CBSE Class 12 PCM Mathematics Board Paper 2022 — Set 3 Solutions

Answer: Given the quadratic equation x² - 2ax + (4b² - a²) = 0, we use the quadratic formula. Here, the coefficient of x² is 1, coefficient of x is -2a, and constant term is (4b² - a²). Calculating the discriminant, we get D = (-2a)² - 4 × 1 × (4b² - a²) = 4a² - 4(4b² - a²) = 4a² - 16b² + 4a² = 8a² - 16b². The roots are given by (2a ± √D) / 2. Thus, the two roots are a plus or minus square root of (2a² - 4b²).

Answer: For the quadratic equation (1 + a²) x² + 2abx + (b² - c²) = 0 to have equal and real roots, its discriminant must be zero. The discriminant is given by b² - 4ac. Here, a = (1 + a²), b = 2ab, and c = (b² - c²). So, the discriminant is (2ab)² - 4(1 + a²)(b² - c²). Simplifying, we get 4a²b² - 4(1 + a²)(b² - c²) = 0. Dividing both sides by 4, a²b² = (1 + a²)(b² - c²). Expanding the right side, a²b² = b² + a²b² - c² - a²c². Canceling a²b² from both sides, 0 = b² - c² - a²c², which implies b² = c²(1 + a²). Hence, proved.

Answer: Given the common difference d = 5 and the 20th term a20 = 135 of an AP, we first find the first term 'a'. The nth term of an AP is given by a + (n - 1) d = a20. So, a + 19 × 5 = 135, which gives a = 135 - 95 = 40. Now, the sum of first n terms of an AP is Sₙ = (n/2) × (2a + (n - 1)d). For n = 20, S20 = (20/2) × (2 × 40 + 19 × 5) = 10 × (80 + 95) = 10 × 175 = 1750. Therefore, the sum of the first 20 terms is 1750.

Answer: The mode of a frequency distribution is the value or class interval that has the highest frequency. In the given data, the class interval 40 - 55 has the maximum frequency of 7 students. Therefore, the mode of this data is the class interval 40 - 55.

Answer: First, find the volume of one marble. Diameter = 1.4 cm, so radius = 1.4 ÷ 2 = 0.7 cm. Volume of one marble = 4/3 × π × (0.7)³ = 1.436 cm³ (approximately). For 150 marbles, total volume = 150 × 1.436 = 215.4 cm³. Diameter of the cylinder = 7 cm, so radius = 3.5 cm. The rise in water level, h = volume of marbles ÷ base area of cylinder = 215.4 ÷ (π × 3.5²) = 215.4 ÷ 38.48 = 5.6 cm approximately. So, the water level rises by about 5.6 cm.

Answer: Each cube has a side of 6 cm, so the area of one face is 6 cm × 6 cm = 36 cm². The total surface area of one cube is 6 × 36 = 216 cm². When three cubes are joined in a line, the combined shape forms a cuboid with dimensions 18 cm (length), 6 cm (width), and 6 cm (height). The total surface area of this cuboid is 2 × (length × width + width × height + height × length) = 2 × (18 × 6 + 6 × 6 + 6 × 18) = 2 × (108 + 36 + 108) = 2 × 252 = 504 cm².

Answer: The first AP has first term a₁ = 9 and common difference d = 7 - 9 = -2. Its nᵗʰ term is given by aₙ = 9 + (n - 1)(-2) = 9 - 2n + 2 = 11 - 2n. The second AP has first term 15 and common difference 12 - 15 = -3. Its nᵗʰ term is aₙ = 15 + (n - 1)(-3) = 15 - 3n + 3 = 18 - 3n. To find n for which the nᵗʰ terms are equal, set 11 - 2n = 18 - 3n. Rearranging, we get n = 7. Therefore, the nᵗʰ terms of both APs are the same when n = 7.

Answer: Given that PQ and PR are tangents to the circle from point P, and ∠OPR = 45°. Since OP and OR are radii to the points of tangency R and Q, angles OQP and ORP are right angles (90°) because the radius is perpendicular to the tangent at the point of contact. In triangle OPR, angle OPR is 45°, and angle ORP is 90°, so the third angle POR equals 45° (because the sum of angles in a triangle is 180°). Since OP = OR (radii of the same circle) and angles at P and R are 90° and 45°, the quadrilateral ORPQ has four right angles, and all sides equal, thus it forms a square.

Answer: To draw a line segment AB of length 8 cm, use a ruler to draw a straight line and mark points A and B such that the distance between them is 8 cm. Next, to locate point P on AB so that AP is to PB as 1 is to 5, divide the segment AB into 6 equal parts (because 1 + 5 = 6 parts). Each part will be 8 cm divided by 6, which is approximately 1.33 cm. Starting from point A, measure 1.33 cm along the segment AB and mark that point as P. This ensures that AP is 1 part and PB is 5 parts, maintaining the ratio 1:5.

Answer: To draw a circle of radius 3 cm with centre O, use a compass set to 3 cm and draw the circle. Mark a point P outside the circle, such that the distance between P and O is 6 cm. To construct tangents from P to the circle, follow these steps: First, join points P and O. With O as the centre and radius 3 cm, the given circle is already drawn. Now, find the midpoint M of the line segment PO. Using M as the centre and MO as radius, draw a circle. This new circle will intersect the given circle at points A and B. Join points P to A and P to B. The line segments PA and PB are the required tangents to the circle from point P. These tangents touch the circle at exactly one point each, and their lengths from P to A and P to B are equal.

Answer: Given two poles of heights 20 meters and 28 meters, connected by a wire inclined at 30° to the horizontal, we need to find the length of the wire and the distance between the poles. First, find the vertical difference between the poles: 28 m - 20 m = 8 m. Consider the triangle formed by the wire, the horizontal distance between the poles, and the vertical difference. The wire acts as the hypotenuse of this right triangle with a vertical side of 8 m and an angle of 30° with the horizontal. Using trigonometric relations, the vertical height (opposite side) = 8 m = wire length × sin 30°. Since sin 30° = 1/2, wire length = 8 ÷ (1/2) = 16 meters. Next, find the horizontal distance between the poles, which is adjacent side = wire length × cos 30°. Cos 30° = √3/2 ≈ 0.866, so horizontal distance = 16 × 0.866 = 13.856 meters approximately. Thus, the length of the wire is 16 meters, and the distance between the two poles is about 13.86 meters.

Answer: To find the mean weight using the assumed mean method, first select an assumed mean (A) from the mid-points of the given class intervals of weights. Then, calculate the deviation (d) of each mid-point from the assumed mean by subtracting A from each mid-point. Multiply each deviation by the frequency (f) of that class to get f × d. Sum all these products to get Σ(f × d). Finally, use the formula: Mean = assumed mean + (Σ(f × d) / total frequency). Here, the total number of animals (N) is 50. Substitute the values into the formula to calculate the mean weight.

Answer:

To find the median of a grouped frequency distribution, follow these steps:

1. Calculate the total number of observations (N) by adding all the frequencies.

2. Find N/2, which indicates the median position in the cumulative frequency.

3. Create a cumulative frequency column by successively adding the frequencies.

4. Identify the median class, which is the class interval where the cumulative frequency is just greater than or equal to N/2.

5. Use the median formula:

Median = L + ( (N/2 - F) / f ) × h

where:

• L = lower boundary of the median class
• F = cumulative frequency before the median class
• f = frequency of the median class
• h = class width of the median class

Substitute the values into the formula to calculate the median.

This method ensures that the median accurately represents the middle value of the grouped data.

Answer: Let us consider two circles with centres O (larger circle) and O' (smaller circle) such that the smaller circle touches the bigger circle internally at A. The radius of the bigger circle is 2r and that of the smaller circle is r. A chord AB of the bigger circle intersects the smaller circle at point C. To prove that C bisects AB, we need to show that AC = CB. Since the smaller circle touches the bigger circle internally at A, point A lies on both circles. Also, since C lies on the smaller circle and chord AB of the bigger circle, C is a point of intersection of chord AB with the smaller circle. By definition of tangent and internal touching, the centres O and O' lie on the same line as the point of contact A. The radius of the bigger circle is twice that of the smaller circle, so O' lies on OA such that OA = 2r and OO' = r. Now, draw the line from centre O to the midpoint M of chord AB. Since OM is perpendicular to AB (in a circle, the perpendicular drawn from the centre to a chord bisects the chord), M is the midpoint of AB. We need to prove C coincides with M, so that C bisects AB. Because the smaller circle intersects the chord AB at C, and this smaller circle has radius r, the only point of intersection on AB at a distance r from O is the midpoint M. Therefore, point C = M. Thus, C bisects the chord AB. Hence, AC equals CB, which proves that C is the midpoint of AB.

Answer:

Given that O is the center of the circle and the radius OA (or OB) is 5 cm, PA and BC are tangents to the circle at points A and B. The length OP is given as 13 cm, where P is the point from which the tangents to the circle are drawn.

We know that the tangent to a circle is perpendicular to the radius at the point of contact. Therefore, OA is perpendicular to PA, and OB is perpendicular to BC. Since the tangents PA and BC are drawn from the same point P outside the circle, the lengths of these tangents from point P are equal (i.e., PA = PC).

To find the length of the tangent PA, we use the right triangle formed by O, A, and P. Here OA is the radius (5 cm), OP is the straight line distance from the center to point P (13 cm), and PA is the tangent segment from P to the circle at A.

Applying the Pythagoras theorem in triangle OAP, we get:

PA² = OP² - OA² = 13² - 5² = 169 - 25 = 144

Thus, PA = sqrt(144) = 12 cm.

Since tangents drawn from an external point are equal in length, the lengths of tangents PA and BC are both 12 cm.

Hence, the length of each tangent PA and BC is 12 cm.

Answer:

Let the height of the tower be h meters and the horizontal distance of the car from the foot of the tower when the angle of depression is 30° be x meters.

From the top of the tower, the angle of depression to the car is 30°, so by trigonometry, tan 30° = h / x. Therefore, x = h / tan 30°.

After 10 seconds, the angle of depression becomes 60°. Let the horizontal distance of the car from the foot of the tower at this time be y meters. Then, tan 60° = h / y, so y = h / tan 60°.

The car travels from distance x to y in 10 seconds, so its speed = (x - y) / 10.

To find time taken to reach the foot of the tower from the point where angle of depression is 60°, distance to cover is y meters.

Time = distance / speed = y / ((x - y) / 10) = 10 * y / (x - y).

Using the values of tan 30° = 1 / √3 and tan 60° = √3,

x = h / (1 / √3) = h√3,

y = h / √3.

Therefore, time = 10 * (h / √3) / (h√3 - h / √3) = 10 * (1 / √3) / (√3 - 1 / √3) = 10 * (1 / √3) / ((3 / √3) - (1 / √3)) = 10 * (1 / √3) / (2 / √3) = 10 * (1 / √3) * (√3 / 2) = 10 * (1 / 2) = 5 seconds.

Hence, the car will take 5 seconds to reach the foot of the tower from the point where the angle of depression is 60°.