CBSE Class 11 Mathematics Question Paper Set 26 with Solutions

Answer: (d) 63. A × B has 3 × 2 = 6 elements, so there are 2⁶ = 64 relations, and removing the empty relation leaves 63.

Teacher's Note:
1. Every subset of A × B is a relation from A to B.
2. Do not forget to subtract the empty relation when the question says non-empty.

Answer: (b) 3. Equal ordered pairs give 2x = y + 3 and y − x = 0, so y = x. Then 2y = y + 3, so y = 3.

Teacher's Note:
1. Equate the first components together and the second components together.
2. Substitute one equation into the other to get a single unknown.

Answer: (b) √3/2. The expression is sin(A − B) = sin(7π/12 − π/4) = sin(π/3) = √3/2.

Teacher's Note:
1. Recognise the pattern sin A cos B − cos A sin B before expanding anything.
2. Convert 7π/12 − 3π/12 = 4π/12 = π/3 carefully.

Answer: (b) √3/2. 31π/3 = 10π + π/3, and 10π is five full rotations, so sin(31π/3) = sin(π/3) = √3/2.

Teacher's Note:
1. Remove multiples of 2π first because sine repeats every 2π.
2. Write the angle as a multiple of 2π plus a standard angle.

Answer: (b) Numerical inequalities. Both sides contain only numbers and no variable.

Teacher's Note:
1. Literal inequalities contain variables such as x < 5.
2. Numerical inequalities compare only numbers, so they are always simply true or false.

Answer: (a) x ∈ (10, ∞). −3x < −30. Dividing by −3 reverses the sign, so x > 10.

Teacher's Note:
1. Reverse the inequality sign when you divide or multiply by a negative number.
2. The inequality is strict, so use a round bracket at 10.

Answer: (c) 7200. Selection: ⁴C₂ × ⁵C₃ = 6 × 10 = 60. Each selection of 5 letters can be arranged in 5! = 120 ways, so the total is 60 × 120 = 7200.

Teacher's Note:
1. First select, then arrange: the word needs both steps.
2. Students often stop at 60 and forget the 5! arrangements.

Answer: (d) 3. If ⁿC_a = ⁿC_b then a = b or a + b = n. Here 3r = r + 3 gives r = 3/2, which is not an integer. So 3r + r + 3 = 15, which gives 4r = 12 and r = 3.

Teacher's Note:
1. Use both conditions a = b and a + b = n, and reject answers that are not whole numbers.
2. Check the answer: ¹⁵C₉ = ¹⁵C₆ is true.

Answer: (a) −1760x⁻³. T₁₀ = ¹²C₉ (2x²)³ (−1/x)⁹ = 220 × 8x⁶ × (−x⁻⁹) = −1760x⁻³.

Teacher's Note:
1. For the (r + 1)th term use ⁿC_r, so the 10th term has r = 9.
2. The power 9 is odd, so the sign of (−1/x)⁹ is negative.

Answer: (a) 4. In the subtraction, terms with even powers of y cancel and terms with odd powers of y double. The odd powers y, y³, y⁵, y⁷ remain, so there are 4 terms.

Teacher's Note:
1. Write both expansions and see which terms survive the subtraction.
2. For (a + b)ⁿ − (a − b)ⁿ with n = 8 the count is n/2.

Answer: (b) m(h − x₁). Slope m = (k − y₁)/(h − x₁), so k − y₁ = m(h − x₁).

Teacher's Note:
1. This is just the slope formula rearranged, and it is the basis of the point-slope form.
2. Keep the same order of subtraction in numerator and denominator.

Answer: (c) x + y = 1. Slope m = tan 135° = −1. Line: y − 2 = −1(x + 1), so y − 2 = −x − 1, giving x + y = 1.

Teacher's Note:
1. Slope is the tangent of the angle made with the positive x-axis.
2. Check your result by substituting (−1, 2): −1 + 2 = 1.

Answer: (d) (0, 0); \( \dfrac{5}{\sqrt{2}} \). Dividing by 2 gives x² + y² = 25/2, a circle with centre (0, 0) and radius √(25/2) = 5/√2.

Teacher's Note:
1. Make the coefficients of x² and y² equal to 1 before reading off the radius.
2. The right side is r², so take a square root at the end.

Answer: (a) √3. The ellipse is x²/1 + y²/(1/4) = 1, so a² = 1 and b² = 1/4. Then c² = a² − b² = 3/4, so c = √3/2. The distance between the foci is 2c = √3.

Teacher's Note:
1. Write the equation in the standard form with 1 on the right and the coefficients in the denominators.
2. The question asks for the distance between the foci, which is 2c and not c.

Answer: (b) (9, 0, 0). Let the point be (x, 0, 0). Then (x − 2)² + 4 + 4 = (x − 5)² + 25 + 16. So x² − 4x + 12 = x² − 10x + 66, which gives 6x = 54 and x = 9.

Teacher's Note:
1. A point on the x-axis has y = 0 and z = 0.
2. Equate the squares of the distances so you avoid square roots.

Answer: (b) YZ-plane. The plane containing the y-axis and z-axis is the YZ-plane, where x = 0.

Teacher's Note:
1. Each coordinate plane is named after the two axes that lie in it.
2. Every point on the YZ-plane has x-coordinate 0.

Answer: (c) 2. 4x² − 1 = (2x − 1)(2x + 1), so the expression equals 2x + 1 for x ≠ 1/2. As x → 1/2, the limit is 2(1/2) + 1 = 2.

Teacher's Note:
1. Direct substitution gives 0/0, so factorise first.
2. Cancel the common factor only after noting x is not equal to 1/2.

Answer: (d) 0. y = x^1/2 + x^−1/2, so dy/dx = (1/2)x^−1/2 − (1/2)x^−3/2. At x = 1 this is 1/2 − 1/2 = 0.

Teacher's Note:
1. Rewrite roots as fractional powers before differentiating.
2. Apply d(xⁿ)/dx = nxⁿ⁻¹ to each term separately.

Answer: (c) not exhaustive. A = {1, 2, 3}, B = {3}, C = {5, 6}. A ∪ B ∪ C = {1, 2, 3, 5, 6} does not contain 4, so the events are not exhaustive. They are also not mutually exclusive because A and B share the outcome 3.

Teacher's Note:
1. Events are exhaustive only if their union is the whole sample space.
2. Write each event as a set of outcomes and then check overlaps and the union.

Answer: (c) 7/36. Total outcomes = 36. A sum of 5 occurs in 4 ways: (1,4), (2,3), (3,2), (4,1). A sum of 10 occurs in 3 ways: (4,6), (5,5), (6,4). Favourable = 7, so P = 7/36.

Teacher's Note:
1. Possible sums are 2 to 12, so only 5 and 10 are divisible by 5.
2. List ordered pairs, because (1,4) and (4,1) are different outcomes.

Answer: 1. Radius r = 40/2 = 20 cm.
2. The chord is 20 cm, equal to the radius, so the triangle formed by the chord and two radii is equilateral.
3. The angle at the centre is θ = 60° = π/3 radian.
4. Arc length l = rθ = 20 × π/3 = 20π/3 cm (about 20.9 cm).

Teacher's Note:
1. Spot that the chord equals the radius to get 60° without any calculation.
2. Always convert the angle into radians before using l = rθ.

Answer: 1. Total arrangements of the 6 digits = 6! = 720.
2. Arrangements with 0 in the first place are not 6-digit numbers: 5! = 120.
3. Required numbers = 720 − 120 = 600.
OR
1. INDEPENDENCE has 12 letters: I once, N 3 times, D 2 times, E 4 times, P once, C once.
2. Arrangements = 12!/(3! × 2! × 4!) = 479001600/288 = 1663200.

Teacher's Note:
1. A number cannot begin with 0, so subtract those cases or fill the first place first (5 × 5! = 600).
2. Count repeated letters carefully and divide by the factorial of each repetition.

Answer: 1. Vertices on the y-axis, so the major axis is along the y-axis and the form is x²/b² + y²/a² = 1 with a = 6.
2. e = c/a = 2/3, so c = 4.
3. b² = a² − c² = 36 − 16 = 20.
4. Equation: x²/20 + y²/36 = 1.

Teacher's Note:
1. Decide the direction of the major axis from the vertices before writing the equation.
2. Use b² = a² − c² for an ellipse; the sign is different for a hyperbola.

Answer: 1. Slope m = (−4 − 3)/(−1 − 0) = (−7)/(−1) = 7.
2. The line passes through (0, 3), so the y-intercept is 3.
3. Equation: y − 3 = 7(x − 0), that is y = 7x + 3 (or 7x − y + 3 = 0).

Teacher's Note:
1. Use the same order of subtraction for the numerator and the denominator of the slope.
2. Verify by substituting the second point: 7(−1) + 3 = −4.

Answer: 1. Group the terms: LHS = [(sin 9x + sin 3x) + (sin 7x + sin 5x)] / [(cos 9x + cos 3x) + (cos 7x + cos 5x)].
2. Using sin A + sin B = 2 sin((A+B)/2) cos((A−B)/2): sin 9x + sin 3x = 2 sin 6x cos 3x and sin 7x + sin 5x = 2 sin 6x cos x.
3. Using cos A + cos B = 2 cos((A+B)/2) cos((A−B)/2): cos 9x + cos 3x = 2 cos 6x cos 3x and cos 7x + cos 5x = 2 cos 6x cos x.
4. LHS = 2 sin 6x (cos 3x + cos x) / [2 cos 6x (cos 3x + cos x)].
5. Cancelling the common factor, LHS = sin 6x / cos 6x = tan 6x = RHS. Hence proved.

Teacher's Note:
1. Pair the terms so that every pair has the same average angle (here 6x).
2. Factor out the common bracket (cos 3x + cos x) and cancel it.

Answer: 1. 1 − cos 4x = 2 sin²2x and 1 − cos 5x = 2 sin²(5x/2).
2. Limit = \( \lim \dfrac{\sin^2 2x}{\sin^2 (5x/2)} = \lim \left[\dfrac{\sin 2x}{2x}\right]^2 \cdot \left[\dfrac{5x/2}{\sin (5x/2)}\right]^2 \cdot \dfrac{(2x)^2}{(5x/2)^2} \).
3. The first two factors tend to 1, so the limit = 4/(25/4) = 16/25.
OR
1. Put t = π/2 − x. As x → π/2, t → 0, and cos x = cos(π/2 − t) = sin t.
2. Limit = lim_t→0 (sin t)/t = 1.

Teacher's Note:
1. Convert 1 − cos kx into 2 sin²(kx/2) so that you can use lim sin θ/θ = 1.
2. For limits at a point other than 0, substitute to move the point to 0.

Answer: 1. Choose 3 bowlers from 5: ⁵C₃ = 10.
2. Choose 1 wicketkeeper from 2: ²C₁ = 2.
3. The remaining 11 − 3 − 1 = 7 players must come from the other 16 − 5 − 2 = 9 players (neither bowlers nor wicketkeepers): ⁹C₇ = 36.
4. Number of ways = 10 × 2 × 36 = 720.

Teacher's Note:
1. Exactly 3 bowlers means the other 7 players must be chosen from the non-bowlers and non-wicketkeepers only.
2. Multiply the independent selections together.

Answer: 1. Multiply both sides by 12: 3(2x/3 + 1) ≥ 4(x − 2).
2. 2x + 3 ≥ 4x − 8.
3. 3 + 8 ≥ 4x − 2x, so 11 ≥ 2x, giving x ≤ 11/2.
4. Solution set: x ∈ (−∞, 11/2].
5. Number line: a filled (solid) dot at 5.5 with the line shaded to the left, towards −∞.

Teacher's Note:
1. Multiply by the LCM of the denominators to clear all the fractions at once.
2. Use a solid dot for ≤ or ≥ and an open dot for strict inequalities.

Answer: 1. f(x) = tan x. By first principle, f′(x) = lim_h→0 [tan(x + h) − tan x]/h.
2. tan(x + h) − tan x = sin h / (cos(x + h) cos x), because sin(A − B) = sin A cos B − cos A sin B gives sin((x + h) − x) = sin(x + h) cos x − cos(x + h) sin x.
3. f′(x) = lim [sin h / h] × [1 / (cos(x + h) cos x)].
4. As h → 0, sin h / h → 1 and cos(x + h) → cos x, so f′(x) = 1/cos²x = sec²x.
OR
1. Let y = u/v with u = sin x + cos x and v = sin x − cos x. Then u′ = cos x − sin x and v′ = cos x + sin x.
2. Quotient rule: dy/dx = (v u′ − u v′)/v².
3. Numerator = (sin x − cos x)(cos x − sin x) − (sin x + cos x)(cos x + sin x) = −(sin x − cos x)² − (sin x + cos x)² = −(2 sin²x + 2 cos²x) = −2.
4. dy/dx = −2/(sin x − cos x)².

Teacher's Note:
1. In the first principle write tan as sin/cos and combine the two fractions into a single sin h term.
2. In the quotient rule keep the order v·u′ − u·v′, because swapping it changes the sign.

Answer: 1. Compare with x²/a² − y²/b² = 1: a² = 36, b² = 4, so a = 6 and b = 2.
2. Transverse axis = 2a = 12; conjugate axis = 2b = 4.
3. Vertices: (±6, 0).
4. c² = a² + b² = 40, so c = 2√10. Foci: (±2√10, 0).
5. Eccentricity e = c/a = √10/3.
6. Latus rectum = 2b²/a = 8/6 = 4/3.
OR
1. Let the line meet the axes at A(a, 0) and B(0, b), with P dividing AB in the ratio 1 : 2 (AP : PB = 1 : 2).
2. By the section formula, x = (1·0 + 2·a)/3 = 2a/3 = −5, so a = −15/2. Also y = (1·b + 2·0)/3 = b/3 = 4, so b = 12.
3. Intercept form: x/a + y/b = 1, so −2x/15 + y/12 = 1.
4. Multiplying by 60: −8x + 5y = 60. The line is 8x − 5y + 60 = 0.
5. Check: P(−5, 4) gives −40 − 20 + 60 = 0.

Teacher's Note:
1. For a hyperbola use c² = a² + b², and remember the transverse axis lies along the term with the positive sign.
2. Using the intercept form for the second option makes the section formula the only real work.

Answer: (a) Different arrangements of the friends in a line are called permutations.
(b) PERMUTATIONS has 12 letters, with T repeated twice. The vowels are E, U, A, I, O (5 vowels).
(i) Fix P at the start and S at the end. The remaining 10 letters include T twice, so the number of ways = 10!/2! = 1814400.
(ii) Treat the 5 vowels as one block. We then have 7 other letters plus the block, which is 8 objects with T repeated twice: 8!/2! = 20160 arrangements. The vowels can be arranged among themselves in 5! = 120 ways. Total = 20160 × 120 = 2419200.

Teacher's Note:
1. For vowels together, arrange the block with the other letters first and then arrange inside the block.
2. Do not forget to divide by 2! because the letter T appears twice.

Answer: (b) From (0, −1): f(0) = b = −1.
(a) From (1, 1): a + b = 1, so a = 1 + 1 = 2. Check: f(2) = 4 − 1 = 3 and f(−1) = −2 − 1 = −3, which match the given pairs.
(c) f(x) = 2x − 1 = 0, so x = 1/2.

Teacher's Note:
1. Substitute a point with x = 0 first because it gives b immediately.
2. Always verify your a and b on one more pair of the function.

Answer: Sample space S has 8 outcomes: HHH, HHT, HTH, THH, HTT, THT, TTH, TTT.
A = {HHT, HTH, THH}; B = {HHH, HHT, HTH, THH}; C = all outcomes except HHH (7 outcomes); D = {HHH}.
(a) A and D have no common outcome, so they are mutually exclusive. Their union has only 4 of the 8 outcomes, so it is not exhaustive. Answer: A and D.
(b) C and D have no common outcome and C ∪ D = S, so they are mutually exclusive and exhaustive. Answer: C and D.

Teacher's Note:
1. List the sample space and write every event as a set of outcomes before comparing them.
2. Mutually exclusive means no common outcome; exhaustive means the union is the whole sample space.