CBSE Class 10 Mathematics Polynomials, Linear Equations, Trigonometry and Statistics Assignment Set 08

Answer: (d) more than 3. Any non-zero constant multiple of (x + 2)(x − 5) has these zeroes, so there are infinitely many.

Teacher's Note:
1. Polynomials k(x + 2)(x − 5) all have the same zeroes.
2. Zeroes do not fix the polynomial uniquely.

Answer: (c) an odd integer. For n = 2k + 1, n² − 1 = 4k(k + 1), and k(k + 1) is always even, so it is divisible by 8.

Teacher's Note:
1. The product of two consecutive integers is even.
2. Check with n = 3: 9 − 1 = 8.

Answer: (a) 1 solution. Here a₁/a₂ = 1/2 and b₁/b₂ = −3/4, which are different, so the lines intersect at one point.

Teacher's Note:
1. If a₁/a₂ ≠ b₁/b₂ the pair has a unique solution.
2. The printed key in the source was wrong for this question.

Answer: (b) 45°. tan θ = 10/10 = 1, so θ = 45°.

Teacher's Note:
1. Height and shadow are equal, so tan θ = 1.
2. Draw the right triangle to see the angle at the ground.

Answer: (c) Range. Range measures the spread of data, not its centre.

Teacher's Note:
1. Mean, median and mode are the three measures of central tendency.
2. Range is a measure of dispersion.

Answer: (c) 100. Mode = 3 Median − 2 Mean gives 80 = 3 Median − 220, so Median = 100.

Teacher's Note:
1. Use the empirical relation 3 Median = Mode + 2 Mean.
2. Median lies between mode and mean here.

Answer: (d) 30°. sin θ = 5/10 = 1/2, so θ = 30°.

Teacher's Note:
1. The ladder is the hypotenuse of the right triangle.
2. Use sine since height and hypotenuse are known.

Answer: length. This is a standard theorem on tangents.

Teacher's Note:
1. The two tangent segments from a point outside the circle are equal.
2. This follows from congruent right triangles formed with the centre.

Answer: tangent. A line meeting the circle at two points is a secant.

Teacher's Note:
1. A tangent is perpendicular to the radius at the point of contact.
2. The common point is called the point of contact.

Answer: two. They touch the circle at the two ends of the diameter perpendicular to the line.

Teacher's Note:
1. Tangents at the ends of a diameter are parallel.
2. So there are exactly two for each direction.

Answer: (b) both negative. The sum of zeroes is −99 and the product is 127 (positive), so both zeroes have the same sign and it is negative.

Teacher's Note:
1. A positive product means same signs.
2. A negative sum then makes both negative.

Answer: (c) infinitely many. The second equation is twice the first, so the lines coincide.

Teacher's Note:
1. Here a₁/a₂ = b₁/b₂ = c₁/c₂ = 1/2.
2. Coincident lines give infinitely many solutions.

Answer: (d) k ≠ 3. A unique solution needs k/6 ≠ (−1)/(−2), that is k ≠ 3.

Teacher's Note:
1. Unique solution needs a₁/a₂ ≠ b₁/b₂.
2. For k = 3 the lines are parallel.

Answer: 1. This is the basic trigonometric identity.

Teacher's Note:
1. It comes from the Pythagoras theorem.
2. It holds for every angle θ.

Answer: cos θ. The sine of an angle equals the cosine of its complement.

Teacher's Note:
1. Complementary angles add up to 90°.
2. Similarly cos (90° − θ) = sin θ.

Answer: (b) 17. Mode = 3 × 15 − 2 × 14 = 45 − 28 = 17.

Teacher's Note:
1. Use Mode = 3 Median − 2 Mean.
2. Here mode is greater than median.

Answer: (d) 16 : 81. The ratio of areas equals the square of the ratio of sides, (4/9)² = 16/81.

Teacher's Note:
1. Area ratio = (side ratio)².
2. Keep the order of the triangles as given.

Answer: (a) 12 cm. AB/LM = 60/40, so AB = 8 × 1.5 = 12 cm.

Teacher's Note:
1. Corresponding sides of similar triangles are in the ratio of the perimeters.
2. The options in the source did not contain the correct value, so they were corrected.

Answer: (a) parallel. An inconsistent pair has no solution, which happens only for parallel lines.

Teacher's Note:
1. Consistent pairs have intersecting or coincident lines.
2. Inconsistent means no common point.

Answer: (a) 3. Each point where the graph cuts the x-axis gives one zero.

Teacher's Note:
1. Zeroes are the x-coordinates of those points.
2. A cubic polynomial can have at most 3 zeroes.

Answer: Let the numbers be x and 27 − x. Then x(27 − x) = 182, so x² − 27x + 182 = 0.
Discriminant = 729 − 728 = 1, so x = (27 ± 1)/2 = 14 or 13.
The numbers are 13 and 14.

Teacher's Note:
1. Form a quadratic with sum and product of the numbers.
2. Check: 13 + 14 = 27 and 13 × 14 = 182.

Answer: tan (A + B) = √3 gives A + B = 60°; tan (A − B) = 1/√3 gives A − B = 30°.
Adding: 2A = 90°, so A = 45°, and B = 60° − 45° = 15°.

Teacher's Note:
1. Use the standard values tan 60° = √3 and tan 30° = 1/√3.
2. Solve the two linear equations in A and B.

Answer: Multiply the first equation by 3: 3x + 3y = 15. Add to 2x − 3y = 4: 5x = 19, so x = 19/5.
Then y = 5 − 19/5 = 6/5. Solution: x = 19/5, y = 6/5.

Teacher's Note:
1. Make the coefficients of y equal in size and opposite in sign.
2. Verify: 2(19/5) − 3(6/5) = 20/5 = 4.

Answer: Let A(3, 0), B(4, 5), C(−1, 4), D(−2, −1). Diagonal AC = √(16 + 16) = 4√2 and diagonal BD = √(36 + 36) = 6√2.
Area = ½ × AC × BD = ½ × 4√2 × 6√2 = 24 square units.

Teacher's Note:
1. Area of a rhombus is half the product of its diagonals.
2. AB = BC = √26 confirms that it is a rhombus.

Answer: Nearer ship: distance = 75/tan 45° = 75 m. Farther ship: distance = 75/tan 30° = 75√3 m.
Required distance = 75(√3 − 1) = 75 × 0.732 ≈ 54.9 m.

Teacher's Note:
1. The angle of depression equals the angle of elevation from the ship.
2. Subtract the two horizontal distances.

Answer: Total: 5 + x + 20 + 15 + y + 5 = 60, so x + y = 15. The median class is 20-30 (cf before it is 5 + x).
Median = 20 + [(30 − (5 + x))/20] × 10 = 28.5, so (25 − x)/2 = 8.5, giving x = 8 and y = 7.

Teacher's Note:
1. Use n/2 = 30 to locate the median class.
2. Check that the cumulative frequency up to 30 is 33, which is at least 30.

Answer: Let the digits be a and b. (10a + b) + (10b + a) = 11(a + b) = 66, so a + b = 6. Also a − b = ±2.
This gives (a, b) = (4, 2) or (2, 4), so the numbers are 42 and 24; there are 2 such numbers.

Teacher's Note:
1. The sum of a number and its reverse is 11 times the digit sum.
2. Consider both a − b = 2 and b − a = 2.

Answer: Let the angles be θ and 90° − θ, and the height h. Then tan θ = h/4 and tan (90° − θ) = cot θ = h/9.
Since tan θ × cot θ = 1, (h/4)(h/9) = 1, so h² = 36 and h = 6 m.

Teacher's Note:
1. For complementary angles, tan θ · tan (90° − θ) = 1.
2. Take the positive root as height cannot be negative.

Answer: Let t = tan θ. LHS = t/(1 − 1/t) + (1/t)/(1 − t) = t²/(t − 1) − 1/[t(t − 1)] = (t³ − 1)/[t(t − 1)] = (t² + t + 1)/t.
This equals t + 1 + 1/t = 1 + (sin²θ + cos²θ)/(sin θ cos θ) = 1 + sec θ cosec θ = RHS.

Teacher's Note:
1. Convert everything to tan θ and use a³ − b³ = (a − b)(a² + ab + b²).
2. Finally write tan θ + cot θ as 1/(sin θ cos θ).

Answer: (x + √3)(x − √3) = x² − 3 is a factor. Dividing x⁴ − 3x³ − x² + 9x − 6 by x² − 3 gives the quotient x² − 3x + 2 = (x − 1)(x − 2).
The zeroes are −√3, √3, 1 and 2.

Teacher's Note:
1. Divide by the factor formed from the known zeroes.
2. Verify by multiplying (x² − 3)(x² − 3x + 2).