CBSE Class 10 Mathematics Real Numbers, Polynomials, Linear and Quadratic Equations, A.P., Triangles and Coordinate Geometry Assignment Set 14

Answer: (b) −5, 7. Put x + 5 = 0 and x − 7 = 0 to get x = −5 and x = 7.

Teacher's Note:
1. A zero is the value of x that makes the polynomial 0.
2. Each linear factor is set to zero separately.

Answer: (a) 30 years. Let the father be x and the son y: x + y = 45 and 3(x − y) = 45, so x − y = 15. Adding gives 2x = 60, so x = 30.

Teacher's Note:
1. Form two linear equations and add them to eliminate y.
2. Check: 30 + 15 = 45 and 3 × 15 = 45.

Answer: (d) ±12. Equal roots need D = k² − 4(4)(9) = 0, so k² = 144 and k = ±12.

Teacher's Note:
1. Equal roots mean the discriminant is zero.
2. Do not forget the negative value of k.

Answer: (b) 35th. Here a = 21, d = −3, so 21 − 3(n − 1) = −81 gives n − 1 = 34 and n = 35.

Teacher's Note:
1. Use aₙ = a + (n − 1)d with d = −3.
2. Substitute back: 21 − 3 × 34 = −81.

Answer: (b) 36 cm². Triangle ADE is similar to triangle ABC, so the ratio of areas is \( \left(\dfrac{2}{3}\right)^2 = \dfrac{4}{9} \), and \( \dfrac{4}{9} \times 81 = 36 \).

Teacher's Note:
1. Areas of similar triangles are in the ratio of the squares of corresponding sides.
2. DE ∥ BC gives the similarity by AA.

Answer: 4025 = 1656 × 2 + 713
1656 = 713 × 2 + 230
713 = 230 × 3 + 23
230 = 23 × 10 + 0
The remainder is now 0, so HCF = 23.

Teacher's Note:
1. Repeat the division until the remainder becomes zero.
2. The last non-zero divisor is the HCF.

Answer: Suppose 4 − 3√2 is rational, say \( \dfrac{p}{q} \) with q ≠ 0 and p, q integers. Then 3√2 = 4 − \( \dfrac{p}{q} \), so √2 = \( \dfrac{4q - p}{3q} \), which is rational. This contradicts the fact that √2 is irrational. Hence 4 − 3√2 is irrational.

Teacher's Note:
1. Use proof by contradiction and the known result that √2 is irrational.
2. A rational number minus a rational multiple of an irrational number cannot stay rational.

Answer: 65 = 5 × 13 and 117 = 3² × 13, so HCF = 13.
65m − 117 = 13 gives 65m = 130, so m = 2.
LCM = 3² × 5 × 13 = 585.

Teacher's Note:
1. Take the common prime factors with the lowest powers for the HCF, and all primes with the highest powers for the LCM.
2. Check: 13 × 585 = 7605 = 65 × 117.

Answer: n³ − n = n(n − 1)(n + 1), the product of three consecutive integers (n − 1), n, (n + 1). Among any three consecutive integers at least one is even and exactly one is a multiple of 3. So the product is divisible by both 2 and 3, and since 2 and 3 are co-prime, it is divisible by 6.

Teacher's Note:
1. Factorise the expression first.
2. A number divisible by two co-prime numbers is divisible by their product.

Answer: Sum of zeroes = 3 + (−4) = −1 and product = 3 × (−4) = −12.
The polynomial is x² − (sum)x + product = x² + x − 12.

Teacher's Note:
1. Use the form x² − (α + β)x + αβ.
2. Any non-zero constant multiple of this polynomial is also correct.

Answer: Adding: 2α = 32, so α = 16 and β = 8.
Polynomial: x² − 24x + 128.
Verification: sum of zeroes = 16 + 8 = 24 = −(−24)/1 and product = 16 × 8 = 128 = 128/1. Both relations hold.

Teacher's Note:
1. Find the two zeroes first by adding and subtracting the given equations.
2. Compare with −b/a for the sum and c/a for the product.

Answer: Divide f(x) by 3x² − 5. The quotient is x² − 3x + 2 and the remainder is k + 10.
For exact divisibility k + 10 = 0, so k = −10.
The quotient x² − 3x + 2 = (x − 1)(x − 2) gives the other two zeroes 1 and 2. (The zeroes of 3x² − 5 are ±√(5/3).)

Teacher's Note:
1. Use the division algorithm and put the remainder equal to zero.
2. The remaining zeroes come from factorising the quotient.

Answer: For no solution, \( \dfrac{3}{2k-1} = \dfrac{1}{k-1} \neq \dfrac{1}{2k+1} \).
3(k − 1) = 2k − 1 gives k = 2.
For k = 2: \( \dfrac{1}{k-1} = 1 \) and \( \dfrac{1}{2k+1} = \dfrac{1}{5} \), which are different, so k = 2.

Teacher's Note:
1. Inconsistent means parallel lines, so a₁/a₂ = b₁/b₂ ≠ c₁/c₂.
2. Always check the third ratio after finding k.

Answer: From the second equation x = y + 1. Then 2(y + 1) + 3y = 12 gives y = 2 and x = 3, so the lines meet at (3, 2).
On the x-axis (y = 0): 2x = 12 gives (6, 0), and x = 1 gives (1, 0).
Base = 6 − 1 = 5 and height = 2, so area = ½ × 5 × 2 = 5 square units.

Teacher's Note:
1. The height of the triangle is the y-coordinate of the intersection point.
2. This is the region you would shade if you drew the graph.

Answer: Let right answers be x and wrong answers y.
4x − y = 70 and 5x − 2y = 80.
From the first, y = 4x − 70. Then 5x − 8x + 140 = 80, so x = 20 and y = 10.
Total questions = 20 + 10 = 30. Using unfair means would violate honesty.

Teacher's Note:
1. Form two linear equations and solve by substitution.
2. Check: 4 × 20 − 10 = 70 and 5 × 20 − 2 × 10 = 80.

Answer: For real roots D ≥ 0: (−3k)² − 4(9)(k) ≥ 0, so 9k² − 36k ≥ 0, that is 9k(k − 4) ≥ 0.
Hence k ≤ 0 or k ≥ 4.

Teacher's Note:
1. Real roots need a non-negative discriminant.
2. The product k(k − 4) is non-negative outside the interval between 0 and 4.

Answer: Let the speed of the stream be x km/h.
\( \dfrac{32}{24-x} - \dfrac{32}{24+x} = 1 \)
32 × 2x = 576 − x², so x² + 64x − 576 = 0, i.e. (x + 72)(x − 8) = 0.
Speed cannot be negative, so the speed of the stream is 8 km/h.

Teacher's Note:
1. Upstream speed is 24 − x and downstream speed is 24 + x.
2. Reject the negative root.

Answer: Let the multiples be 7n and 7(n + 1). Then 49n² + 49(n + 1)² = 637, so n² + (n + 1)² = 13, i.e. n² + n − 6 = 0.
(n + 3)(n − 2) = 0 gives n = 2 or n = −3.
The multiples are 14 and 21 (or −21 and −14).

Teacher's Note:
1. Divide by 49 early to keep the numbers small.
2. Both positive and negative pairs satisfy the condition.

Answer: For three terms in A.P., 2 × middle = sum of the other two.
2(4x − 1) = (x − 2) + (5x + 2), so 8x − 2 = 6x and x = 1/3.
Check: the terms are −5/3, 1/3, 11/3 with common difference 2.

Teacher's Note:
1. Equal common differences give 2b = a + c.
2. Verify by listing the three terms.

Answer: Let the angles be a − d, a, a + d. Their sum is 3a = 180°, so a = 60°.
a + d = 2(a − d) gives 60 + d = 120 − 2d, so d = 20°.
The angles are 40°, 60° and 80°.

Teacher's Note:
1. The middle term of three angles in A.P. is always 60°.
2. Use the angle sum property of a triangle.

Answer: a = 121 and d = −4, so aₙ = 125 − 4n.
aₙ < 0 requires 4n > 125, so n > 31.25 and the smallest integer is n = 32.
The 32nd term is the first negative term, and its value is −3.

Teacher's Note:
1. Write aₙ as an inequality and solve for n.
2. n must be a whole number.

Answer: a + 16d = 2(a + 7d) + 5 gives a = 2d − 5.
a + 10d = 43 gives 12d − 5 = 43, so d = 4 and a = 3.
aₙ = 3 + (n − 1)4 = 4n − 1.

Teacher's Note:
1. Express a in terms of d and substitute in the second condition.
2. Check: a₁₇ = 67 and a₈ = 31, and 2 × 31 + 5 = 67.

Answer: Statement: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, it divides the two sides in the same ratio.
Proof: In triangle ABC let DE ∥ BC with D on AB and E on AC. Join BE and CD, and draw EM ⊥ AB and DN ⊥ AC.
ar(ADE)/ar(BDE) = (½ × AD × EM)/(½ × DB × EM) = AD/DB. Similarly ar(ADE)/ar(CED) = AE/EC.
Triangles BDE and CED are on the same base DE and between the same parallels DE and BC, so ar(BDE) = ar(CED).
Therefore AD/DB = AE/EC. Hence proved.

Teacher's Note:
1. The proof compares areas of triangles with a common height.
2. Triangles on the same base and between the same parallels have equal area.

Answer: Statement: In a right triangle the square of the hypotenuse equals the sum of the squares of the other two sides.
Proof: In right triangle ABC with the right angle at B, draw BD ⊥ AC. Triangles ADB and ABC are similar (AA), so AD/AB = AB/AC, giving AB² = AD × AC. Triangles BDC and ABC are similar, so DC/BC = BC/AC, giving BC² = DC × AC.
Adding: AB² + BC² = AC(AD + DC) = AC × AC = AC². Hence proved.

Teacher's Note:
1. Drop the perpendicular from the right-angle vertex to the hypotenuse.
2. The two smaller triangles are each similar to the whole triangle.

Answer: Let CD = x, so DB = 3x and BC = 4x.
In right triangle ADB, AB² = AD² + 9x². In right triangle ADC, AC² = AD² + x².
Subtracting: AB² − AC² = 8x².
Now BC² = 16x² = 2 × 8x² = 2(AB² − AC²), so 2AB² = 2AC² + BC². Hence proved.

Teacher's Note:
1. Apply Pythagoras theorem in both right triangles.
2. Express everything in terms of CD.

Answer: Let A(3, 0), B(6, 4), C(−1, 3).
AB² = 3² + 4² = 25, AC² = 4² + 3² = 25, BC² = 7² + 1² = 50.
AB = AC, and AB² + AC² = 50 = BC², so the triangle is right-angled at A and isosceles.

Teacher's Note:
1. Use the distance formula for all three sides.
2. Check the equal sides and Pythagoras together.

Answer: Let P = (2y, y). PQ² = PR² gives (2y − 2)² + (y + 5)² = (2y + 3)² + (y − 6)².
5y² + 2y + 29 = 5y² + 45, so y = 8 and x = 16.
P = (16, 8). Check: PQ² = 14² + 13² = 365 and PR² = 19² + 2² = 365.

Teacher's Note:
1. Write P in terms of one variable.
2. Equating squared distances avoids square roots.

Answer: Let the line divide AB in the ratio k : 1. The point is \( \left(\dfrac{3k+2}{k+1}, \dfrac{7k-2}{k+1}\right) \).
Putting it in 2x + y = 4: 2(3k + 2) + (7k − 2) = 4(k + 1), so 13k + 2 = 4k + 4 and k = 2/9.
The ratio is 2 : 9. The point is \( \left(\dfrac{24}{11}, -\dfrac{4}{11}\right) \), and 2(24/11) − 4/11 = 4 confirms it.

Teacher's Note:
1. Use the section formula with ratio k : 1.
2. Substitute the dividing point into the equation of the line.

Answer: Let the point be (x, 0). Then (x − 2)² + 25 = (x + 2)² + 81.
−4x + 29 = 4x + 85, so −8x = 56 and x = −7.
The point is (−7, 0).

Teacher's Note:
1. A point on the x-axis has y-coordinate 0.
2. Equate the squares of the two distances.