CBSE Class 10 Mathematics Rational Expressions MCQs Set 15

Answer: (b) \( 2x^2+7x-3 \)

Explanation:
1. Required = (7x² − 1) − (5x² − 7x + 2)
2. = 2x² + 7x − 3.

Teacher's Note:
1. To find what must be added, subtract the given expression from the target.
2. Be careful with the sign of each term inside the bracket.

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

Explanation:
1. Required = x³ − (1 − x + x² − 2x³)
2. = x³ − 1 + x − x² + 2x³ = 3x³ − x² + x − 1.

Teacher's Note:
1. Subtract the given polynomial from the result.
2. Collect like terms at the end.

Answer: (b) \( 3-9x-7x^2 \)

Explanation:
1. Sum = 7x² + 9x − 2
2. Required = 1 − (7x² + 9x − 2) = 3 − 9x − 7x².

Teacher's Note:
1. First add the two expressions, then subtract from 1.
2. Keep the terms in order of powers.

Answer: (a) \( -4x^3-x^2+16x-8 \)

Explanation:
1. Required = (6x³ − 2x² + 9x + 2) − (10x³ − x² − 7x + 10)
2. = −4x³ − x² + 16x − 8.

Teacher's Note:
1. If A minus X equals B, then X = A − B.
2. Subtract term by term.

Answer: (a) \( \dfrac{x^2+x-3}{x+1} \)

Explanation:
1. Required = \( \dfrac{(x^2+2)(x+1)-(x^2+6x-1)}{x^2-1} \)
2. \( =\dfrac{x^3-4x+3}{x^2-1}=\dfrac{(x-1)(x^2+x-3)}{(x-1)(x+1)} \)
3. = \( \dfrac{x^2+x-3}{x+1} \).

Teacher's Note:
1. Bring both fractions to the common denominator x² − 1 and subtract.
2. Then cancel the common factor x − 1.

Answer: (d) \( \dfrac{x+5}{x+1} \)

Explanation:
1. Each bracket is \( \dfrac{x+k+1}{x+k} \) for k = 1 to 4
2. The product telescopes: \( \dfrac{x+2}{x+1}\cdot\dfrac{x+3}{x+2}\cdot\dfrac{x+4}{x+3}\cdot\dfrac{x+5}{x+4} \)
3. = \( \dfrac{x+5}{x+1} \).

Teacher's Note:
1. Combine each bracket into a single fraction.
2. Intermediate numerators and denominators cancel in pairs.

Answer: (a) \( \dfrac{x+2}{x-2} \)

Explanation:
1. Numerator: x² + 8x + 12 = (x + 2)(x + 6)
2. Denominator: x² + 4x − 12 = (x + 6)(x − 2)
3. Cancel x + 6 to get \( \dfrac{x+2}{x-2} \).

Teacher's Note:
1. Factorise the numerator and the denominator.
2. Cancel only common factors, never common terms.

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

Explanation:
1. x² − 9x + 14 = (x − 2)(x − 7) and x² − 3x + 2 = (x − 1)(x − 2)
2. Cancel (x − 1), (x − 7) and one (x − 2)
3. Result: x − 2.

Teacher's Note:
1. Factorise each quadratic first.
2. List the common factors before cancelling.

Answer: (c) \( \dfrac{y}{x-y} \)

Explanation:
1. First bracket: \( \dfrac{x^2-(x^2-y^2)}{x^2-y^2} \) = \( \dfrac{y^2}{x^2-y^2} \)
2. Second bracket: \( \dfrac{x-y+2y}{y} \) = \( \dfrac{x+y}{y} \)
3. Product = \( \dfrac{y^2(x+y)}{y(x-y)(x+y)} \) = \( \dfrac{y}{x-y} \).

Teacher's Note:
1. Simplify each bracket separately.
2. Then multiply and cancel the common factors.

Answer: (a) \( \dfrac{y}{x} \)

Explanation:
1. Factorise: \( \dfrac{(x-y)(x+y)}{(x+y)^2}\times\dfrac{y(x+y)}{x(x-y)} \)
2. Cancel (x − y) and (x + y) to get \( \dfrac{y}{x} \).

Teacher's Note:
1. Use a² − b² and (a + b)² factorisations.
2. Cancel every common factor.

Answer: (c) \( \dfrac{(1+x)(2-x)}{2x-1} \)

Explanation:
1. Combine: \( \dfrac{(a-1)+(2a+1)}{a^2-1} \) = \( \dfrac{3a}{(a-1)(a+1)} \)
2. With a = (1 + x)/(2 − x): a + 1 = 3/(2 − x) and a − 1 = (2x − 1)/(2 − x)
3. So the value is \( \dfrac{3a}{(a-1)(a+1)} \) = \( \dfrac{(1+x)(2-x)}{2x-1} \).

Teacher's Note:
1. First combine the two fractions over the denominator a² − 1.
2. Then substitute a and simplify.

Answer: (b) 1

Explanation:
1. \( x^2+y^2=\dfrac{4T^2+(1-T^2)^2}{(1+T^2)^2} \)
2. \( =\dfrac{(1+T^2)^2}{(1+T^2)^2} \) = 1.

Teacher's Note:
1. This is the identity for sin 2θ and cos 2θ in terms of tan θ.
2. Expand (1 − T²)² and add 4T².

Answer: (b) \( \dfrac{x-3}{x-5} \)

Explanation:
1. Factorise: \( \dfrac{(x-2)(x-3)}{(x-4)(x-5)}\times\dfrac{(x-1)(x-4)}{(x-1)(x-2)} \)
2. Cancel (x − 2), (x − 4), (x − 1) to get \( \dfrac{x-3}{x-5} \).

Teacher's Note:
1. Factorise every quadratic.
2. Cancel across the multiplication sign.

Answer: (c) 1 and 3

Explanation:
1. Statement 1 is true because a polynomial p(x) = p(x)/1.
2. Statement 2 is false: 1/x is a rational expression but not a polynomial.
3. Statement 3 is true. So 1 and 3 are correct.

Teacher's Note:
1. A rational expression is a ratio of polynomials with a non-zero denominator.
2. Polynomials are the special case with denominator 1.

Answer: (a) The sum of two rational expressions is a rational expression.

Explanation:
1. Rational expressions are closed under addition, subtraction, multiplication and division by a non-zero expression
2. So only statement (a) is correct; the others deny closure.

Teacher's Note:
1. Closure holds for all four operations (division by non-zero).
2. Compare with rational numbers, which behave the same way.

Answer: (c) \( \dfrac{a^2+b^2}{ab} \)

Explanation:
1. \( \dfrac{1}{x} \) = (a + b)/a and \( \dfrac{1}{y} \) = (a − b)/b
2. Sum = \( \dfrac{b(a+b)+a(a-b)}{ab} \) = \( \dfrac{a^2+b^2}{ab} \).

Teacher's Note:
1. Take reciprocals first.
2. Then add using the common denominator ab.

Answer: (a) \( \dfrac{4}{x^2+4} \)

Explanation:
1. \( \sqrt2x-\sqrt8=\sqrt2(x-2) \) and \( \sqrt8x+\sqrt{32}=2\sqrt2(x+2) \)
2. Product \( \dfrac{\sqrt2(x-2)\cdot2\sqrt2(x+2)}{(x-2)(x+2)(x^2+4)} \) = \( \dfrac{4}{x^2+4} \).

Teacher's Note:
1. Factor out the surd coefficient first.
2. Use x² − 4 = (x − 2)(x + 2).

Answer: (d) 1

Explanation:
1. Factorise: \( \dfrac{(x-3)(x-4)}{(x-7)(x+5)}\times\dfrac{(x-6)(x-7)}{(x-3)(x+2)}\times\dfrac{(x+2)(x+5)}{(x-4)(x-6)} \)
2. Every factor in a numerator also appears in a denominator, so everything cancels and the value is 1.

Teacher's Note:
1. Factorise every quadratic into linear factors.
2. Check that each factor cancels exactly once.

Answer: (d) 1

Explanation:
1. \( \dfrac{1-x^4}{1-x} \) = (1 + x)(1 + x²)
2. Value = \( (1+x)(1+x^2)\cdot\dfrac{x}{1+x^2}\cdot\dfrac{1}{x(1+x)} \) = 1.

Teacher's Note:
1. Factorise 1 − x⁴ as (1 − x)(1 + x)(1 + x²).
2. Dividing by a fraction means multiplying by its reciprocal.

Answer: (c) −1

Explanation:
1. Numerator = (x − y)/y and denominator = (y − x)/y
2. Ratio = (x − y)/(y − x) = −1.

Teacher's Note:
1. Take y as the common denominator.
2. Note that y − x = −(x − y).

Answer: (d) \( \dfrac{4}{x-y} \)

Explanation:
1. Numerator = 4xy and denominator = xy(x − y)
2. Value = \( \dfrac{4}{x-y} \).

Teacher's Note:
1. Use (a + b)² − (a − b)² = 4ab.
2. Factor xy from the denominator.

Answer: (b) 0

Explanation:
1. Factorise: x² − x − 6 = (x − 3)(x + 2), 2x² + 5x − 3 = (2x − 1)(x + 3), x² + 5x + 6 = (x + 2)(x + 3)
2. Terms become \( \dfrac{1}{x+2}+\dfrac{1}{x+3}-\dfrac{2x+5}{(x+2)(x+3)} \)
3. = \( \dfrac{(x+3)+(x+2)-(2x+5)}{(x+2)(x+3)} \) = 0.

Teacher's Note:
1. Cancel the common factor in each fraction first.
2. Then combine over (x + 2)(x + 3).

Answer: (b) 1

Explanation:
1. First term: \( \dfrac{(x-y+z)(x+y-z)}{(x+z-y)(x+z+y)} \) = \( \dfrac{x+y-z}{x+y+z} \)
2. Second term \( \dfrac{y-x+z}{x+y+z} \) and third term \( \dfrac{z+x-y}{x+y+z} \) in the same way
3. Add: \( \dfrac{(x+y-z)+(y-x+z)+(z+x-y)}{x+y+z} \) = \( \dfrac{x+y+z}{x+y+z} \) = 1.

Teacher's Note:
1. Factorise each numerator and denominator as a difference of two squares.
2. The three numerators add up to x + y + z.

Answer: (a) (x − 3)

Explanation:
1. x³ − 27 = (x − 3)(x² + 3x + 9)
2. Cancel the quadratic factor to get x − 3.

Teacher's Note:
1. Use a³ − b³ = (a − b)(a² + ab + b²) with a = x, b = 3.
2. Cancel the common factor.