)(1).