Haryana Board Practice Paper -XI 2024-25 Maths PDF

BOARD OF SCHOOL EDUCATION HARYANA Practice Paper -XI (2024-25) Marking Scheme MATHEMATICS...

BOARD OF SCHOOL EDUCATION HARYANA Practice Paper -XI (2024-25) Marking Scheme MATHEMATICS CODE: 835  Important Instructions: All answers provided in the Marking scheme are SUGGESTIVE Examiners are requested to accept all possible alternative correct answer(s). SECTION – A (1Mark × 20Q) Q. No. EXPECTED ANSWERS Marks Question 1. If X= {a, b, c, d, e} and Y= {d, e, f, g} then (X-Y) ∩ (X+Y) is Solution: (B) {a, b, c} 1 Question 2 If A = {a, d}, B = {b, c, e}, C = {b, c, f}, then A × (B –C) is Solution: (A) {(a, e), (d, e)} 1 Question 3 75° in radian measure is Solution: (B) 5π/12 1 Question 4. a + ib form of i-35 is : Solution: (A) i 1 Question 5. 1 1 𝑋 If + = then value of x is: 8! 9! 10! Solution: (A) 100 1 Question 6. The G.M. between 1 and 64 is : Solution: (C) 8 1 Question 7. The value of x for which the numbers -3/11, x, -11/3 are in G.P Solution: (B) ±1 1 Question 8. The derivative of sin (x + a) is: Solution: (A) cos (x + a) 1 Question 9. If the variance of a data is 25, then its standard deviation is: Solution: (B) 5 1 Question10. If P(A∪B) = P(A∩B) for any two events A and B, then Solution: (C) P(A) = P(B) 1 Question11. Find the number of terms in the expansion of (3x + 9)9. Solution: 9 + 1 = 10 1 Question12. Find the centre and radius of the circle x² + y² + 8x + 10y - 8 = 0. Solution: Centre ( -4 , -5 ) and Radius is 7 1 Question13. xn − an Write the value of lim. x →a x − a Solution: xn − an 1 lim = n.an-1 x →a x − a Question14. Find the mean deviation about the mean for the following data: 6, 7, 10, 12, 13, 4, 8, 12. Solution: 6+7+10+12+13+4+8+12 1 Mean of the given data is x̅ = =9 8 Deviations from mean ( xi - x̅ ) are -3, -2, 1, 3, 4, -5, -1, 3 Absolute deviations i.e. | xi - x̅ | are 3, 2, 1, 3, 4, 5, 1, 3 ∑8i=1 | xi − x̅ | 3+2+1+3+4+5+1+3 22 Mean Deviation = = = = 2.75 n 8 8 Question15. Let U = {1, 2, 3, 4, 5, 6}, A = {2, 3} and B = {3, 4, 5}, then (A ∪ B)’ = …………. Solution: (A ∪ B)’ = {2, 3, 4, 5}’ = {1} 1 Question16. cos (A - B) is equal to ……………... Solution: cos (A - B) = cosA. cosB + sinA. sinB 1 Question17. If C(n, a) = C(n, b), then either a = b or n = a + b. (True/ False) Solution: True 1 Question18. A die is rolled. Let A be the event of getting a multiple of 2 and B be the event of getting a multiple of 3. Then A and B are mutually exclusive events. (True/ False) Solution: False 1 Question19. Assertion (A): If (x+1, y-2) = (3, 1), then x = 3 and y = 2. Reason (R) : Two ordered pairs are equal if their corresponding elements are equal. Solution: (D) Assertion (A) is false and Reason (R) is true. 1 Question20. Assertion (A): The point (-5, 2, 0) lies on the XY plane. Reason(R):The coordinates of a point P(x, y, z) in XY plane are (0, 0, z). Solution: (C) Assertion (A) is true and Reason (R) is false. 1 SECTION – B (2Marks × 5Q) Question21. If A = {3,5,7,9, 11 }, B = {7, 9, 11, 13}, C = {11, 13, 15} and D = {15.17}; find (AUD) ∩ ( BUC ) Solution: A U D = {3, 5, 7, 9, 11, 13 } B U C = {7, 9, 11, 13, 15 } 1 1 ⸫ (AUD) ∩ ( BUC ) = {7, 9, 11, 13 } Question22. Find the multiplicative inverse of 4 − 3i. Solution: 1 Multiplicative Inverse of 4 – 3i = 4 − 3i 1 4 + 3i  M.I. = × 1 4 − 3i 4 + 3i 4 + 3i  = (4)2 − (3i)2 4 + 3i  = 16 − 9i2 4 + 3i 4 3i  = = 25 + 1 16+ 9 25 OR (3−2i)(2+3i) Find the conjugate of Question22. (1+2i)(2−i) Solution: (3−2i)(2+3i) 6 + 9i − 4i − 6i2 Given = (1+2i)(2−i) 2 − i + 4i − 2i2 6 + 5i + 6  =2 + 3i + 2 12 + 5i = 4 + 3i 12 + 5i 4 − 3𝑖 = × 4 + 3i 4 − 3𝑖  48 − 36i + 20i − 15i2 48 − 16i + 15 63 16i 𝟏 = = = − 1𝟐 16 − 12i + 12i − 9i2 16 + 9 25 25 𝟏 (3−2i)(2+3i) 63 16i ⸫ Conjugate of (1+2i)(2−i) = 25 + 𝟐 25 Question23. 5−2x x Solve the inequality ≤ 6 – 5 and show the graph of the solution 3 on number line. Solution: 5−2x x We have ≤6–5 3 5−2x x−30  ≤ 3 6 Multiply on both side by 6, we have  2(5 – 2x ) ≤ x – 30  10 – 4x ≤ x – 30  -5x ≤ -40  5x ≥ 40  x≥8 𝟏 1𝟐 Graph of the solution on number line 𝟏 𝟐 Question24. Find the 12th term of a G.P. whose 8th term is 192 and the common ratio is 2. Solution: We have, a8 = 192 r=2 𝟏  ar7 = 192 𝟐  a(2)7 =192 192 3 𝟏  a = 128 = 2 𝟐 3 ⸫ a12 = a.r11 = 2. (2)11 1 a12 = 3. (2)10 = 3. (1024) = 3072 Question25. Find the coordinates of the focus, axis, the equation of directrix and the length of the latus rectum of the parabola y2 = 12x. Solution: Equation of parabola is y2 = 12x Comparing with y2 = 4ax , we have 4a = 12 => a = 3  The coefficient of x is +ve so it is a right handed parabola. 𝟏 This parabola is symmetrical about x-axis as it involves y2 𝟐 𝟏 Thus the focus is (3, 0) 𝟐 Equation of directrix x = -3 𝟏 𝟐 Length of latus rectum is 4a = 4 × 3 = 12 𝟏 𝟐 OR Find the equation of the ellipse, whose length of the major axis is 20 Question25. and foci are (0, ±5). Solution: Since the foci are on y-axis, the major axis is along the y-axis. x2 y2 So equation of ellipse is of the form b2 + =1 a2 Given that a = semi major axis =20/2 = 10 1 And the relation c 2 = a2 − b2 , where c = 5 from foci (0, ±5) gives 52 = 102 − b2 i.e. b2 = 75 x2 y2 Therefore, the equation of the ellipse is + =1 75 100 1 SECTION – C (3Marks × 6Q) Question26. Draw appropriate Venn Diagram for (A ∪ B)’ and A’ ∪ B’. Solution: Venn Diagram of (A ∪ B)’ 𝟏 1𝟐 Venn Diagram of A’ ∪ B’ 𝟏 1𝟐 Question27. Find the domain and Range of the function √9 − x 2. Solution: Here y = √9 − x 2 y will have real values if 9 − x 2 ≥ 0 ⇒ x2 − 9 ≤ 0 ⇒ (x-3) (x + 3) ≤ 0 𝟏 ⇒ -3 ≤ x ≤ 3 ⇒ x ∈ [ -3, 3] 𝟏 𝟐 Domain = [-3, 3] Also, y2 = 9 – x2 ⇒ x2 = 9 – y2 ⇒ x = ± √9 − y 2 Clearly x is defined when 9 - y² ≥ 0 i.e., when y² - 9 ≤ 0 ⇒ (y-3)(y + 3) ≤ 0 ⇒ -3 ≤ y ≤ 3 ⇒ у є [-3, 3] But y = √9 − x 2 ≥ 0 for all x ∈ [-3, 3] i.e., y attains only non- negative values. 𝟏 𝟏 ⸫ y ∈ [0, 3] for all x ∈ [-3, 3] 𝟐 ⸫ Range = [0, 3]. Question28. 2 x 5 Expand: (x − ) ;x≠0 2 Solution: 2 x 5 2 5 −x 0 2 4 −x 1 2 3 −x 2 (x − ) = 5C0 (x) ( 2 ) + 5C1 (x) ( 2 ) + 5C2 (x) ( 2 ) + 5C3 2 𝟏 2 2 −x 3 2 1 −x 4 2 0 −x 5 𝟏 𝟐 (x ) ( 2 ) + 5C4 (x) ( 2 ) + 5C5(x) ( 2 ) 32 16 −x 8 x2 4 −x3 2 x4 x5 = x5 + 5.( x4 ) ( 2 ) + 10 (x3 ) ( 4 ) + 10 (x2 ) ( ) + 5 (x) (16) − 32 8 𝟏 32 40 20 5x3 x5 𝟏 𝟐 = x5 − x3 + −5x + − 32 x 8 OR Compute (98)5. Question28. Solution: (98)5 = (100 − 2)5 = 5C0 (100)5 (2)0 + 5C1 (100)4 (2)1 + 5C2 (100)3 (2)2 + 5C3(100)2 (2)3 + 5C4 (100)1 (2)4 + 5C5(100)0 (2)5 2 = 10000000000 + 1000000000 + 40000000 + 800000 +8000 + 32 = 11040808032 1 Question29. Find the sum of the sequence 7, 77, 777, 7777, …….. to n terms. Solution: This is not a GP., however, we can relate it to a GP. by writing the terms as Sn = 7+77+777 + 7777 +... to n terms 7 = 9 [ 9 + 99 + 999 + 9999 +... to n term] 7 1 = 9 [ (101 - 1) + (102 - 1) + (103 - 1) + (104 - 1) +...n terms] 7 = 9 [(10+102+103+...n terms) - (1+1+1+...n terms)] It is a G.P. where a = 10 and r = 10 > 1 1 a(rn − 1) ⸫ Sn = r−1 7 10(10n − 1) 7 10(10n − 1) =9[ − n] = 9 [ − n] 1 10 − 1 9 OR 39 The sum of first three terms of a G.P. is 10 and their product is 1. Find Question 29 the common ratio and the terms. Solution: a Let three terms in G.P. are r , a , ar a ⸫ × a ×ar = 1 => a3 = 1 => a = 1 1 r 1 ⸫ three terms now are , 1,r r 1 39 𝟏 A.T.Q. + 1 + r = 10 𝟐 r 1+r+r2 39 => = 10 r => 10r + 10r +10r2 = 39r => 10r2 − 29r +10 = 0 => 10r2 − 25r − 4r +10 = 0 => (10r – 2)( r – 5 ) = 0 1 => r = 5 or 5 1 1 ⸫ if common ratio r = 5 , term are 5, 1, 5 𝟏 1𝟐 1 if common ratio r = 5 , terms are 5 , 1, 5 Question30. Find the equation of the set of the points which are equidistant from the points (1, 2, 3) and (3, 2, -1) Solution: Let P(x, y, z) be any point which is equidistant from the points A(1, 2, 3) and B(3, 2, -1). ⸫ PA = PB  PA2 = PB2 1  (x - 1)2 + ( y - 2)2 + ( z - 3)2 = (x -3)2 + ( y - 2)2 + ( z + 1)2  x2 + 1 – 2x + y2 + 4 - 4y + z2 + 9 -6z = x2 + 9 – 6x + y2 + 4 - 4y + z2 + 1 + 2z  – 2x – 6z = – 6x + 2z  4x – 8z = 0 2  x – 2z =0 Question31. In an entrance test that is graded on the basis of two examinations, the probability of a randomly chosen student passing the first examination is 0.8 and the probability of passing the second examination is 0.7. The probability of passing atleast one of them is 0.95. What is the probability of passing both? Solution: 𝟏 Let P(A) be the probability of passing the first exam => P(A) = 0.8 𝟐 𝟏 Let P(B) be the probability of passing the first exam => P(B) = 0.7 𝟐 𝟏 ⸫ Probabilty of passing atleast one of them = P(A ∪ B) = 0.95 𝟐 ⸫ Probabilty of passing both = P(A ∩ B) 𝟏 We know, P(A ∪ B) = P(A) + P(B) – P(A ∩ B) 𝟐 0.95 = 0.8 + 0.7 – P(A ∩ B) P(A ∩ B) = 1.5 – 0.95 = 0.55 1 ⸫ Probabilty of passing both = P(A ∩ B) = 0.55  SECTION – D (5Marks × 4Q) Question32. (cos 7x + cos 5x) (i) Prove that: = cot x (sin 7x − sin 5x) (ii) Prove that: sin x + sin 3x + sin 5x + sin 7x = 4cos x.cos 2x.sin 4x Solution: (i) (cos 7x + cos 5x) = cot x (sin 7x − sin 5x) (cos 7x + cos 5x) L.H.S. = (sin 7x − sin 5x) C+D C−D Using cosC + cosD = 2cos ( ).cos ( ) 2 2 C+D C−D 1 and sinC – sinD = 2cos( ).sin( ) , we have 2 2 7x+5x 7x−5x 2cos ( ) cos( ) 2 2  = 7x+5x 7x−5x 2cos ( 2 ) sin( 2 ) 2.cos 6x.cos x cos x  = = = cot x 1 2.cos 6x.sin x sin x  L.H.S. = R. H.S. Solution:(ii) L.H.S. = sin x + sin 3x + sin 5x + sin 7x => = (sin 7x + sin x ) + (sin 5x + sin 3x ) [rearranging] C+D C−D 𝟏 Using sinC + sinD = 2sin( ).cos( ) 2 2 𝟐 We have, 7x+x 7x−x 5x+3x 5x−3x = [ 2sin( ).cos( ) ] + [ 2sin( ).cos( )] 2 2 2 2 = 2sin 4x.cos 3x + 2sin 4x.cos x = 2sin4x ( cos3x + cosx ) C+D C−D Using cosC + cosD = 2cos ( ).cos ( ) 𝟏 2 2 𝟏 𝟐 3x+x 3x−x We have, = 2sin4x [ 2cos( ).cos( )] 2 2 = 2sin4x [ 2cos2x.cosx ] = 4sin4x.cos2x.cosx 1 = R.H.S. Question33. Find the equation of the right bisector of the line segment joining the points (3, 4) and (-1, 2). Solution: Let the given points be A (3, 4) and B (-1, 2). Let M be the midpoint of AB. 1 3−1 4+2 ⸫ Coordinates of M = ( , ) = (1, 3) 2 2 2−4 1 And, slope of AB = − 1 − 3 = 2 1 Let m be the slope of the right bisector of the line joining the points (3, 4) and (-1, 2). 1 ⸫ m × Slope of AB = - 1 1 m×2=-1 𝟏  m=-2 𝟐 So, the equation of the line that passes through M (1, 3) and has slope -2 is y - 3 = - 2(x - 1)  2x + y - 5 = 0 Hence, the equation of the right bisector of the line segment joining the 𝟏 points (3, 4) and (-1, 2) is 2x + y - 5 = 0 1𝟐 OR Find the equation of the line passing through (-3, 5) and perpendicular Question33. to the line through the points (2, 5) and (-3, 6). Solution: Slope of the line passing through the points A(2. 5) and B(-3, 6) y2 − y1 m₁ = x2 − x1 6−5 1 m₁ = −3−2 = −5 𝟏 1 m1 = − 5 𝟐  P(-3,5) A(2, 5) B(-3,6) 1 If PL is perpendicular to AB from the point P(-3, 5), then consider its slope as m2. Lines PL and AB are mutually perpendicular if, Slope of PL x Slope of AB = -1 => m1 × m2 = -1 1 1 In other words, m2 × (− 5) = -1 ⸫ m2 = 5 1 The slope of PL is 5 and it passes through P(-3, 5), then the equation of PL is, ( y – y1 ) = m2 ( x – x1 ) or ( y – 5 ) = 5 ( x + 3 ) y – 5 = 5x + 15 𝟏 5x – y + 20 = 0 1𝟐 Question34. Find the derivative of tan x from first principle. Solution: Let f(x) = tanx , the d(f(x)) f( x+h )– f( x ) = lim dx h →0 h tan( x+h )– tan( x ) = lim h 1 h →0 1 sin( x+h ) sin( x ) = lim h [cos( x+h ) − ] h →0 cos( x) 1 sin( x+h )cos( x)−cos( x+h )sin( x ) = lim h [ ] h →0 cos( x+h )cos( x) 1 sin( x+h−x ) 2 = lim h [cos( x+h )cos( x) ] [ Using formula for sin(A - B) ] h →0  1 sin( h ) = lim h [cos( x+h )cos( x) ] h →0 sin( h ) 1 = lim × lim [cos( x+h )cos( x) ] h →0 h h →0 1 = 1. [cos2 x] 2 = sec2x OR a + bx, x1 possible values of a and b? Solution: Here, limit exist at x → 1 i.e., LHL = RHL = f(1) = 4...(1) LHL at x → 1 = lim f(x) x→1− = lim f(1 − h) h→0 = lim [ a + b(1 − h)] h→0 = a + b (1-0) =a+b...(2) 𝟏 1𝟐 RHL at x→1 = lim f(x) x→1+ = lim f(1 + h) h→0 = lim [ b − a(1 + h)] h→0 = b - a (1+ 0) 𝟏 1𝟐 =b-a …(3)  From (1) and (2) a+b=4 From (1) and (3) b-a=4 1 Adding both a + b + b - a = 4 + 4 2b = 8 b=4 Also, a+b=4 a+4=4 1 a=0 Question35. Calculate mean, variance and standard deviation for the following distribution. Classes 30- 40 40-50 50-60 60-70 70-80 80-90 90-100 Frequency 3 7 12 15 8 3 2 Solution: From the given data, we construct the following table. Class Frequency Midpoint fixi (xi - 𝐱̅ )2 fi(xi - 𝐱̅ )2 fi xi 30 – 40 3 35 105 729 2187 40 – 50 7 45 315 289 2023 50 – 60 12 55 660 49 588 60 – 70 15 65 975 9 135 70 – 80 8 75 600 169 1352 80 – 90 3 85 255 529 1587 90 – 100 2 95 190 1089 2178 𝟏 3𝟐 50 3100 10050  1 Thus Mean x̅ = ∑𝑖=7 𝑖=1 𝑓𝑖 𝑥𝑖 N 3100 = = 62 𝟏 50 𝟐 1 Variance ( 𝜎 2 ) = ∑𝑖=7 𝑖=1 𝑓𝑖 (𝑥𝑖 − 𝑥̅ ) 2 N 𝟏 10050 𝟐 = = 201 50 𝟏 and Standard deviation(𝜎) = √201 = 14.18 𝟐 SECTION – E (4Marks × 3Q) Question36. To demonstrate the compound angle formulae in trigonometry, Mahesh and Siraj selected two angles ‘A’ and ‘B’ such that A, B ∈ π 3 9 (0, 2 ) and sin A = 5 , cos B = 41. Based on the above information, answer the following questions. (i) Find the value of sin B + cos A. (2) (ii) Find the value of cos (A + B). (2) Solution: 3 9 Given, sin A = and cos B = 5 41 we know, cos A = √1 – sin2 A 3 2 16 4 So, cos A= √1 – (5) = √25 = 5 Also sin B = √1 – cos 2 B 9 2 1681−81 1600 40 So, sin B= √1 – (41) = √ = √1681 = 41 1681 40 So, sin B = 41 1 40 4 Thus sin B + cos A = 41 + 5 200+164  sin B + cos A = 205  364  sin B + cos A = 205 1 𝟏 ii) cos (A + B) = cos A × cos B - sin A × sin B 𝟐 3 9 4 40 = (5)(41) - (5)(41) 1 27 160 = - 205 205 27− 133 =- 205 𝟏 133 =- 𝟐 205 Question37. The assembly incharge of a school wants to generate signals for calling classes for the assembly. He has got 5 coloured flags viz., Yellow, Red, Orange, Green and Blue to make signals. Based on the above information answer the following questions: (i) How many different signals can be generated by using all 5 flags? (1) (ii) To call the middle section for the assembly, he has to generate different signals by using 2 flags only. How many such 𝟏 arrangements are possible? (1𝟐 ) (iii)To call the senior section for the assembly, he has to generate different signals by using 4 flags only. How many such 𝟏 arrangements are possible? (1𝟐 ) Solution: (i) Total number of different flags given = 5 Number of ways to generate a signal of 5 flags together = 5P5 5! = (5−5)! = 5! = 120ways 1  To call the middle section for the assembly, a signal of only two flags is to be generated. Number of ways to generate a signal of 2 flags together = 5P2 ( ii) 5! = (5−2)! 5! = 3! 𝟏 5.4.3! 1𝟐 = = 5.4= 20ways 3! To call the senior section for the assembly, a signal of only four flags is to be generated. Number of ways to generate a signal of 4 flags together = 5P4 ( iii) 5! = (5−4)! 5! = 1! 𝟏 5.4.3! 1𝟐 = = 5! = 120ways 1! Question Due to heavy storm, an electric wire got broken and fell on the ground 38. and is bent taking a shape of a mathematical figure as shown below. Based on the above information, answer the following questions. (i) Name of the shape in which wire is bent. (a) circle (b) parabola (c) ellipse (d) hyperbola (ii)The equation of the shape so formed is: x2 y2 x2 y2 x2 y2 (a) + = 1 (b) + =1 (c) − =1 (d) none of 9 4 4 9 9 4 these (iii) The eccentricity of the shape so formed is: 2 √x √5 √5 (a) 3 (b) (c) (d) √3 3 4 (iv) The length of the latus rectum of the shape so formed is: 8 (a) 9 (b) 3 (c) -4 (d) none of these. Solution: (i) (c) ellipse 1 (ii) x2 y2 1 (a) + =1 9 4 (iii) (c) Here a = 3 and b = 2 √a2 − b2 Eccentricity e = a √32 − 22 √5  e= = 1 3 3 2 (iv) (b) The length of the latus rectum = 2b a 2(2)2 8 1 = =3 3