IMO Maths Olympiad Class 10 Chapter 02 Polynomials MCQs with Answers

Answer: (a) 3, 5

Explanation:
1. Sum of zeroes of 5x² − (3 + k)x + 7 is \( \dfrac{3+k}{5} \) = 0, so k = −3.
2. Then 2x² − 2(−3 + 11)x + 30 = 2x² − 16x + 30 = 2(x² − 8x + 15) = 2(x − 3)(x − 5).
3. So the zeroes are 3, 5.

Teacher's Note:
1. For ax² + bx + c, sum of zeroes = −b/a.
2. Take out the common factor 2 before factorising.

Answer: (d) 7 or −7

Explanation:
1. a + b = −p and ab = 12.
2. (a − b)² = (a + b)² − 4ab ⇒ 1 = p² − 48 ⇒ p² = 49.
3. So p = 7 or −7 (roots −3, −4 or 4, 3; both have a − b = 1).

Teacher's Note:
1. (a − b)² = (a + b)² − 4ab links the difference of roots to the coefficients.
2. Do not drop the negative square root.

Answer: (c) x³ + 3x² + 8x − 4

Explanation:
1. A cubic with zeroes α, β, γ is x³ − (α + β + γ)x² + (αβ + βγ + γα)x − αβγ.
2. Substituting: x³ − (−3)x² + 8x − 4.
3. So the polynomial is x³ + 3x² + 8x − 4.

Teacher's Note:
1. The signs alternate: −(sum), +(pair sum), −(product).
2. Any non-zero multiple of this polynomial has the same zeroes.

Answer: (b) 1 − c

Explanation:
1. f(x) = x² + kx − (k + c), so p + q = −k and pq = −(k + c).
2. (p − 1)(q − 1) = pq − (p + q) + 1 = −k − c + k + 1.
3. So (p − 1)(q − 1) = 1 − c.

Teacher's Note:
1. First write the polynomial in standard form ax² + bx + c.
2. Shortcut: (p − 1)(q − 1) = f(1)/1 = 1 + 0 − c.

Answer: (b) −38x − 60

Explanation:
1. Divide: 6x³ + 11x² − 39x − 65 = (x² + x − 1)(6x + 5) + (−38x − 60).
2. Subtracting the remainder leaves an exact multiple of x² + x − 1.
3. So −38x − 60 must be subtracted.

Teacher's Note:
1. The quantity to subtract is always the remainder.
2. Keep track of signs when subtracting in long division.

Answer: (c)

Explanation:
1. The real zeroes of y = p(x) are the x-coordinates of the points where the graph meets the x-axis.
2. (a) touches the x-axis at 2 points, (b) meets it at 3 points and (d) cuts it at only 1 point.
3. Graph (c) cuts the x-axis at 4 points, so it has more than three distinct real roots.

Teacher's Note:
1. Count the distinct points where the curve meets the x-axis.
2. A point where the curve only touches the axis is still one root (repeated).

Answer: (b) 5

Explanation:
1. The highest power of x is 5, so the polynomial has degree 5 (taking a ≠ 0).
2. A polynomial of degree n has at most n zeroes.
3. So it has at most 5 zeroes.

Teacher's Note:
1. The number of zeroes is limited by the degree, not by the number of terms.
2. A degree-n graph can cut the x-axis at most n times.

Answer: (c) 6, 25

Explanation:
1. Dividing x⁴ + αx² + β by x² + 2x + 5 gives quotient x² − 2x + (α − 1) and remainder (12 − 2α)x + (β − 5α + 5).
2. For a factor the remainder must be zero: 12 − 2α = 0 ⇒ α = 6, and β − 30 + 5 = 0 ⇒ β = 25.
3. Check: (x² + 2x + 5)(x² − 2x + 5) = x⁴ + 6x² + 25, so α = 6, β = 25.

Teacher's Note:
1. Since there is no x³ or x term, the other factor must be of the form x² − 2x + m.
2. Comparing coefficients is quicker than long division here.

Answer: (d) \( \dfrac{b^3+3abc}{a^2c} \)

Explanation:
1. α + β = \( -\dfrac{b}{a} \) and αβ = \( -\dfrac{c}{a} \).
2. \( \dfrac{\alpha^2}{\beta} + \dfrac{\beta^2}{\alpha} = \dfrac{\alpha^3+\beta^3}{\alpha\beta} = \dfrac{(\alpha+\beta)^3 - 3\alpha\beta(\alpha+\beta)}{\alpha\beta} \).
3. Numerator \( = -\dfrac{b^3}{a^3} - \dfrac{3bc}{a^2} = -\dfrac{b^3+3abc}{a^3} \); dividing by \( -\dfrac{c}{a} \) gives \( \dfrac{b^3+3abc}{a^2c} \).

Teacher's Note:
1. Note the constant term is −c, so αβ = −c/a, not c/a.
2. α³ + β³ = (α + β)³ − 3αβ(α + β).

Answer: (b) (−6x² + 9) m

Explanation:
1. Perimeter = 2(length + breadth), so length + breadth = 2x³ − x² + 2.
2. Breadth = (2x³ − x² + 2) − (2x³ + 5x² − 7).
3. Breadth = (−6x² + 9) m.

Teacher's Note:
1. Halve the perimeter first, then subtract the length.
2. Change every sign of the length when subtracting.

Answer: (b) (2x² + 3) L

Explanation:
1. 2x³ + 2x² + 3x + 3 = 2x²(x + 1) + 3(x + 1) = (2x² + 3)(x + 1).
2. 4x³ − 2x² + 6x − 3 = 2x²(2x − 1) + 3(2x − 1) = (2x² + 3)(2x − 1).
3. The biggest common measure is the H.C.F. = (2x² + 3) L.

Teacher's Note:
1. Factorise by grouping pairs of terms.
2. The biggest measure that divides both is the HCF.

Answer: (d) 1683 m²

Explanation:
1. When x = 3, length = 3 × 9 + 6 = 33 m and breadth = 2 × 27 − 3 = 51 m.
2. Area = 33 × 51 = 1683 m².

Teacher's Note:
1. Substitute first, then multiply: it is quicker than expanding.
2. Expanding (6x⁵ + 4x⁴ − 9x² − 6x) at x = 3 gives the same 1683.

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

Explanation:
1. Statement I is false: the division algorithm needs r(x) = 0 or degree of r(x) less than degree of g(x).
2. Statement II: 4x⁵ + 3x³ + 2x² + 8 = (4x² + 2x + 1)(x³ − \( \dfrac{1}{2} \)x² + \( \dfrac{3}{4} \)x + \( \dfrac{1}{4} \)) + (\( -\dfrac{5}{4} \)x + \( \dfrac{31}{4} \)).
3. The remainder \( -\dfrac{5}{4}x + \dfrac{31}{4} \) has degree 1, so Statement II is true.

Teacher's Note:
1. In the division algorithm the remainder always has a smaller degree than the divisor.
2. A quadratic divisor leaves a remainder of degree 1 or less.

Answer: (b) (P) → (ii); (Q) → (iii); (R) → (i)

Explanation:
1. (P) Product of zeroes = \( \dfrac{4}{k^2+4} = 1 \) ⇒ k² = 0 ⇒ k = 0 → (ii).
2. (Q) Sum of zeroes = \( -\dfrac{k}{2} = 3 \) ⇒ k = −6 → (iii).
3. (R) Dividing ax³ + bx − c by x² + bx + c gives remainder (b − ac + ab²)x + (abc − c); abc − c = 0 gives ab = 1 → (i).

Teacher's Note:
1. Reciprocal zeroes means product = 1.
2. For exact division set each remainder coefficient to zero.

Answer: (c) (i) 0, (ii) 0, (iii) 0

Explanation:
1. Putting x = 1: L + M + N + R + P = 0 …(1); putting x = −1: L − M + N − R + P = 0 …(2).
2. Adding: 2(L + N + P) = 0, so L + N + P = 0. Subtracting: 2(M + R) = 0, so M + R = 0.
3. M³ + R³ = (M + R)³ − 3MR(M + R) = 0. Answer: 0, 0, 0.

Teacher's Note:
1. p(1) is the sum of the coefficients; p(−1) alternates their signs.
2. If M + R = 0 then R = −M, so M³ + R³ = 0.