RBSE Class 11th 2024 Mathematics-APS (1136)-2024 Previous Year Papers

The letters in “TRIGONOMETRY” are: T, R, I, G, O, N, M, E, Y. The set is {T, R, I, G, O, N, M, E, Y}.

5π/6 radians is already in radian measure. In degrees: (5π/6) × (180°/π) = 150°.

(–5i)(⅓ i) = –5 × ⅓ × i² = –(5/3) × (–1) = 5/3. So, a + ib = 5/3 + 0i.

Using binomial theorem: (a + b)⁵ = a⁵ + 5a⁴b + 10a³b² + 10a²b³ + 5ab⁴ + b⁵.

Equation: (x – h)² + (y – k)² = r² ⇒ (x + 3)² + (y – 2)² = 16.

Point (4, –2, 3): x > 0, y < 0, z > 0 ⇒ Octant IV. Point (–4, 2, 5): x < 0, y > 0, z > 0 ⇒ Octant II.

Variance = (standard deviation)² = 6² = 36.

Sample space S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}.

A ∪ B = {1, 2, 3, 4, 6, 8}. So, (A ∪ B)′ = U – (A ∪ B) = {5, 7, 9}.

P × P × P = {(1,1,1), (1,1,2), (1,2,1), (1,2,2), (2,1,1), (2,1,2), (2,2,1), (2,2,2)}.

7x + 3 < 5x + 9 ⇒ 7x – 5x < 9 – 3 ⇒ 2x < 6 ⇒ x < 3. On number line, all points to the left of 3 (open circle at 3).

ⁿC₉ = ⁿC₈ ⇒ n = 9 + 8 = 17. So, ⁿC₁₇ = ¹⁷C₁₇ = 1.

(x – 2y)⁵ = x⁵ – 10x⁴y + 40x³y² – 80x²y³ + 80xy⁴ – 32y⁵.

First term a = 5/2, common ratio r = (5/4)/(5/2) = 1/2. 20th term = a r¹⁹ = (5/2) × (1/2)¹⁹ = 5 / 2²⁰.

lim_x→0 (sin 5x) / (bx) = (5/b) × lim_x→0 (sin 5x)/(5x) = (5/b) × 1 = 5/b.

Sample space: {HH, HT, TH, TT}. Favorable outcomes (at least one tail): {HT, TH, TT} = 3. Probability = 3/4.

Relation: 3x – y = 0 ⇒ y = 3x. For x ∈ A, y must also be in A. So, x = 1, 2, 3, 4 (since 3×4 = 12 ≤ 14, but 3×5 = 15 > 14). Thus, R = {(1,3), (2,6), (3,9), (4,12)}. Domain = {1, 2, 3, 4}. Codomain = A = {1, 2, ..., 14}. Range = {3, 6, 9, 12}.

sin(π/6) = 1/2, sec(π/3) = 2, sin(5π/6) = sin(π – π/6) = sin(π/6) = 1/2, cot(π/4) = 1. LHS = 3 × (1/2) × 2 – 4 × (1/2) × 1 = 3 – 2 = 1 = RHS.

The given complex number is already in the form \(a + ib\), where \(a = 5\) and \(b = \sqrt{2}\).

Thus, \(5 + \sqrt{2}i\) is the required form.

Multiply both sides by 4 (LCM of 2 and 4):

\(2(3x - 4) \ge x + 1\)

\(6x - 8 \ge x + 1\)

\(6x - x \ge 1 + 8\)

\(5x \ge 9\)

\(x \ge \frac{9}{5}\)

The solution set is \([\frac{9}{5}, \infty)\).

Graph on number line: A closed circle at \(\frac{9}{5}\) and a ray extending to the right towards infinity.

Write \(102 = 100 + 2\). Using Binomial theorem:

\((100 + 2)^5 = \sum_{k=0}^{5} \binom{5}{k} (100)^{5-k} (2)^k\)

= \(\binom{5}{0}(100)^5 + \binom{5}{1}(100)^4(2) + \binom{5}{2}(100)^3(2)^2 + \binom{5}{3}(100)^2(2)^3 + \binom{5}{4}(100)(2)^4 + \binom{5}{5}(2)^5\)

= \(1 \times 10^{10} + 5 \times 10^8 \times 2 + 10 \times 10^6 \times 4 + 10 \times 10^4 \times 8 + 5 \times 100 \times 16 + 1 \times 32\)

= \(10,000,000,000 + 1,000,000,000 + 40,000,000 + 800,000 + 8,000 + 32\)

= \(11,040,808,032\)

The median through R passes through R and the midpoint of PQ.

Midpoint M of PQ: \(\left(\frac{2 + (-2)}{2}, \frac{1 + 3}{2}\right) = (0, 2)\)

Points R(4, 5) and M(0, 2). Slope of RM = \(\frac{5 - 2}{4 - 0} = \frac{3}{4}\)

Equation of line through R(4,5) with slope \(\frac{3}{4}\):

\(y - 5 = \frac{3}{4}(x - 4)\)

\(4(y - 5) = 3(x - 4)\)

\(4y - 20 = 3x - 12\)

\(3x - 4y + 8 = 0\)

Distance formula: \(d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}\)

Here, \(A = 3, B = -4, C = -26\), point \((3, -5)\):

\(d = \frac{|3(3) + (-4)(-5) - 26|}{\sqrt{3^2 + (-4)^2}} = \frac{|9 + 20 - 26|}{\sqrt{9 + 16}} = \frac{|3|}{\sqrt{25}} = \frac{3}{5}\)

Let \(P(x, y, z)\) be any point equidistant from A(1,2,3) and B(3,2,-1).

\(PA = PB\)

\(\sqrt{(x-1)^2 + (y-2)^2 + (z-3)^2} = \sqrt{(x-3)^2 + (y-2)^2 + (z+1)^2}\)

Squaring both sides:

\((x-1)^2 + (y-2)^2 + (z-3)^2 = (x-3)^2 + (y-2)^2 + (z+1)^2\)

\(x^2 - 2x + 1 + z^2 - 6z + 9 = x^2 - 6x + 9 + z^2 + 2z + 1\)

\(-2x - 6z + 10 = -6x + 2z + 10\)

\(-2x + 6x - 6z - 2z = 0\)

\(4x - 8z = 0\)

\(x - 2z = 0\)

\(\lim_{x \to 0} \frac{x(a + \cos x)}{b \sin x} = \frac{1}{b} \lim_{x \to 0} \frac{x}{\sin x} \cdot \lim_{x \to 0} (a + \cos x)\)

We know \(\lim_{x \to 0} \frac{x}{\sin x} = 1\) and \(\lim_{x \to 0} (a + \cos x) = a + 1\)

Thus, the limit = \(\frac{1}{b} \times 1 \times (a + 1) = \frac{a+1}{b}\)

First, simplify numerator: \((3 - 2i)(2 + 3i) = 6 + 9i - 4i - 6i^2 = 6 + 5i + 6 = 12 + 5i\)

Denominator: \((1 + 2i)(2 - i) = 2 - i + 4i - 2i^2 = 2 + 3i + 2 = 4 + 3i\)

Now, \(\frac{12 + 5i}{4 + 3i}\). Multiply numerator and denominator by conjugate of denominator \((4 - 3i)\):

= \(\frac{(12 + 5i)(4 - 3i)}{(4 + 3i)(4 - 3i)} = \frac{48 - 36i + 20i - 15i^2}{16 - 9i^2} = \frac{48 - 16i + 15}{16 + 9} = \frac{63 - 16i}{25}\)

= \(\frac{63}{25} - \frac{16}{25}i\)

(i) \(B \cap D = \{7,9,11,13\} \cap \{15,17\} = \emptyset\) (empty set)

(ii) \(B \cup C = \{7,9,11,13,15\}\); \(A \cap (B \cup C) = \{3,5,7,9,11\} \cap \{7,9,11,13,15\} = \{7,9,11\}\)

(iii) \(A \cap C = \{3,5,7,9,11\} \cap \{11,13,15\} = \{11\}\)

(iv) \(B \cup D = \{7,9,11,13,15,17\}\); \(A \cap (B \cup D) = \{3,5,7,9,11\} \cap \{7,9,11,13,15,17\} = \{7,9,11\}\)

Let \(f(x) = \tan x\). By first principle:

\(f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = \lim_{h \to 0} \frac{\tan(x+h) - \tan x}{h}\)

= \(\lim_{h \to 0} \frac{\frac{\sin(x+h)}{\cos(x+h)} - \frac{\sin x}{\cos x}}{h}\)

= \(\lim_{h \to 0} \frac{\sin(x+h)\cos x - \cos(x+h)\sin x}{h \cos(x+h)\cos x}\)

= \(\lim_{h \to 0} \frac{\sin[(x+h) - x]}{h \cos(x+h)\cos x} = \lim_{h \to 0} \frac{\sin h}{h} \cdot \frac{1}{\cos(x+h)\cos x}\)

= \(1 \times \frac{1}{\cos x \cdot \cos x} = \sec^2 x\)

Total persons = 2 + 3 = 5. Number of ways to choose 3 persons from 5 = \(\binom{5}{3} = 10\).

Committees with 1 man and 2 women: Choose 1 man from 2 men = \(\binom{2}{1} = 2\) ways; choose 2 women from 3 women = \(\binom{3}{2} = 3\) ways.

Total such committees = \(2 \times 3 = 6\).

Let the common ratio be \(r\). Then \(b = ar, c = ar^2, d = ar^3\).

LHS: \((a^2 + a^2r^2 + a^2r^4)(a^2r^2 + a^2r^4 + a^2r^6) = a^2(1 + r^2 + r^4) \cdot a^2r^2(1 + r^2 + r^4) = a^4r^2(1 + r^2 + r^4)^2\)

RHS: \((a \cdot ar + ar \cdot ar^2 + ar^2 \cdot ar^3)^2 = (a^2r + a^2r^3 + a^2r^5)^2 = [a^2r(1 + r^2 + r^4)]^2 = a^4r^2(1 + r^2 + r^4)^2\)

Thus, LHS = RHS. Hence proved.

\(f(x) = \sum_{n=1}^{100} \frac{x^n}{n} + 1\)

\(f'(x) = \sum_{n=1}^{100} \frac{n x^{n-1}}{n} = \sum_{n=1}^{100} x^{n-1} = 1 + x + x^2 + \cdots + x^{99}\)

\(f'(1) = 1 + 1 + 1 + \cdots\) (100 terms) = 100

\(f'(0) = 1\) (since all terms except the first become 0)

Thus, \(f'(1) = 100 = 100 \times 1 = 100 f'(0)\). Hence proved.

Using sum-to-product formulas:

\(\sin 5x + \sin 3x = 2 \sin\left(\frac{5x+3x}{2}\right) \cos\left(\frac{5x-3x}{2}\right) = 2 \sin 4x \cos x\)

\(\cos 5x + \cos 3x = 2 \cos\left(\frac{5x+3x}{2}\right) \cos\left(\frac{5x-3x}{2}\right) = 2 \cos 4x \cos x\)

Thus, \(\frac{\sin 5x + \sin 3x}{\cos 5x + \cos 3x} = \frac{2 \sin 4x \cos x}{2 \cos 4x \cos x} = \frac{\sin 4x}{\cos 4x} = \tan 4x\)

Hence proved.

Solution:

Given equation: 9x² + 49y² = 36

Divide both sides by 36:

x²/4 + y²/(36/49) = 1

So, a² = 4 ⇒ a = 2, b² = 36/49 ⇒ b = 6/7

Since a > b, the major axis is along the x-axis.

c² = a² – b² = 4 – 36/49 = (196 – 36)/49 = 160/49 ⇒ c = √(160)/7 = (4√10)/7

Vertices: (±a, 0) = (±2, 0)

Foci: (±c, 0) = (±4√10/7, 0)

Length of major axis: 2a = 4

Length of minor axis: 2b = 12/7

Eccentricity: e = c/a = (4√10/7)/2 = (2√10)/7

Solution:

Given: x²/16 – y²/9 = 1

Here, a² = 16 ⇒ a = 4, b² = 9 ⇒ b = 3

c² = a² + b² = 16 + 9 = 25 ⇒ c = 5

Vertices: (±a, 0) = (±4, 0)

Foci: (±c, 0) = (±5, 0)

Eccentricity: e = c/a = 5/4

Length of latus rectum: 2b²/a = 2×9/4 = 18/4 = 9/2

Solution:

First, find the mean (x̄):

Class midpoints: 10, 25, 35, 45, 55, 65, 75

Σf = 5+9+11+15+8+6+4 = 58

Σfx = (5×10)+(9×25)+(11×35)+(15×45)+(8×55)+(6×65)+(4×75) = 50+225+385+675+440+390+300 = 2465

Mean x̄ = 2465/58 ≈ 42.5

Now, calculate |xᵢ – x̄| and f|xᵢ – x̄|:

|10–42.5|=32.5, f×|d| = 5×32.5=162.5

|25–42.5|=17.5, f×|d| = 9×17.5=157.5

|35–42.5|=7.5, f×|d| = 11×7.5=82.5

|45–42.5|=2.5, f×|d| = 15×2.5=37.5

|55–42.5|=12.5, f×|d| = 8×12.5=100

|65–42.5|=22.5, f×|d| = 6×22.5=135

|75–42.5|=32.5, f×|d| = 4×32.5=130

Σf|xᵢ – x̄| = 162.5+157.5+82.5+37.5+100+135+130 = 805

Mean deviation about mean = 805/58 ≈ 13.88

Solution:

Class midpoints (xᵢ): 35, 45, 55, 65, 75, 85, 95

Σf = 3+7+12+15+8+3+2 = 50

Σfx = (3×35)+(7×45)+(12×55)+(15×65)+(8×75)+(3×85)+(2×95) = 105+315+660+975+600+255+190 = 3100

Mean (x̄) = 3100/50 = 62

Now, calculate (xᵢ – x̄)² and f(xᵢ – x̄)²:

(35–62)² = 729, f× = 3×729 = 2187

(45–62)² = 289, f× = 7×289 = 2023

(55–62)² = 49, f× = 12×49 = 588

(65–62)² = 9, f× = 15×9 = 135

(75–62)² = 169, f× = 8×169 = 1352

(85–62)² = 529, f× = 3×529 = 1587

(95–62)² = 1089, f× = 2×1089 = 2178

Σf(xᵢ – x̄)² = 2187+2023+588+135+1352+1587+2178 = 10050

Variance (σ²) = 10050/50 = 201

Standard deviation (σ) = √201 ≈ 14.18

Solution:

Let M = event that a student studies Mathematics, B = event that a student studies Biology.

Given: P(M) = 40% = 0.4, P(B) = 30% = 0.3, P(M ∩ B) = 10% = 0.1

We need P(M ∪ B) = P(M) + P(B) – P(M ∩ B)

= 0.4 + 0.3 – 0.1 = 0.6

Probability = 0.6 or 60%

Solution:

Sample space S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}, total outcomes = 8

(i) 3 heads: Only {HHH} → 1 outcome. Probability = 1/8

(ii) At least 2 heads: {HHH, HHT, HTH, THH} → 4 outcomes. Probability = 4/8 = 1/2

(iii) Exactly two tails: {HTT, THT, TTH} → 3 outcomes. Probability = 3/8

(iv) At most two tails: All outcomes except {TTT} → 7 outcomes. Probability = 7/8