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ISC Class XII Sample / Model Paper 2026 : Mathematics

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Shambhu Dutta
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ISC 2025 EXAMINATION Sample Question Paper - 2 Mathematics Time Allowed: 3 hours Maximum Marks: 80 General Instructions: This Question Paper consists of three sections A, B and C. Candidates are required to attempt all questions from Section A and all questions EITHER from Section B OR Section C. Section A: Internal choice has been provided in two questions of two marks each, two questions of four marks each and two questions of six marks each. Section B: Internal choice has been provided in one question of two marks and one question of four marks. Section C: Internal choice has been provided in one question of two marks and one question of four marks. All working, including rough work, should be done on the same sheet as, and adjacent to the rest of the answer. The intended marks for questions or parts of questions are given in brackets [ ]. Mathematical tables and graph papers are provided. SECTION A - 65 MARKS 1. In subparts (i) to (x) choose the correct options and in subparts (xi) to (xv), answer the questions as [15] instructed. (a) (b) If A is a skew-symmetric matrix of order n, and C is a column matrix of order n 1, then CAC is a) An identity matrix of order n b) A unit matrix of order one c) A unit matrix of order Zero d) A zero matrix of order one [1] 1/2 cos x log( 1+x 1 x )dx [1] is equal to 1/2 a) b) 0 5 c) 1 (c) (d) If tan 1 x + tan 1y = a) 3 c) 2 4 5 b) d) 5 Find the particular solution for 2xy + y c) y = If P(A) = 2x 1 log|x| 2x 1+log|x| 4 5 [1] then cot 1 x + cot 1 y equals 5 a) y = (e) d) 1 2 2 2x dy dx 5 = 0; y = 2 (x 0, x e) b) y = (x 0, x e) d) y = , and P(A B) = 7 10 then P(B|A) is equal to when x = 1 3x 1 log|x| 5x 1+log|x| (x 0, x e) (x 0, x e) [1] [1] a) 7 8 c) (f) (g) 1 10 b) 17 d) 1 20 8 Let A = {1, 2, 3, ...n} and B = {a, b}. Then the number of surjections from A into B is a) 2n + 1 b) nP2 c) 2n - 2 d) 2n - 1 If y = a) 1/x e e dy then dx = [1] [1] ? 1/x b) e d) 2e 1/x log x 2 x c) (h) (1/x 1) e If y = sec-1 ( a) c) (i) 1 x 1 ) 2 2x 1 then dy log x [1] ? = dx 1/x 2 b) 2 (1 x ) 2 d) 2 (1+x ) 2 1 x2 2 1 x2 Which of the following is not correct? a) |kA| = k3|A|, where A = [a ij ] 3 3 [1] b) If A is a skew-symmetric matrix of odd order, then |A| = 0 c) a + b a c+ d = g+ h e e + f (j) Assertion (A): [ 5 1 6 7 ][ b c + g f 2 1 3 4 1 2 3 Reason (R): 0 1 0 1 1 0 ] = [ 1 1 0 1 2 ij ] 2 1 3 4 0 3 d) |A| = |AT|, where A = [a d h ][ 1 4 a) Both A and R are true and R is the 1 6 7 ] 1 0 1 3 [1] 1 2 0 1 2 3 1 0 1 0 4 1 1 0 b) Both A and R are true but R is not the correct explanation of A. correct explanation of A. c) A is true but R is false. 2. 5 3 3 d) A is false but R is true. (k) Let f : R R : f(x) = x-2 + 1. Find : f-1{10} (l) Solve for x given that [ 2 3 1 1 ][ x ] = [ 4 1 [1] [1] ] 3 (m) Prove that the greatest integer function f: R (n) If P(A) = 0.8, P (B) = 0.5 and P(B|A) = 0.4, find P(A|B) [1] (o) Given two independent events A and B such that P (A) = 0.3, P (B) = 0.6. Find P(A and B). [1] Show that the function f(x) = { 1 + x, if x 2 5 x, if x > 2 R , given by f(x) = [x], is neither one-one nor onto. [1] [2] is not differentiable at x = 2. OR Find the equation of the normal at the point (am2, am3) for the curve ay2 = x3. 3. Evaluate the Integral: ( 4. Find the absolute maximum value and the absolute minimum value of the function f(x) = ( 1 tan x x+log cos x [2] ) dx 1 2 - x)2 + x3 in the [2] given interval [ -2, 25]. 5. Evaluate: [2] 5 (|x| + |x + 1| + |x - 5|) dx 1 OR Evaluate: sin3 x cos x dx 6. If f(x) = 7. Prove that tan { 8. Evaluate: 9. 2x 2 (1+x ) then show that f(tan ) = sin 2 4 + 1 2 1 2e 2x +3e x 1 cos ( a b )} + tan{ 4 [2] 1 2 1 cos ( a b )} = 2b a . [4] dx +1 If x = a cos3 and y = a sin3 , find [4] d 2 y 2 dx . Also, find its value at = [4] 6 OR Find 10. dy dx when y = (sin x)cos x + (cos x)sin x [4] Read the text carefully and answer the questions: There are two antiaircraft guns, named as A and B. The probabilities that the shell fired from them hits an airplane are 0.3 and 0.2 respectively. Both of them fired one shell at an airplane at the same time. (a) How is Bayes' theorem different from conditional probability? (b) Write the rule of Total Probability. (c) What is the probability that the shell fired from exactly one of them hit the plane? (d) If it is know that the shell fired from exactly one of them hit the plane, then what is the probability that it was fired from B? OR Read the text carefully and answer the questions: [4] A building contractor undertakes a job to construct 4 flats on a plot along with parking area. Due to strike the probability of many construction workers not being present for the job is 0.65. The probability that many are not present and still the work gets completed on time is 0.35. The probability that work will be completed on time when all workers are present is 0.80. Let: E1: represent the event when many workers were not present for the job; E2: represent the event when all workers were present; and E: represent completing the construction work on time. (a) What is the probability that all the workers are present for the job? (b) What is the probability that construction will be completed on time? (c) What is the probability that many workers are not present given that the construction work is completed on time? (d) What is the probability that all workers were present given that the construction job was completed on time? 11. Read the text carefully and answer the questions: [6] A manufacture produces three stationery products Pencil, Eraser and Sharpener which he sells in two markets. Annual sales are indicated below: Market Products (in numbers) Pencil Eraser Sharpener I 10,000 2,000 18,000 II 6,000 20,000 8,000 If the unit Sale price of Pencil, Eraser and Sharpener are 2.50, 1.50 and 1.00 respectively,Based on the information given above, answer the following questions: (a) What is the total revenue collected from Market-I? (b) What is the total revenue collected from Market-II? (c) What is the gross profit from both markets considering the unit costs of the three commodities as 2.00, 1.00, and 50 paise respectively? 12. Find the equation of a curve passing through the point (1, 1). If the tangent drawn at any point P(x, y) on the [6] curve meets the co-ordinate axes at A and B such that P is the mid-point of AB. OR Find the particular solution of the following differential equation : (x2 + xy)dy = (x2 + y2)dx, given that y=0,when x=1. 13. Show that the rectangle of maximum perimeter which can be inscribed in a circle of radius 10 cm is a square of side 10 2 cm [6] . OR Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is 4r 3 14. Read the text carefully and answer the questions: Mr. Ajay is taking up subjects of mathematics, physics, and chemistry in the examination. His probabilities of getting a grade A in these subjects are 0.2, 0.3, and 0.5 respectively. (a) Find the probability that Ajay gets Grade A in all subjects. [6] (b) Find the probability that he gets Grade A in no subjects. (c) Find the probability that he gets Grade A in two subjects. (d) Find the probability that he gets Grade A in at least one subject. SECTION B - 15 MARKS 15. In subparts (i) and (ii) choose the correct options and in subparts (iii) to (v), answer the questions as [5] instructed. (a) ^ ^ ^ ^ The value of p for which the vectors 2 i + p j + k and 4 i ^ j + 26k are perpendicular to each other, [1] ^ 6 is: a) b) 3 17 3 c) -3 d) 17 3 (b) Find the distance between the planes 2x - y + 2z = 5 and 5x - 2.5y + 5z = 20. [1] (c) Find the value of p for which the vectors [1] (d) Find the coordinates of the foot of the perpendicular drawn from the origin to 2x + 3y + 4z 12 = 0 a) ( c) ( (e) 27 29 24 29 , , 36 29 36 29 , , 48 29 48 29 and ^ ^ ^ 3 i + 2j + 9k ) b) ( ) d) ( 24 29 24 29 , , ^ ^ ^ i 2pj + 3k 39 29 36 29 , , 48 29 49 29 are parallel. [1] ) ) Find the vector equation of a plane which is at a distance of 5 units from the origin and its normal [1] ^ vector is 2^i 3^j + 6k . 16. Find a unit vector in the direction of the vector a = ^ ^ ^ i + 2j + 3k [2] . OR Show that (a b) (a + b) = 17. 2 (a b) , a and b Show that the distance d from point P to the line l having equation r = a + b is given by d = |b P Q| | b| where Q [4] is any point on the line l. OR Show that the lines 18. x 1 y 2 = 2 = z+3 3 and x 2 2 y 6 = 3 = z 3 4 intersect and find their point of intersection. Find the area under the given curves and given lines: y = x2, x = 1, x = 2 and x - axis. [4] SECTION C - 15 MARKS 19. In subparts (i) and (ii) choose the correct options and in subparts (iii) to (v), answer the questions as [5] instructed. (a) If C(x) and R(x) are respectively Cost function and Revenue function, then profit function P(x) is [1] given by (b) a) P(x) = R(x) C(x) b) P(x) = R(x) c) P(x) = C(x) + R(x) d) P(x) = R(x) - C(x) The value of objective function Z = 2x + 3y at comer point (3, 2) is a) 9 b) 5 c) 15 d) 12 [1] (c) Find the coefficient of correlation from the regression lines: x - 2y + 3 = 0 and 4x - 5y + 1 = 0 (d) The total cost and total revenue functions of a commodity are given by C(x) = x + 40 and R(x) = 10x - [1] 0.2x2. Find the break-even point. [1] (e) The demand function for a certain commodity is given by p = 1000 - 15x - x, 0 < x < 25. What is the [1] price per unit and the total revenue from the sale of 2 units? 20. A monopolist has a demand function x = 106 - 2p and the average cost function AC = 5 + x 50 , where p is the [2] price per unit output and x is the number of units of output. If the total revenue is R = px, determine the most profitable output and the maximum profit. OR The total cost function is given by C = x + 2x3 - 3.5x2, find the marginal average cost function (MAC). Also, find the points where the MAC curve cuts the x-axis and y-axis. 21. The following table shows the mean and standard deviation of the marks of Mathematics and Physics scored by [4] the students in a school: Mathematics Physics Mean 84 81 Standard Deviation 7 4 The correlation coefficient between the given marks is 0.86. Estimate the likely marks in Physics if the marks in Mathematics are 92. 22. Two tailors A and B, earn Rs 300 and Rs 400 per day respectively. A can stitch 6 shirts and 4 pairs of trousers [4] while B can stitch 10 shirts and 4 pairs of trousers per day. To find how many days should each of them work and if it is desired to produce at least 60 shirts and 32 pairs of trousers at a minimum labour cost, formulate this as an LPP. OR A dietician has to develop a special diet using two foods P and Q. Each packet (containing 30 g) of food P contains 12 units of calcium, 4 units of iron, 6 units of cholesterol and 6 units of vitamin A. Each packet of the same quantity of food Q contains 3 units of calcium, 20 units of iron, 4 units of cholesterol and 3 units of vitamin A. The diet requires atleast 240 units of calcium, atleast 460 units of iron and at most 300 units of cholesterol. How many packets of each food should be used to minimise the amount of vitamin A in the diet? What is the minimum amount of vitamin A? Solution SECTION A - 65 MARKS 1. In subparts (i) to (x) choose the correct options and in subparts (xi) to (xv), answer the questions as instructed. (i) (d) A zero matrix of order one Explanation: { By definition of skew-symmetric matrix. (ii) (b) 0 Explanation: { 1/2 f (x) = cos x log( 1+x 1 x )dx 1/2 1/2 f ( x) = cos( x) log( 1 x 1+x )dx = f (x) 1/2 Hence, f(x) is odd function. (iii) (b) 5 Explanation: { We know that, tan-1 x + cot-1 x = 2 We have, tan 1 x + tan 1 y = 4 /5 (1) Let, cot 1x + cot 1 y = k (2) Adding (1) and (2) tan-1 x + tan-1 y + cot-1 x + cot-1 y = Now, tan 1 A + cot 1 A = 2 4 5 + k ......(3) for all real numbers. So, (tan 1 x + cot 1 x) + (tan 1y + cot 1 y) = (4) From (3) and (4), we get, + k = 4 5 (iv) k= - k= 4 5 5 (a) y = 2x 1 log|x| (x 0, x e) Explanation: { Let y = vx dy dx dv = v + x dx Question becomes v + x x x dv dx dv dx 2v+v v v 2 2 = 2 2v+ v 2v 2 = v 2 2 = 2 = dv 2 dv dx dx x = logx + c When x=1 y=2 we get 2x y 2 2 = logx + c = log1 + c 2x y y = = logx 1 2x 1 log|x| c = 1 2v+v 2 (v) (a) 7 8 Explanation: { 4 P (A) = 5 , P (A B) = P (A B) P (B/A) = (vi) 7/10 = P (A) 7 10 4/5 = 7 8 (c) 2n - 2 Explanation: { Given, A = (1, 2, 3,..., n] and B = {a, b} Since, the number of surjections from A to B = total number of functions from A to B - number of functions from A to B whose, images are proper subset of B and total number of functions from a set with p number of elements into a set with q number of elements = qp. Number of surjections from A into B = 2n - 2 (vii) (a) e 1/x x2 Explanation: { 1 Here y = e Taking log both sides, we get x loge y = (Since log abc = cloga b) 1 x Differentiating with respect to x, we obtain 1 dy y dx 2 x 2 1 x2 dy dx = dx 1 = 2 y x dy Therefore, (viii) (b) or 1 = 1 1 e 2 x x Explanation: { y = sec 1 ( 1 2 2x 1 1 sec y = ) 2 2x 1 cos y = 2x2 - 1 y = cos-1 (2x2 - 1) Put x = cos ,we get y =cos-1 (2cos2 - 1) y =cos-1 (cos 2 ) y = 2 But = cos-1 x, we get (ix) d(cos dy dx = 1 x) d(cos dy dx dy dx dy dx = 2 1 x) dx = 2 ( = dx 1 1 x2 ) 2 1 x2 a+ b a c + d = g + h e (c) e+ f b c + g f d h Explanation: { For adding the determinants, we need to find the value of the determinants and add them.We cannot apply the method applicable for matrix addition. (x) (d) A is false but R is true. Explanation: { Assertion: [ 2 1 [ 5 5 1 6 7 4 1 3 4 10 3 5 4 12 + 21 6 + 28 ] = [ 1 ][ 3 2 ][ 10 + 6 2 + 7 ] = [ 6 7 16 5 39 25 ] = [ 15 + 24 7 1 33 34 ] = [ 3 + 28 ] ] 5 Hence, [ 1 2 1 ][ 6 7 2 3 4 1 2 3 1 0 = = = 1 2 0 + ( 1) + 0 1 + 0 + 0 1 1 + 0 5 8 0 1 1 14 0 0 1 2 0 3 7 ] 0 1 4 0 + 2 + 12 0 + 1 + 0 1 2 3 1 0 1 1 0 2 + 1 + 0 3 + 0 + 0 0 + 0 + 1 0 1 + 1 0 + 0 + 0 2 + 0 + 4 4 + 3 + 4 1 1 3 1 0 0 6 3 Hence, 0 1 0 1 0 1 2 6 + 0 + 0 1 1 0 4 11 0 + 1 + 0 1 + 0 + 0 6 1 1 6 1 1 1 3 0 + 0 + 0 0 1 0 1 2 + 9 1 4 1 1 + 0 + 6 and = 1 5 ][ 3 Reason: Here, 0 1 ] [ 1 1 0 1 2 3 0 1 1 0 1 2 3 1 4 0 1 2 3 1 0 1 0 1 0 4 1 (xi)We have, f(x) = x2 + 1 Let f-1(10) = x. Then,we have, f(x) = 10 x2 + 1 = 10 x2 = 9 x = 3. f-1{10} = {-3, 3} (xii) 2x 3y [ 1 ]= [ x + y ] 3 2x 3y = 1 x + y = 3 x = 3 y 2 (3 y) 3y = 1 -5y = -5 y = 1 x = 3 1 x = 2 (xiii) For not one-one: 1.1, 1.2 R(domain) now,1.1 1.2 but f(1.1) = f(1.2) = 1 f is not one-one. For not onto: Let R (co-domain), but [x] = is not possible for x in domain. 1 1 2 2 so, f is not onto. (xiv) Given: P(A) = 0.8, P(B) = 0.5 and P(B|A) = 0.4 By definition of conditional probability P (B|A) = P (A B) = P (B|A) P (A) = 0.32 N ow, P (A|B) = 0.32 0.5 = 0.64 P (A B) P (A) P(A|B) = 0.64 (xv)P (A and B) = P (A).P(B) = 0.3 0.6 = 0.18 2. RHD at x = 2 = Rf = lim h 0 h h (2) = limh 0 f (2+h) f (2) = limh 0 ( 1) = 1 And, LHD at x=2 = Lf '(2) = lim = lim h 0 h h 0 f (2 h) f (2) h [ 5 (2+h) 3 h ] 1+(2 h) 3 = lim [ h 0 ] = limh 0 h h = limh 0 1 = 1 h Thus, Rf (2) Lf (2) . Hence, f(x) is not differentiable at x = 2. OR ay2 = x3 It is given that Now, differentiating both sides with respect to x, we get dy 2ay. dy dx 2 = 3x dx 2 3x = 2ay Then, the slope of the tangent to the given curve at (am2, am3) is 2 3(am ) dy dx ] 2 = 2 2 3 2a(am ) (am ,am ) 2 4 2 3 3a m = 2a m 3m = 2 Then, slope of normal at (am2, am3) = 1 = Slope of the tangent at (am2 ,am2 ) 2 3m Therefore, equation of the normal at (am2, am3) is given by: 3 y am = 2 2 3m (x am ) 3am4 = 3my - 2x + 3my - am2(2 + 3m2) = 0 3. Let I = ( Since 1 -2x + 2am2 1 tan x x+log(cos x) ) dx dx = log |x| + c x We have, I = ( 1 tan x x+log(cos x) ) dx .... (i) Let x + log (cos x) = t 1 ( sin x) 1 + dt = cos x 1 tan x = dt dx dx (1 - tan x) dx = dt Putting this value in equation (i), we get I = dt t I = log |t| + c I = log |x + log (cos x)| + c 4. We have, f(x) = ( 2 1 x) 2 f'(x) = -2( + x3, where x [-2, 25]. - x) + 3x2 = -1 + 2x + 3x2 1 2 At the points of local maximum and local minimum, we must have f'(x) = 0 3x2 + 2x - 1 = 0 (3x - 1) (x + 1) = 0 x = 1 3 , -1 The values of f(x) at these points and also at the end - points of the interval are computed as given below. f(-2) = ( 1 f(-1) = ( 1 2 + 2) 2 2 + 1) 2 + (-2)3 = 25 + (-1)3 = 5 4 7 1 8 3 - 8 = - , f( 4 and, f(2.5) = ( ) = ( 1 2 Of these values, the maximum value of f(x) is Thus, the absolute maximum = 157 8 1 2 2 2.5) 157 8 1 3 2 ) + ( 1 3 + (2.5)2 = 3 157 8 5 0 (|x| + |x+1| + |x-5|) dx = 1 =[ 1 (6 0 x) 2 ] 2 + [ 1 = 13 2 + 85 2 x ( +6) 2 1 27 8 7 and, the absolute minimum = - . 8 5 (6 - x)dx + (x + 6)dx 0 ] + 7 5 2 1 36 and the minimum value is - 5. If x [-1, 0] f(x) = -x + x + 1 - x + 5 = 6 - x If x [0, 5] f(x) = x + x + 1 - x + 5 = x + 6 = ) 0 = 49 OR sin3 x Let I = cos x dx Now let sin x - t. Then, d (sin x) = dt cos x dx - dt dx = dt cos x = 7 108 , Put sin x = t and dx = I= sin3 x , we get dt cos x t3 cos cos x dx = x dt 2x 6. Here we are given that, f(x) = 4 t 3 = t dt = cos x 4 4 + C = sin x 4 + C 2 (1+ x ) Need to prove: f(tan ) = sin 2 f(tan ) = 2 tan 1+ tan2 [as 1 + tan2 = sec2 ] 2 tan f(tan ) = 2 se c f(tan ) = 2 f(tan ) = 2 sin cos = sin 2 [Proved] sin cos cos 2 [as tan = 7. Here we need to prove that tan( Let LHS = tan( Put cos a 1 LHS = tan( tan 4 = 1 tan 4 2 4 ) + tan( a 1 b 1 4 ) + tan( cos 2 1 a b 4 ) 1 2 a b 4 ) 2 2 tan 2 2 2 ) + (1 tan 2 ) 2 cos 2 ] 2 1+tan (1 tan b 1 + 4 1 cos 2 2 = a 1 tan 4 1+tan 1 tan + 2 (1+tan cos ) + tan( 2 + 2 tan 2 tan 2 1 tan + 4 1+tan = +tan 1 + 4 = cos = b and sec = sin cos )(1+tan 2 ) 2 2 1+tan 2 = 2( 2 ) = 2 1 tan 2 = cos a/b 2b = a 2 RHS Hence proved. 8. We have, = 1 2e 2x +3e x = dx 1 e 2x + e dx = 2 +1 2+3e 3 e x 2x x +e dx + C 2x +1 Let e-x = t. -e-x dx = dt dx = dt e x tdt I = t = 2 2+3t+t dt 2 t +3t+2 Let t = (2t + 3) + Comparing the coefficients of like powers of t, we get 2 = 1, 3 + = 0 = , and = (2t+3)+ I = dt 2 1 3 2 2 t +3t+2 I = 2t+3 2 t +3t+2 dt 1 dt 2 t +3t+2 By using the respective values of and I= I= 1 2 1 2 2t+3 dt + 2 t +3t+2 3 2 1 2 2 2 log t + 3t + 2 + 3 2 1 I= 1 2 2x log e x + 3e 2( 1 + 2 + 3 2 t+ log ) log t+ 2 dt (t+3/2) (1/2) e e x x 3 2 3 2 +1 +2 + 1 2 1 2 + C 9. According to the question, x = a cos3 and y = a sin3 We are required to find the value of d 2 y dx2 at = 6 Therefore,on differentiating both sides of x and y w.r.t ,we get, dx d = 3a cos = 3a cos 2 and d d d (cos ) ( sin ) = 3a cos dy 2 2 sin = 3a sin 2 d d (sin ) cos 1 a b ) = 2b a = 3a sin dy Now, 2 dy/d = ( dx 2 3a sin = (cos ) = 3a sin dx/d cos 3a cos 2 ) 2 cos = tan sin Again, On differentiating both sides w.r.t x, we get, d 2 y d = 2 dx = sec = sec 2 2 dx 1 ( At = ( d 2 y 2 6 ) 1 9 4 ) sin 1 4 ) (sin ) 6 ) 2 1 3a( 1 3a cos 6 ( 2 = ( 4 3 3a( ) = sin 3a(cos 6 1 = 2 = at = d dx ) dx (tan ) d d 3a cos d ( tan ) = dx )( 16 1 32 = 27a ) 2 OR We have, y = (sin x)cos x + (cos x)sin x c os x log(sin x) y = e sin x log(cos x) + e y = ecos x log sin x + esin x log cos x Differentiating with respect to x, dy dx = d dx cos x log sin x (e cos x log sin x = e d dx log(sin x) cos x = (sin x) dx sin x log cos x (e ) sin x log cos x (cos x log sin x) + e c os x = e d )+ [cos x [cos x d dx log sin x + log sin x 1 d sin x dx d dx d dx (sin x log cos x) sin x (cos x)] log(cos x) +e (sin x) + log sin x ( sin x)] [sin x sin x +(cos x) d dx [sin x log cos x + log cos x 1 d cos x dx d dx (sin x)] (cos x) + log cos x (cos x)] = (sin x)cos x[cot x cos x - sin x log sin x] + (cos x)sin x[tan x(-sin x) + cos x log cos x] = (sin x)cos x[cot x cos x - sin x log sin x] + (cos x)sin x[cos x log cos x - sin x tan x] 10. Read the text carefully and answer the questions: There are two antiaircraft guns, named as A and B. The probabilities that the shell fired from them hits an airplane are 0.3 and 0.2 respectively. Both of them fired one shell at an airplane at the same time. (i) Bayes' theorem defines the probability of an event based on the prior knowledge of the conditions related to the event whereas in case of the condition probability, we find the reverse probabilities using Bayes' theorem. (ii) Consider on event E which occurs via two different events A and B. The probability of E is given by the value of total probability as: P(E) = P(A E) + P(B E) P(E) = P(A) P( E A ) + P(B)( E B ) (iii)Let P be the event that the shell fired from A hits the plane and Q be the event that the shell fired from B hits the plane. The following four hypotheses are possible before the trial, with the guns operating independently: E1 = PQ, E2 = P Q , E3 = P Q, E4 = PQ Let E = The shell fired from exactly one of them hits the plane. P(E1) = 0.3 0.2 = 0.06, P(E2) = 0.7 0.8 = 0.56, P(E3) = 0.7 0.2 = 0.14, P(E4) = 0.3 0.8 = 0.24 P( E E1 ) = 0, P( E E2 P(E) = P(E1) P( P( E E3 ) +0, P( ) E + P(E4) P( ) = 1, P( + P(E2) P( ) E1 E E3 E E4 E E2 E E4 ) ) = 1 + P(E3). ) = 0.14 + 0.24 = 0.38 (iv)By Bayes' Theorem, E3 P( E ) 0.14 0.38 = 7 19 E E 1 E )+P ( E2 ) P ( E 2 ) E3 = P ( E1 ) P ( = E P ( E3 ) P ( )+P ( E3 )P ( E E 3 )+P ( E4 ) P ( E ) E 4 NOTE: The four hypotheses form the partition of the sample space and it can be seen that the sum of their probabilities is 1. The hypotheses E1 and E2 are actually eliminated as P( E E1 ) = P( E E2 ) = 0 Alternative way of writing the solution: i. P(Shell fired from exactly one of them hits the plane) = P[(Shell from A hits the plane and Shell from B does not hit the plane) or (Shell from A does not hit the plane and Shell from B hits the plane)] = 0.3 0.8 + 0.7 0.2 = 0.38 P(Shell fired from B hit the plane Exactly one of them hit the plane) ii. P = P = P(Exactly one of them hit the plane) (Shell from only B hit the plane ) (Exactly one of them hit the plane ) 0.14 0.38 = 7 19 OR Read the text carefully and answer the questions: A building contractor undertakes a job to construct 4 flats on a plot along with parking area. Due to strike the probability of many construction workers not being present for the job is 0.65. The probability that many are not present and still the work gets completed on time is 0.35. The probability that work will be completed on time when all workers are present is 0.80. Let: E1: represent the event when many workers were not present for the job; E2: represent the event when all workers were present; and E: represent completing the construction work on time. (i) P(E2) = 1 - P(E1) = 1 0.65 = 0.35 (ii) P(E) = P(E1) P ( E ) E1 + P(E2) P ( = 0.65 0.35 + 0.35 0.8 = 0.35 1.45 = 0.51 E (iii) P E P E E P( ) = E ( 1 ) ( 1 P ( E2 E P(E 1 ) P( ) = E )+ P(E 2 ) P( 1 P ( E2 ) P ( P ( E1 ) P ( E E 1 ) ) = 1 E (iv) E E2 E E E 0.51 = 0.45 ) 2 ) = E 2 )+P ( E2 ) P ( 0.65 0.35 E ) 0.35 0.8 0.51 = 0.55 E 2 11. Read the text carefully and answer the questions: A manufacture produces three stationery products Pencil, Eraser and Sharpener which he sells in two markets. Annual sales are indicated below: Market Products (in numbers) Pencil Eraser Sharpener I 10,000 2,000 18,000 II 6,000 20,000 8,000 If the unit Sale price of Pencil, Eraser and Sharpener are 2.50, 1.50 and 1.00 respectively,Based on the information given above, answer the following questions: (i) Let A be the 2 3 matrix representing the annual sales of products in two markets. x y A = 10000 6000 z 2000 20000 18000 8000 Let B be the column matrix representing the sale price of each unit of products x, y, z. 2.5 B = 1.5 1 Now, revenue = sale price number of items sold =[ 10000 2000 2.5 ] 1.5 6000 =[ 18000 20000 8000 25000 + 3000 + 18000 ] 1 =[ 15000 + 30000 + 8000 46000 ] 53000 Therefore, the revenue collected from Market I = 46000 (ii) Let A be the 2 3 matrix representing the annual sales of products in two markets. x y A = 10000 6000 z 2000 20000 18000 8000 Let B be the column matrix representing the sale price of each unit of products x, y, z. 2.5 B = 1.5 1 Now, revenue = sale price number of items sold =[ 10000 2000 2.5 ] 1.5 6000 =[ 18000 20000 8000 25000 + 3000 + 18000 ] 15000 + 30000 + 8000 1 =[ 46000 ] 53000 The revenue collected from Market II = 53000. (iii)Let C be the column matrix representing cost price of each unit of products x, y, z. Then, C = 2 1 0.5 Total cost in each market is given by 10000 AC = [ 2000 ] 6000 20000 8000 20000 + 2000 + 9000 =[ 2 18000 ] 1 =[ 12000 + 20000 + 4000 0.5 31000 ] 36000 Now, Profit matrix = Revenue matrix - Cost matrix 46000 =[ 31000 ] [ 53000 ] =[ 36000 15000 ] 17000 Therefore, the gross profit from both the markets = 15000 + 17000 = 32000 12. The below figure obtained by the given information Let the coordinate of the point P is (x, y). It is given that, P is mid-point of AB. So, the coordinates of points A and B are (2x, 0) and (0, 2y) respectively. 0 2y Slope of AB = = 2x 0 y x Since, the segment AB is a tangent to the curve at P. dy dx dy y = = y x dx x On integrating both sides, we get log y = log x + log c log y = log ...(i) C x Since, the given curve passes through (1, 1). C log 1 = log 1 0 = log C C = 1 log y = log 1 y = 1 x x xy = 1 OR Given, (x2 + xy)dy = (x2 + y2)dx dy dx 2 = ( x +y 2 ...........(i) ) 2 x +xy This is a homogeneous differential equation. dy On putting y = vx v + x dv dx x ( 1+v 1 v 2 = ( dv dx 2 dx 2 x +v x ) 2 = v 1 + x dv dx in Eq. (i), we get, x +x xv 2 = 2 1+v v = 1+v ) dv = 1 x dx 2 1+ v v v 1+v = 1 v 1+v On integrating both sides, we get ( 1+v 1 v 1 ) dv = [ 1 + x dx 2 1 v ] dv = log |x| + log C -v - 2 log (1 - v) = log | x | + log C -v = 2log(1 - v) + log | x | + log C -v = log(1 - v)2 + log {C |x|} [ log m + log n = log mn] -v = log{C|x|(1-v)2} C |x|(1 y x 2 ) y/x = e [ v = y x ] ...(ii) On putting x = 1 and y = 0 in Eq. (ii), we get C 1(1 - 0) = e0 C = 1 Thus, the required solution is y |x|(1 x 2 = e 2 y/x y/x ) (x y ) = |x|e 13. AB = 2x; BC = 2y In ABC, (2x)2 + (2y)2 = (20)2 4x2 + 4y2 = 400 x2 + y2 = 100 ...(i) P = 4x + 4y = 4x + 4 100 x = 4 = 0 [from (i)] 2 dP 4x 100 x2 dx x= d 2 5 2 cm P 4 100 < 0 3 100 x2 = 100 x2 = 4 dx2 x ( x ) 100 x2 2 (100 x ) 2 Hence, perimeter is maximum, when x = 5 2 y = 5 2 [from (i)] x = y ABCD is square of side 10 2 cm OR 1 V = 1 = 2 1 2 . (r 3 dv = dx dv = dx 1 = 1 3 2 2 [r 2 1 2 2 x )(r + x) [ R [(r [r 3 2 = r 2 x ] 2 x )(1) + (r + x)(0 2x)] 2 x 2 2rx 2x ] 2 2rx 3x ] 2 3rx + rx 3x ] [r(r 3x) + x(r 3x)] 3 1 = 3 2 1 = 1 [r 3 = R H R . (r + x) 3 = 3 2 [(r 3x)(r + x)] 3 For critical points let dv dx = 0 r = 3x r = x 3 d 2 v = dx2 d 2 v 2 3 ] dx = 1 x= 1 3 [0 2r 6x] r = 1 3 [ 2r 6 r 3 ] 3 [ 4r] = - tive maximum Altitude = r + x = = r 3 + r 4r 3 14. Read the text carefully and answer the questions: Mr. Ajay is taking up subjects of mathematics, physics, and chemistry in the examination. His probabilities of getting a grade A in these subjects are 0.2, 0.3, and 0.5 respectively. (i) P(Grade A in Maths) = P(M) = 0.2 P(Grade A in Physics) = P(P) = 0.3 P(Grade A in Chemistry) = P(C) = 0.5 P(Grade A in all subjects) = P(M P C) = P(M) P(P) P(C) P(Grade A in all subjects) = 0.2 0.3 0.5 = 0.03 (ii) P(Grade A in Maths) = P(M) = 0.2 P(Grade A in Physics) = P(P) = 0.3 P(Grade A in Chemistry) = P(C) = 0.5 P(Grade A in no subjects) = P (M P C ) = P (M ) P (P ) P (C ) P(Grade A in no subjects) = 0.8 0.7 0.5 = 0.280 (iii)P(Grade A in Maths) = P(M) = 0.2 P(Grade A in Physics) = P(P) = 0.3 P(Grade A in Chemistry) = P(C) = 0.5 P(Grade A in 2 subjects) = P (M P C ) + P (P C M ) + P (M C P ) P(Grade A in 2 subjects) = 0.2 0.3 0.5 + 0.3 0.5 0.8 + 0.2 0.5 0.7 = 0.03 + 0.12 + 0.07 = 0.22 P(Grade A in 2 subjects) = 0.22 (iv) P(Grade A in Maths) = P(M) = 0.2 P(Grade A in Physics) = P(P) = 0.3 P(Grade A in Chemistry) = P(C) = 0.5 P(Grade A in atleast one subject) = 1 P(grade A in no subject) = 1 P (M P(Grade A in atleast one subjects) = 1 0.280 = 0.72 SECTION B - 15 MARKS P C) 15. In subparts (i) and (ii) choose the correct options and in subparts (iii) to (v), answer the questions as instructed. (i) (b) 3 Explanation: { 3 (ii) We need to find the distance between the given two planes,2x - y + 2z = 5 and 5x - 2.5y + 5z = 20. Now, 2x - y + 2z = 5 2x - y + 2z - 5 = 0....(i) =5[x - 0. 5y + z] = 20 ( simplifying it) x 1 2 y + z= 4 and 5x - 2.5y + 5z = 20 = 2x - y + 2z - 8 = 0.....(ii) Clearly, planes (i) and (ii) are parallel. Distance between two parallel planes, d = d2 d1 a2 + b2 + c2 = 8 ( 5) 2 2 ( 2 +( 1) + 2 2 [ :d2 = -8, d1= -5, a = 2, b = -1 and c = 2] = 8+5 4+1+4 = 3 9 = | 1| = 1 ^ ^ (iii)3^i + 2^j + 9k and ^i 2p^j + 3k are two parallel vectors, so their direction ratios will be proportional. 3 1 = 2 2p = 9 3 2 2p = 3 1 6p = 2 p = (iv) (c) ( 24 29 , 36 29 , 48 29 ) 2 6 p = 1 3 Explanation: { Direction ratio s of the line are ( 2, 3, 4 ). Therefore, the equation of the line is: x 0 2 y 0 = = 3 z 0 = 4 Thus, the coordinates of any point P on the above line are P ( 2 , 3 , 4 ). But , this point P also lies on the given plane: 2(2 ) + 3(3 ) + 4(4 ) 12 = 0. 29 = 12 = 12 29 Therefore, the coordinates of the foot of perpendicular are given by: (2 12 29 ,3 12 29 ,4 ^ (v) Here n = 2^i 3^j + 6k , then |n | = 4 + 9 + 36 = 49 = 7 units ^ = n n | n | = 2 7 ^ i 3 7 ^ j + 6 7 ^ k So, the required equation is r ( 2 7 ^ i 3 7 ^ j + 6 7 ^ k) = 5 2 2 2 + 3 = 14 ^ 16. We have,a = ^i + 2^j + 3k |a | = 1 + 2 Unit vector in the direction of a is given by ^ = a ^ ^ ^ ( i +2 j +3k) a = | a | =( 14 1 14 ^ i + 2 14 ^ j + 3 14 ^ k) . . OR L.H.S b) = (a (a + b) = a a + a b b a b b . = 0 + a b b a 0 [a a = b b = 0 ] = a b + a b [a b = b a ] = 2(a b) 17. Let PM be perpendicular from P to line l and Q be point on it such that PQ makes an angle with l. In right triangle PMQ, we have sin = PM PQ sin = d PQ d = PQ sin [by cross multiplication ] d|b| = |P Q||b| sin d|b| = |b P Q| [Multiplying both sides by |b| ] [ b is parallel to line /. So, the angle between b and P Q is also ] d = | b P Q | | b| OR The coordinates of any point on the first line are given by x 1 = y 2 2 = z+3 3 = x= y = 2 + 2 z = 3 - 3 The coordinates of a general point on the first line is ( , 2 + 2, 3 3) 12 29 ) And the coordinates of any point on the second line are given by x 2 y 6 = 2 3 = z 3 4 = x = 2 + 2 y = 3 + 6 z = 4 + 3 The coordinates of a general point on the second line are If the lines intersect, then they have a common point. So, for some values of and we must have = 2 + 2, 2 + 2 = 3 + 6, 3 3 = 4 + 3 2 = 2 .....(i) 2 3 = 4 ....(ii) 3 4 = 6 ....(iii) (2 + 2, 3 + 6, 4 + 3) Substituting = 2 and = 0 in (iii), we get LHS = 3 4 = 3(2) - 4(0) = 6 = RHS Therefore = 2 and = 0 satisfy the third equation, the required given lines intersect at (2, 6, 3). 18. Equation of the curve (parabola) is y = x2 ...(i) Required area bounded by curve (i), vertical line x = 1, x = 2 and x - axis 2 = ydx 1 2 2 = x dx 1 2 3 = ( = 8 3 x 3 ) 1 1 3 = 7 3 sq units SECTION C - 15 MARKS 19. In subparts (i) and (ii) choose the correct options and in subparts (iii) to (v), answer the questions as instructed. (i) (d) P(x) = R(x) - C(x) Explanation: { P(x) = R(x) - C(x) (ii) (d) 12 Explanation: { Z = 2x + 3y Z (3, 2) = 2 3 + 3 2 = 6 + 6 = 12 (iii)Let the line of regression of y on x be x - 2y + 3 = 0 2y = x + 3 y = + x 3 2 bxy = 1 2 2 Let the line of regression of x on y be 4x - 5y + 1 = 0 4x = 5y - 1 x= x- 5 1 4 bxy = 5 4 r2 = =( 1 2 4 byx bxy 5 )( 4 5 ) = 8 < 1 Hence, our assumption of regression equation is correct. r= = 0.79 5 8 (iv)The break-even points are given by C(x) = R(x) Now, C(x) = R(x) x + 40 =10x - 0.2x2 0.2x2 - 9x + 40 = 0 2x2 - 90x + 400 = 0 x2 - 45x + 200 = 0 (x - 40) (x - 5) = 0 x = 5, 40 Hence, the break-even points are x = 5 and x = 40 (v) Let R(x) be the revenue function. Then, R(x) = px R(x) = 1000x - 15x2 - x3 When x = 2, we get p = 1000 - 15 2 - 22 = 966 and, R = 2000 - 15 4 - 8 = 1932 20. x = 106 - 2p p = (106 x) 1 2 Revenue, R = px = (106 x)x 1 2 2 = 53x - x 2 AC = 5 + x 50 C = AC(x) = (5 + =5+ x 50 2 x 50 )x P = Revenue - cost dP dx dP dx 13(2x) = 48 13(2x) d = 0 25 48 = 2 = 0 48 x= 25 13 2x 25 13 25 13 2 = 46.1538 2 P dx2 (13) = 0 25 , negative since d 2 P dx2 is negative Profit is maximum at x = 46 units Profit = 48x - x 13 2 25 When x = 46 Profit = 48 46 = 2208 - 27508 25 13 25 46 46 = 2208 - 1100.32 = Rs 1107.68 OR Given, total cost function, C(x) = x + 2x3 - 3.5x2 Here, AC = = 1 x C x 2x3 - (x + 3.5x2) = 1 + 2x2 - 3.5x = 2x2 - 3.5x + 1 We know that, marginal average cost (MAC) is MAC = = d dx d dx 2 (2x (AC ) 3.5x + 1) MAC = 4x - 3.5 Since, MAC is linear, it is a straight line. It will cut x-axis at (0.875, 0) and y-axis at (0, -3.5) = 84, y = 81 21. Given, x = 7, = 4 and r = 0.86 x y byx = r = 0.86 4 7 y x = 0.49 Regression equation of y on x ) y - y = byx(x x y - 81 = 0.49(x - 84) y - 81 = 0.49x - 41.16 y = 0.49x + 39.84 Putting x = 92, y = 45.08 + 39.84 = 84.92 Hence, the likely marks in Physics are 84.92. 22. According to the given situation , the given data can be tabularised as following Tailor A Tailor B Minimum Total No. No. of Shirts 6 10 60 No. of Trousers 4 4 32 Rs 300/day Rs 400/day Wage Let tailor A and tailor B work for x days and y days, respectively.Given that the minimum number of shirts that can be stitched per day is 60. The inequality representing the information is given as 6x + 10y 60 ( shirt constraint) ( dividing by 2 we get) 3x + 5y 30 Given that the minimum number of trousers that can be stitched per day is 32. 4x + 4y 32 ( trouser constraint) ( dividing throughout by 4 we get x+y=8 x 0, y 0 ( non negative constarint which restricts the feasible region in the first quadrant only , since it is real world situation and the variables cannot take negative values. Let z be the objective function representing the total labour cost. Hence the equation for the cost function is given as z = 300x + 400y So, the given L P.P. is designed as z = 300x + 400y x 0, y 0, 3x + 5y 30 and x + y 8 OR Let x and y be the number of packets of food P and Q respectively. Obviously x 0, y 0 . Mathematical formulation of the given problem is as follows: Minimise Z = 6x + 3y (vitamin A) subject to the constraints 12x + 3y 240 (constraint on calcium), i.e. 4x + y 80 ......(i) 4x + 20y 460 (constraint on iron), i.e. x + 5y 115 ......(ii) 6x + 4y 300 (constraint on cholesterol), i.e. 3x + 2y 150 ......(iii) ......(iv) Let us graph the inequalities (i) to (iv). The feasible region (shaded) determined by the constraints (i) to (iv) is shown in Figure and note that it is bounded. x 0, y 0 The coordinates of the corner points L, M and N are (2, 72), (15, 20) and (40, 15) respectively. Let us evaluate Z at these points: Corner Point Z = 6x + 3y (2, 72) 228 (15, 20) 150 Minimum (40, 15) 285 From the table, we find that Z is minimum at the point (15, 20). Hence, the amount of vitamin A under the constraints given in the problem will be minimum, if 15 packets of food P and 20 packets of food Q are used in the special diet. The minimum amount of vitamin A will be 150 units.

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