IMO Maths Olympiad Class 10 Chapter 04 Quadratic Equations MCQs with Answers

Answer: (d) x = 2

Explanation:
1. The expression under the first root is 2 + x itself, so \( x = \sqrt{2+x} \).
2. Squaring: x² = 2 + x ⇒ x² − x − 2 = 0 ⇒ (x − 2)(x + 1) = 0.
3. Since x > 0, x = 2.

Teacher's Note:
1. For an endless nested root, replace the repeating part by x itself.
2. Reject the negative root because a square root is never negative.

Answer: (d) \( \dfrac{c(a-b)}{a(b-c)} \)

Explanation:
1. Let the other root be α.
2. Product of roots = \( \dfrac{\text{constant term}}{\text{coefficient of } x^2} \), so \( \alpha \times 1 = \dfrac{c(a-b)}{a(b-c)} \).
3. So the other root is \( \dfrac{c(a-b)}{a(b-c)} \).

Teacher's Note:
1. When one root is known, the product of roots gives the other root in one step.
2. Here the coefficients add up to 0, which is why 1 is a root.

Answer: (b) 2a = b + c

Explanation:
1. The sum of the coefficients is (a − b) + (b − c) + (c − a) = 0, so x = 1 is a root.
2. The roots are equal, so both roots are 1 and their product is 1: \( \dfrac{c-a}{a-b} = 1 \).
3. So c − a = a − b, i.e. 2a = b + c.

Teacher's Note:
1. If the coefficients of a quadratic add up to zero, 1 is a root.
2. This shortcut avoids expanding the discriminant (b − c)² − 4(a − b)(c − a).

Answer: (c) 5

Explanation:
1. Write the equation as \( \sqrt{x-1} - \sqrt{x+1} = -1 \) …(1)
2. Rationalising, \( \dfrac{(x-1)-(x+1)}{\sqrt{x-1}+\sqrt{x+1}} = -1 \Rightarrow \sqrt{x-1}+\sqrt{x+1} = 2 \) …(2)
3. Adding (1) and (2): \( 2\sqrt{x-1} = 1 \Rightarrow x - 1 = \dfrac{1}{4} \), so \( x = \dfrac{5}{4} \) and 4x = 5.

Teacher's Note:
1. The product of \( \sqrt{x-1}-\sqrt{x+1} \) and its conjugate is the constant −2, which makes the conjugate easy to find.
2. Check: √(1/4) − √(9/4) + 1 = 0.5 − 1.5 + 1 = 0.

Answer: (d) None of these

Explanation:
1. α + β = −3 and αβ = \( \dfrac{a}{2} \), so \( \dfrac{\alpha}{\beta}+\dfrac{\beta}{\alpha} = \dfrac{(\alpha+\beta)^2-2\alpha\beta}{\alpha\beta} = \dfrac{9-a}{a/2} = \dfrac{2(9-a)}{a} \).
2. We need \( \dfrac{2(9-a)}{a} \lt 2 \), i.e. \( \dfrac{9-a}{a} - 1 = \dfrac{9-2a}{a} \lt 0 \).
3. This holds when a < 0 or \( a \gt \dfrac{9}{2} \).
4. No option gives this complete set, so the answer is None of these.

Teacher's Note:
1. Do not multiply an inequality by a without knowing its sign; split into a > 0 and a < 0.
2. Option (a) gives only part of the solution set.

Answer: (b) (a + 1), −(a + 2)

Explanation:
1. Split the middle term: x = (a + 2)x − (a + 1)x.
2. x² + (a + 2)x − (a + 1)x − (a + 1)(a + 2) = [x − (a + 1)][x + (a + 2)] = 0.
3. So the roots are (a + 1) and −(a + 2).

Teacher's Note:
1. Look for two numbers whose difference is 1 (the coefficient of x) and whose product is (a + 1)(a + 2).
2. Check: sum = −1 and product = −(a + 1)(a + 2).

Answer: (b) b² = ac

Explanation:
1. Equal roots ⇒ D = 0: 4b²(a + c)² − 4(a² + b²)(b² + c²) = 0.
2. Expanding: a²b² + 2ab²c + b²c² − a²b² − a²c² − b⁴ − b²c² = 0.
3. So b⁴ − 2ab²c + a²c² = 0 ⇒ (b² − ac)² = 0 ⇒ b² = ac.

Teacher's Note:
1. After expanding, look for a perfect square.
2. b² = ac means a, b, c are in G.P.

Answer: (a) 1, −8

Explanation:
1. Let y = x^1/3; then y² + y − 2 = 0 ⇒ (y − 1)(y + 2) = 0.
2. y = 1 or y = −2, i.e. x^1/3 = 1 or x^1/3 = −2.
3. Cubing: x = 1 or x = −8.

Teacher's Note:
1. x^2/3 = (x^1/3)², so the substitution turns it into a quadratic.
2. Remember to convert back from y to x; option (b) stops at y.

Answer: (b) \( -\dfrac{1}{2} \)

Explanation:
1. Cross-multiplying: m(x² − x) − m(m + 1) = x(x − 1)(m − 1).
2. Simplifying: x² − m² − x − m = 0 ⇒ (x + m)(x − m − 1) = 0, so the roots are −m and m + 1.
3. Equal roots: −m = m + 1 ⇒ \( m = -\dfrac{1}{2} \).

Teacher's Note:
1. Clear the fractions first and collect terms.
2. m = 0 and m = 1 are not allowed since they make a denominator zero.

Answer: (b) 4 km/hr

Explanation:
1. Let the speed be x km/hr. Then \( \dfrac{48}{x} - \dfrac{48}{x+2} = 4 \).
2. \( \dfrac{96}{x(x+2)} = 4 \) ⇒ x² + 2x − 24 = 0 ⇒ (x + 6)(x − 4) = 0.
3. Speed cannot be negative, so the speed is 4 km/hr.

Teacher's Note:
1. Time = distance ÷ speed; the difference of the two times is 4 hours.
2. Reject the negative root in a real-life problem.

Answer: (a) 50

Explanation:
1. Let the number of persons be x. Then \( \dfrac{6500}{x} - \dfrac{6500}{x+15} = 30 \).
2. \( \dfrac{6500 \times 15}{x(x+15)} = 30 \) ⇒ x² + 15x − 3250 = 0 ⇒ (x − 50)(x + 65) = 0.
3. So the number of persons is 50.

Teacher's Note:
1. Each share = total ÷ number of persons; form the equation from the difference of shares.
2. Check: 6500 ÷ 50 = 130 and 6500 ÷ 65 = 100, a difference of 30.

Answer: (b) 2 km/hr

Explanation:
1. Let the speed of the stream be s km/hr: upstream speed 5 − s, downstream speed 5 + s.
2. \( \dfrac{5.25}{5-s} - \dfrac{5.25}{5+s} = 1 \Rightarrow 10.5s = 25 - s^2 \) ⇒ 2s² + 21s − 50 = 0.
3. (s − 2)(2s + 25) = 0, so the speed of the stream is 2 km/hr.

Teacher's Note:
1. Upstream speed = boat speed − stream speed; downstream speed = boat speed + stream speed.
2. Check: 5.25 ÷ 3 − 5.25 ÷ 7 = 1.75 − 0.75 = 1 hour.

Answer: (c) Statement - I is false but Statement - II is true.

Explanation:
1. Statement I is false: if b² − 4ac = 0 the roots are equal, so distinct real roots need b² − 4ac > 0.
2. Statement II: D = 4(a + b)² − 8(a² + b²) = −4a² − 4b² + 8ab = −4(a − b)².
3. For a ≠ b, D < 0, so there are no real roots: Statement II is true.

Teacher's Note:
1. ‘Distinct’ rules out D = 0.
2. A discriminant of the form −(perfect square) is never positive.

Answer: (d) (i) F, (ii) F, (iii) T, (iv) T

Explanation:
1. (i) Let x be the value: \( x = 2 + \dfrac{1}{x} \) ⇒ x² − 2x − 1 = 0 ⇒ \( x = 1 + \sqrt{2} \) (positive root), not √2: F.
2. (ii) Let AC = x, CB = 2 − x: x² = 2(2 − x) ⇒ x² + 2x − 4 = 0 ⇒ \( x = \sqrt{5} - 1 \), so \( CB = 3 - \sqrt{5} \) m, not \( 3 + \sqrt{5} \): F.
3. (iii) A quadratic has at most two roots: T. (iv) This is the definition of a root: T.
4. So the answer is F, F, T, T.

Teacher's Note:
1. For an endless continued fraction, the repeating part equals the whole value.
2. A part of a 2 m segment cannot be longer than 2 m, which rules out 3 + √5 at once.

Answer: (a) \( \dfrac{3}{7} \)

Explanation:
1. Let the fraction be \( \dfrac{x}{2x+1} \). Then \( \dfrac{x}{2x+1} + \dfrac{2x+1}{x} = \dfrac{58}{21} \).
2. 21(x² + 4x² + 4x + 1) = 58x(2x + 1) ⇒ 11x² − 26x − 21 = 0 ⇒ (x − 3)(11x + 7) = 0.
3. x = 3 (the other root is not a natural number), so the fraction is \( \dfrac{3}{7} \).

Teacher's Note:
1. Convert the mixed number to an improper fraction first: 2 16/21 = 58/21.
2. 7/3 also gives the same sum, but its denominator is not one more than twice its numerator.