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ISC Class XII Sample / Model Paper 2026 : Mathematics

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Ronak Ajitsaria
Sanskriti the Gurukul, Kamrup
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ISC 2026 EXAMINATION Sample Question Paper - 5 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 [15] questions as instructed. (a) If AT 3 4 = 1 a) 2 0 1 4 0 1 4 3 3 (b) 1 2 1 a) 1 2 c) log tan( 3 (c) (d) cos 1 (cos( a) c) 3 1 2 3 ] , then b) AT - 3 d) is ________. 4 0 2 4 3 3 2 [1] 1 x BT 3 dx log tan( 1 0 cos x+ 3 sin x 2 2 c) and B = [ 1 3 3 1 0 1 is equal to + + )) x 2 3 [1] b) )+C log tan( b) d) 3 = 2xy 3 + )+C x 4 )+C [1] 3 dx 3 is equal to Solution of (x + 1) 2 d) 2log tan( )+C dy x is 2 3 2 3 [1] a) log|y| = 2(x log|1 + x|) + c c) log y = {x + log|x|} + C (e) b) log|y| = 2(x + log|1 x|) + c d) log y = {x log|x|} + C A problem in Statistics is given to three students A, B and C whose chances of solving it independently are 1 , 2 and 1 3 1 4 [1] , respectively. The probability that the problem will be solved, is a) c) (f) 1 12 1 2 b) 11 d) 3 12 4 Let R be any relation in the set A of human beings in a town at a particular time. If R = { [1] [x, y ) : x is exactly 7 cm taller than y], then R is (g) (h) (i) a) an equivalence relation b) reflexive c) not symmetric d) symmetric but not transitive If y = x x x x e e e +e , then dy [1] is equal to dx a) y2 + 1 b) 1 + y2 c) 1 - y2 d) y2 - 1 2 If y = Ae5x + Be-5x, then d y 2 dx [1] is equal to a) 5 y b) -25 y c) 15 y d) 25 y If A and B are invertible matrices, then which of the following is not correct? a) adj. A = |A|. A 1 b) (A + B) 1 = B 1 + A 1 c) (AB) 1 = B 1 A 1 d) det(A) 1 = [det (A)] 1 (j) 2 x 3 Assertion (A): If A = 3 2 4 x 2 1 1 5 [1] [1] is a symmetric matrix, then x = 6. Reason (R): If A is a symmetric matrix, then A = A'. a) Both A and R are true and R is the b) Both A and R are true but R is not correct explanation of A. the correct explanation of A. c) A is true but R is false. (k) (l) d) A is false but R is true. Let f and g be two real function defined by f(x) = and g(x) = (x + 4)3 find the 1 x+4 Find the matrix X such that 2A - B + X = 0, where A =[ 3 1 ] 0 2 and B =[ 2 1 f . 1 ] 0 Show that the function f : N N : f(x) = x3 is one - one into (n) Let A and B be the events such that P(A) = (o) A die is thrown. If E is the event the number appearing is a multiple of 3 and F be the 7 , P(B) = 9 13 and P(A B) = [1] 3 (m) 13 [1] [1] 4 13 find P(B/A). [1] [1] event the number appearing is even then find whether E and F are independent? 2. If f (x) 2 = 2x 1 and y = f(x2), then find dy dx at x = 1. OR [2] Find the interval in which the function f(x) = x 8 + 6x 2 is increasing or decreasing. 3. Evaluate 0 (2 log cos x - log sin 2x) dx [2] 4. Show that f(x) = loga x, 0 < a < 1 is a decreasing function for all x> 0. [2] 5. Evaluate: [2] 2 1 (x 1)(x+1)(x+2) dx OR Evaluate the definite integral: 0 /4 6. x 2 sin xdx Let A be the set of all triangles in a plane. Show that the relation R = {( 1 , 2 ) : 1 2 } is [2] an equivalence relation on A. 7. Prove that tan 8. Evaluate: 9. Find , if y = tan dx Find dy 1 dy cos x ( 1+sin x dx x(x 2)(x 4) 1 ) = 4 x 2 , x ( 2 , 2 ). . [4] [4] 1+x2 1 x [4] . OR 10. dx when sin x sin 2x sin 3x sin 4x Read the text carefully and answer the questions: [4] To hire a marketing manager, it's important to find a way to properly assess candidates who can bring radical changes and has leadership experience. Ajay, Ramesh and Ravi attend the interview for the post of a marketing manager. Ajay, Ramesh and Ravi chances of being selected as the manager of a firm are in the ratio 4 : 1 : 2 respectively. The respective probabilities for them to introduce a radical change in marketing strategy are 0.3, 0.8, and 0.5. If the change does take place. (a) Find the probability that it is due to the appointment of Ajay (A). (b) Find the probability that it is due to the appointment of Ramesh (B). (c) Find the probability that it is due to the appointment of Ravi (C). (d) Find the probability that it is due to the appointment of Ramesh or Ravi. OR 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 [4] the same time. (a) What is the probability that the shell fired from exactly one of them hit the plane? (b) If it is known that the shell fired from exactly one of them hit the plane, then what is the probability that it was fired from B? (c) What is the probability that the shell was fired from A? (d) How many hypotheses are possible before the trial, with the guns operating independently? Write the conditions of these hypotheses. 11. Read the text carefully and answer the questions: [6] Two farmers Ankit and Girish cultivate only three varieties of pulses namely Urad, Massor and Mung. The sale (in ) of these varieties of pulses by both the farmers in the month of September and October are given by the following matrices A and B. September sales (in ): U rad A = 10000 50000 M asoor 20000 30000 M ung 30000 Ankit 10000 Girish October sales (in ): A= 12. U rad M asoor M ung 5000 10000 6000 20000 30000 10000 Ankit Girish (a) Find the combined sales of Masoor in September and October, for farmer Girish. (b) Find the combined sales of Urad in September and October, for farmer Ankit. (c) Find a decrease in sales from September to October. Solve the following differential equation: e x y (1 x y y ) + (1 + e x ) dx dy = 0, when x = 0, y = 1 [6] OR Solve the differential equation: 13. dy dx 3y cot x = sin 2x, given y = 2 when x = 2 . Show that the radius of a closed right circular cylinder of given surface area and maximum volume is equal to half of its height. OR Show that the rectangle of maximum perimeter which can be inscribed in a circle of radius 'a' is a square of side 2a . [6] 14. Read the text carefully and answer the questions: [6] Family photography is all about capturing groups of people that have family ties. These range from the small group, such as parents and their children. New-born photography also falls under this umbrella. Mr Ramesh, His wife Mrs Saroj, their daughter Sonu and son Ashish line up at random for a family photograph, as shown in figure. (a) Find the probability that daughter is at one end, given that father and mother are in the middle. (b) Find the probability that mother is at right end, given that son and daughter are together. (c) Find the probability that father and mother are in the middle, given that son is at right end. (d) Find the probability that father and son are standing together, given that mother and daughter are standing together. SECTION B - 15 MARKS 15. In subparts (i) and (ii) choose the correct options and in subparts (iii) to (v), answer the [5] questions as instructed. (a) ^ i + 4k is to be written as the sum of a vector parallel to a = ^ i +^ j The vector b = 3^ [1] and a vector perpendicular to a . Then = 2 c) 1 3 2 (i + j) b) 3 (i + j) d) 1 2 3 (i + j) (i + j) (b) Find the direction cosines of x, y and z-axis. [1] (c) ^ ^ ^ ^ ^ Find |a b | , if a = 2^ i + j + 3k and b = 3 i + 5 j 2k . [1] (d) The distance between the planes; 2x + 2y - z + 2 = 0 and 4x + 4y - 2z + 5 = 0 is [1] (e) 16. a) a) 1 c) 1 2 8 b) 1 d) 1 6 4 Find the distance of the point (2, 1, -1) from the plane x - 2y + 4z = 9. Find a b if | a | = 2, | b| = 5 and | a b| = 8. [1] [2] OR If a and b are perpendicular vectors, |a + b| = 13 and |a | = 5, find the value of |b| . 17. Find the shortest distance between the pairs of lines whose vector equations are: ^ ^ ^ r = ( 1) i + ( + 1) j (1 + ) k [4] ^ ^ and r = (1 ) ^ i + (2 1) j + ( + 2) k OR Reduce the equation 2x - 3y - 6z = 14 to the normal form and hence find the length of perpendicular from the origin to the plane. Also, find the direction cosines of the normal to the plane. 18. Find the area of the region lying in the first quadrant and enclosed by the x-axis, the line y = x and the circle x2 + y2 = 32. [4] SECTION C - 15 MARKS 19. In subparts (i) and (ii) choose the correct options and in subparts (iii) to (v), answer the [5] questions as instructed. (a) Law of variable proportion explain three stages of production. In the first stage of [1] production: (b) a) AP falls b) MP is zero c) MP rises d) Both MP and AP rise The corner points of the feasible region for a Linear Programming problem are P(0, 5), [1] Q(1, 5), R(4, 2) and S(12, 0). The minimum value of the objective function Z = 2x + 5y is at the point. (c) a) R b) Q c) S d) P = 18, y = 100, x = 14, y = 20 and correlation coefficient rxy = 0.8, find the If x [1] regression equation of y on x. (d) The demand function is x = 24 2p 3 , where x is the number of units demanded and p is [1] the 3 price per unit. Find the revenue function R in terms of p. (e) If P = 100 q+2 - 4 represents the demand function for a product where p is the price per [1] unit when q units are sold, find the marginal revenue. 20. Given the total cost function for x units of a commodity as C(x) = 1 3 x 3 [2] + x2 - 8x + 5, find: i. the marginal cost function ii. average cost function iii. slope of average cost function OR A company produced a commodity with 20,000 fixed costs. The variable cost is estimated to 35% of the total revenue. When it is sold at a rate of 6 per unit. Find the total revenue, total cost and profit function of the revenue when it is sold. 21. Find the line of regression of y on x from the following table: [4] x 1 2 3 4 5 y 7 6 5 4 3 Hence, estimate the value of y when x = 6. 22. A company manufactures two types of novelty souvenirs made of plywood. Souvenirs of type A require 5 minutes each for cutting and 10 minutes each for assembling. Souvenirs of type B require 8 minutes each for cutting and 8 minutes each for assembling. There are 3 hours and 20 minutes available for cutting and 4 hours available for assembling. The profit is 50 each for type A and 60 each for type B souvenirs. How many souvenirs of each type should the company manufacture in order to maximize profit? Formulate the above LPP and solve it graphically and also find the maximum profit. OR Maximise Z = 3x + 4y, subject to the constraints: x + y 1, x 0, y 0 . [4] 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. (a) 4 3 (c) 3 0 1 2 Explanation: Given, A 3 T = 1 B 1 2 2 B T (b) (a) 2 4 = 1 log tan( Explanation: x 3 1 I = 1 1 dx 1 2 cos x 2 I = I = I = (c) 1 6 ln tan( x 2 1 1 2 3 ] 1 3 1 2 1 2 2 1 1 4 2 = 3 3 1 3 0 2 dx sin x dx ) sec(x 2 2 2 cos(x 1 1 3 1 2 + )+C cos x+ 3 sin x I = 2 2 0 + 2 = 1 = [ 1 3 1 3 A T ] 1 and B 1 = [ T 2 0 1 T 4 6 + )dx 3 ) +c (a) 3 Explanation: cos (d) 1 (cos( 3 )) =cos 1 (cos 3 ) = 3 , because, cos is positive in fourth quadrant. (a) log|y| = 2(x log|1 + x|) + c Explanation: (x + 1) 1 y dy dx = 2xy dy = 2 1 y x 1+x dx dy = 2 (1 1 1+x )dx log|y| = 2(x log|1 + x|) + c (e) (d) 4 3 Explanation: P (problem will be solved) = 1 - P (problem will not solved by A, B and C) = 1 - {(1 12 ) (1 13 ) (1 14 )} =1(f) 1 2 2 3 3 4 = 1 - 14 (c) not symmetric Explanation: = 3 4 Here, R is not reflexive as x is not 7 cm taller than x. R is not symmetric as if x is exactly 7 cm taller than y, then y cannot be 7 cm taller than x and R is not transitive as if x is exactly 7 cm taller than y and y is exactly 7 cm taller than z, then x is exactly 14 cm taller than z. (g) (c) 1 - y2 Explanation: Solution. x dy x ( e +e = d = dx x dx x )( e +e ( x x x x e e ) e +e x x x ) ( e e x x ( e +e x )( e e x ) = 2 ) x ( e +e x x ( e +e 2 ) 2 ) x x x x ( e e ( e +e 2 ) 2 = 1 y ) Which is the required solution. (h) (d) 25 y Explanation: We have, y = ae5x + be 5x On differentiating w.r.t x, we get d 2 = 5ae5x 5be 5x y 2 dx d 2 = 25(ae5x+be 5x) y 2 dx d 2 y = 25 y 2 dx Hence, this is the answer. (i) (b) (A + B) 1 = B 1 + A 1 Explanation: Given A and B are invertible matrices. Now, (A B)B 1 A 1 = A(BB-1) A-1 = AIA-1 = (AI)A-1 = AA-1 = I 1 1 1 (AB ) = B A A A-1 = I |A A-1| = |I| |A||A-1| = 1 1 1 A = |A| 1 det(A ) 1 = [det(A) ] Also we know that A adj. A = |A| A 1 (A + B ) But B 1 = +A 1 |A+B| 1 1 (A + B ) (j) 1 = 1 |A| adj. (A + B) 1 |B| B = adj A adj. B + 1 +A 1 |A| adj. A 1 (a) Both A and R are true and R is the correct explanation of A. Explanation: Explanation: Here, A = A' 2 x 3 3 2 4 1 x 2 2 1 = x 3 5 On comparing, we get x - 3 = 3 or x = 6 x 2 3 2 1 4 1 5 2 . (k) We observe that f(x) = 0 for any x ( 1 f 1 ) (x) = f (x) 1 = 1/(x+4) = R { 4} Therefore, 1 f : R { 4} R (x + 4) (l) We have to find X, Given 2A B + X = 0 2 ([ 3 1 0 2 ]) [ 2 2 1 0 3 1 X = [ +X=0 ] 3 1 0 2 ] 2 ([ 0 2 3 1 = [ 6 2 0 4 ] [ 0 3 8 1 0 1 = [ ]) ] ] (m)f : Z Z : f(x) = x3 is one - one into f(x) = x3 Since the function f(x) is monotonically increasing from the domain Z Z f(x) is one one Range of f(x) = ( - , ) Z(codomain) f(x) is into f : Z Z : f(x) = x3 is one - one into. (n) We have, P(B/A) = = = 4 13 . 4 7 P(A B) P(A) 7 13 (o) Two event A and B are independent if P (A B) = P (A). P (B) Sample space of the experiment is, S = {1, 2, 3, 4, 5, 6} Now E = {3, 6}, F = { 2, 4, 6} and E F = {6} Then P(E) = 26 = 13 , P(F) = 36 = 12 and P(E F) = 16 Clearly P(E F) = P(E). P(F) = 1 6 Hence E and F are independent events. 2. We have, f (x) = 2x2 1 and y = f(x2) Differentiating 'y' w.r.t. 'x' dy dx dy dx = d dx = f 2 f (x ) 2 (x ) d dx 2 (x ) is given by dy dx dy dx = f = f 2 ( x ) 2x 2 ( x ) 2x Putting x = 1, we get, dy dx = 2(1) f dy dx dy dx dy dx 2 (1 ) = 2 f (1) 2 = 2 1 [ f (1) = 2(1) 1 = 2 1 = 1] = 2 Given: f(x) d P (x) = = x (8x dx 8 7 + 6x 2 OR + 12x) + 12x To find the critical point of f(x), we must have f (x) = 0 6 4x (2x + 3) = 0 7 f (x) = 8x 6 x (2x + 3) = 0 3 6 x = 0, 2 Since 3 , is a complex number, therefore we only check for x = 0, on both of its sides. 6 2 Clearly, f (x) > 0 if x > 0 and f (x) < 0 if x < 0 Thus, f(x) increases on (0, ) and f(x) is decreasing on interval x ( , 0) x 3. Let I = 2 = 0 0 2 0 log log 2 = 0 log 2 = 0 cos 0 2 x log sin 2x) dx x dx sin x cos 2 x 2 sin x cos x cos x 2 sin x dx dx (log cos x log sin x log 2)dx 2 = 2 (log cos = , then we have (2 log cos x log sin 2x)dx 2 = 2 0 x 2 log cos xdx 0 x log sin xdx 0 2 log 2 We know that 0 log cos x dx = 0 log sin x dx .....(i) Hence from equation (i), we have 2 2 I = 0 2 log 2 = 2 log 2 4. Given: f(x) = loga x , 0 < a < 1 Theorem: Let f be a differentiable real function defined on an open interval (a,b), then i. If f (x) > 0 for all x (a, b) , then f(x) is increasing on (a, b) ii. If f (x) < 0 for all x (a, b) , then f(x) is decreasing on (a, b) For the value of x obtained in (i) f(x) is increasing and for remaining points in its domain it is decreasing. Here we have, f(x) = loga x, 0 < a < 1 f (x) = f(x) = d dx ( loga x) 1 x log a As given 0 < a < 1 log(a) < 0 and for x > 0 = 1 x log < 0 a Therefore f (x) is 1 = < 0 x log a f (x) < 0 Hence, it is the condition for f(x) to be decreasing Thus f(x) is decreasing for all x > 0. 1 5. Let I = (x 1)(x+1)(x+2) Using partial fractions, 1 = A + x 1 (x 1)(x+1)(x+2) B + x+1 C x+2 1 = A(x + 1) (x + 2) + B(x - 1) (x + 2) + C (x2 - 1) Put x = 1 1 1 = 6A A = 6 Put x = -1 1 1 = -2B B = - 2 Put x = -2 1 = 3C C = So, 1 I= 6 I= dx 1 x 1 Let I = 0 /4 2 = x 1 2 3 dx x+1 log |x 1| 6 x 1 x 2 2 2 1 dx 3 x+2 log |x +1 + sin xdx sin xdx = x 2 1 + 1 3 log x + 2| + c OR , where sin xdx 2x ( sin xdx) dx = x cos x + 2 [x cos xdx ( cos xdx) dx] = x 2 cos x + 2 [x sin x sin xdx] x 0 1 = [ 1 2 [ 0 4 x 2 sin xdx = [ x 2 + 2 + 16 = 2 + 2 16 2 = x 4 cos x + 2x cos xdx 2 2 2 2 2 + 2] 2 + 2] 2 2 16 2 2 cos x + 2x sin x + 2 cos x] 4 0 sin xdx = 2 + 6. Let R = {( 1 , 2 ) Now, For reflexivity: 2 2 2 : 1 2 } 2 2 16 2 be a relation defined on A. R is Reflexive if ( , ) R A We observe that for each A we have, since, every triangle is similar to itself. = ( , ) R A R is reflexive. For symmetry: R is Symmetric if ( 1 , 2 ) Let 1 2 = 1 2 = 2 1 ( 2 , 1 ) R R is symmetric For Transitivity: R ( 2 , 1 ) R, 1 , 2 A R is Transitive if ( 1 , 2 ) R and (( ( 1 , 2 ) R 2 and 1 3 1 2 1 2 and (Q 2 , 3 ) R ( 1 , 3 ) R A1 , 2 , A3 A , 3 ) R 1 , 2 , 3 A 3 3 = R is transitive. Since R is reflexive, symmetric and transitive, it is an equivalence relation on A 7. tan 1 ( RHS = cos x x 4 2 ,x ( 2 , 2 ) x 4 ) = 1+sin x 2 LHS = tan 1 ( 1+sin x ) cos x 1 = tan cos ( cos x 2 2 [ cos x = cos 2 x 2 [ 1 = sin 2 x 2 x x 2 [ 2 2 2 sin 1 = tan ( cos x 2 x 2 x sin +sin 2 cos 2 x ) 2 ] 2 x ] 2 x )(cos x x ] 2 2 (cos cos 2 x x 2 +2 sin sin cos x 2 sin x + cos 2 (cos = tan 2 +sin [ sinx = 2 sin 1 x 2 +sin 2 x 2 +sin x 2 ) ] 2 ) x 2 x ) 2 Dividing the numerator and Denominator by cos 2 , cos LHS = tan 1 x 2 cos x sin cos 2 cos cos x sin 1 1 = tan 4 1 tan ( 1+tan [tan( 4 x 2 + cos 2 = tan x 2 x 2 x 2 x 2 x 2 x ) 2 x 2 )] x 2 1 [ tan (tan ) = ; ( = RHS Hence proved 8. Let the given integral be, I= 2 , 2 )] dx x(x 2)(x 4) Now using partial fractions we have, which implies, A(x - 2)(x - 4) + Bx(x - 4) + Cx(x - 2) = 1 Now put x - 2 = 0 Therefore, x = 2 A(0) + B 2(2 - 4) + C(0) = 1 B 2(-2) = 1 1 B = 4 Now put x - 4 = 0 Therefore, x = 4 A(0) + B (0) + C 4(4 - 2) = 1 C 4(2) = 1 1 C = 8 Now put x = 0 1 x(x 2)(x 4) = A x + B x 2 + C x 4 ... (i) A(0 - 2)(0 - 4) + B(0) + C(0)=1 1 A = 8 Now From equation (1) we get 1 x(x 2)(x 4) dx 1 8 8 1 = x(x 2)(x 4) = 1 = 1 log |x| 1 1 8 Put, x = tan 2 1+x 1 = tan 1 [ sec 1 = tan 1 [ 1 cos tan sin ] ] 2 2 sin = tan 1 [tan y = dy = dx 2 = 2 2 1 2 4 8 1 x 2 1 x 4 dx + 1 8 1 x 4 dx log |x 4| + c 1 2 cos ] 1 8 = tan-1x tan 2 sin 1 1 + x 2 = tan 1 [ x 2 dx x 1+tan 1 1 4 log |x 2| + 4 9. Given, y = tan 1 y = tan 1 x ] 2 1 tan x 1 2 2(1+x ) OR We have, y = sinx sin 2x sin 3x sin 4x ...(i) Taking log on both sides log y = log(sin x sin 2x sin 3x sin 4x) log y = log sin x + log sin 2x + log sin 3x + log sin 4x Differentiating with respect to x using chain rule, 1 dy y dx = 1 dy y dx 1 dy y dx 1 dy y dx dy dx dy dx d dx = = (log sin x) + 1 d sin x dx 1 sin x d dx (log sin 2x) + (sin x) + (cos x) + 1 d sin 2x dx 1 sin 2x d dx (log sin 3x) + (sin 2x) + (cos 2x) d dx 1 d sin 3x dx (2x) + 1 sin 3x d dx (log sin 4x) (sin 3x) + (cos 3x) d dx 1 d sin 4x dx (3x) + (sin 4x) 1 sin 4x (cos 4x) d dx (4x) = [cot x + cot 2x(2) + cot 3x (3) + cot 4x (4)] = y[cot x + 2 cot 2x + 3 cot 3x + 4 cot 4x] = (sin x sin 2x sin 3x sin 4x) [cot x + 2 cot 2x + 3 cot 3x + 4 cot 4x] [Using equation (i)] The differentiation of the given function y is as above. 10. Read the text carefully and answer the questions: To hire a marketing manager, it's important to find a way to properly assess candidates who can bring radical changes and has leadership experience. Ajay, Ramesh and Ravi attend the interview for the post of a marketing manager. Ajay, Ramesh and Ravi chances of being selected as the manager of a firm are in the ratio 4 : 1 : 2 respectively. The respective probabilities for them to introduce a radical change in marketing strategy are 0.3, 0.8, and 0.5. If the change does take place. (a) Let E1: Ajay (A) is selected, E2: Ramesh (B) is selected, E3: Ravi (C) is selected Let A be the event of making a change 4 P ( E1 ) = 7 , P ( E2 ) = 1 7 P(A/E1) = 0.3, P(A/E2) = 0.8, P(A/E3) = 0.5 7 7 = 0.3+ 1.2 3 = P ( E1 ) P (A/E1 )+P ( E2 )P (A/E2 )+P ( E3 )P (A/E3 ) 4 4 7 P ( E1 ) P (A/E1 ) P(E1/A) = = 2 , P ( E3 ) = 1 7 1.2 0.3 0.8+ 12 30 = 2 7 = 0.5 1.2 7 1.2 7 + 0.8 7 + 1 7 = 3 7 7 2 5 (b) Let E1: Ajay(A) is selected, E2: Ramesh(B) is selected, E3: Ravi (C) is selected Let A be the event of making a change 4 P ( E1 ) = 7 , P ( E2 ) = 1 7 P(A/E1) = 0.3, P(A/E2) = 0.8, P(A/E3) = 0.5 7 7 = 0.3+ 0.8 3 = P ( E1 ) P (A/E1 )+P ( E2 )P (A/E2 )+P ( E3 )P (A/E3 ) 1 4 7 P ( E1 ) P (A/E1 ) P(E2/A) = = 2 , P ( E3 ) = 1 7 0.8 0.8 0.8+ 8 30 = 2 7 = 0.5 0.8 7 1.2 7 + 0.8 7 + 1 7 = 3 7 7 4 15 (c) Let E1: Ajay (A) is selected, E2: Ramesh (B) is selected, E3: Ravi (C) is selected Let A be the event of making a change P ( E1 ) = 4 7 , P ( E2 ) = 1 7 2 , P ( E3 ) = P(A/E1) = 0.3, P(A/E2) = 0.8, P(A/E3) = 0.5 P(E3/A) = P ( E1 ) P (A/E1 ) 7 4 7 0.3+ P ( E1 ) P (A/E1 )+P ( E2 )P (A/E2 )+P ( E3 )P (A/E3 ) 2 = 7 1 7 1 0.5 0.8+ 2 7 = 0.5 7 1.2 7 + 0.8 7 + 1 = 1 3 7 (d) Let E1: Ajay (A) is selected, E2: Ramesh (B) is selected, E3: Ravi (C) is selected Let A be the event of making a change P ( E1 ) = 4 7 , P ( E2 ) = 1 7 2 , P ( E3 ) = P(A/E1) = 0.3, P(A/E2) = 0.8, P(A/E3) = 0.5 Ramesh or Ravi P(E2/A) + P(E3/A) = 4 15 + 1 3 = 9 15 = 7 3 5 OR 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) 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 = P Q 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 E P(E) = P(E1) P ( E E (b) 1 ) = 0 2 ,P (E E ) = 1 3 ) + P ( E2 ) P ( E E2 ,P (E E ) = 1 4 ) +P ( E3 ) P ( E E3 ) + P ( E4 ) P ( E E4 ) = 0.14 + 0.24 + = 0.38 By Bayes Theorem, P ( = 0.14 0.38 = 7 19 E3 E P ( E3 ) P ( = ) P ( E1 ) P ( E E 1 )+P ( E2 ) P ( E E 2 E E 3 ) )+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 ( (c) E E1 ) = P ( E E2 ) = 0 By Bayes Theorem, P ( = 0.24 0.38 = E4 E P ( E4 ) P ( = ) P ( E1 ) P ( E E 1 )+P ( E2 ) P ( E E 2 E E 4 ) )+P ( E3 ) P ( E E 3 )+P ( E4 ) P ( E E 4 ) 12 19 (d) 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 = P Q 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 11. Read the text carefully and answer the questions: Two farmers Ankit and Girish cultivate only three varieties of pulses namely Urad, Massor and Mung. The sale (in ) of these varieties of pulses by both the farmers in the month of September and October are given by the following matrices A and B. September sales (in ): U rad A = 10000 50000 M asoor M ung 20000 30000 Ankit 30000 10000 Girish October sales (in ): A= (a) U rad M asoor M ung 5000 10000 6000 20000 30000 10000 A+B=[ 10000 20000 Ankit Girish 30000 ] 50000 30000 10000 15000 30000 36000 70000 40000 20000 = [ +[ 5000 10000 6000 20000 10000 10000 ] ] The combined sales of Masoor in September and October, for farmer Girish 40000. (b) 10000 A+B=[ 20000 30000 ] 50000 30000 10000 15000 30000 36000 70000 40000 20000 = [ 10000 6000 20000 10000 10000 ] A-B=[ =[ 10, 000 20, 000 30, 000 -[ 10, 000 5000 10, 000 6000 20, 000 10, 000 10, 000 ] 10, 000 5000 20, 000 10, 000 30, 000 6000 50, 000 20, 000 30, 000 10, 000 10, 000 10, 000 ] 5000 10, 000 30, 000 y 20, 000 y x (1 ) + (1 + e y Substitute x = vy v dy ) dv + v v+e Girish = 0 dx dy dv dy (v + ev) + (1 + ev)y ) x 0 Ankit = v + y dy dx ev - vev + v + vev +(1 + ev)y 1+e 24, 000 ] x 12. Given, e 30, 000 ] A-B=[ dv dy =0 = 0 dv dy = 0 y Integrating v 1+e ( v v+e ) dv = - dy y log|v + ev|= -log|y| + log c log|v + ev| + log|y| = log c log|{(v + ev)y}| = log c log {( x y x y +e ) y} = log c x = c When x = 0 and y = 1 c = 1 y x +e y x y x +e y = 1 OR Given, dy 3y cot x dx = sin 2 x This is a linear equation of the form where, p = -3cot x, Q = sin 2x I.F. = e I.F. = e = e = (sin x ) + Py = Q = e 3 cot xdx 3 cot xdx 3 log | sin x| dy dx dx 3 3 = e log | sin x| 1 = 3 sin x Solution is given as, y ( I.F. ) = Q ( I.F. )dx y 1 3 sin 3 y sin 3 sin = 3 sin 2 2 sin x cos x 3 x x sin dx x dx x = -2 cosec x + c Putting y = 2, x = 2 3 sin 2 = 2 cosec( 2 ) +c [ sin 2x = 2 sinx cosx] = 2 cosec x cot x dx 3 sin x y 1 = sin 2x x 1 y ] 50, 000 ( 5000 The combined sales of Urad in September and October, for farmer Ankit is 15000. (c) +[ 2 = -2 + c c = 4 y 2 = 3 sin +4 sin x x y = -2 sin2x + 4 sin3x 13. Given: Total surface area of cylinder 2 = 2 r + 2 rh 2 A = 2 r + 2 rh 2 ...(i) A 2 r h = 2 r [Where r is the radius and h is the height of the cylinder] Volume of cylinder (V) = r2h 2 (A 2 r ) 2 V = r V = 2 = dr A d Ar dr 2 2 3 r 2 3 r dV = dr (A 2 r ) 2 Now, dV 2 Ar V = [from (i)] 2 r r r d dr 3 ( r ) For maximum/minimum; dV = 0 dr A 2 A 2 3 r 2 = 3 r 2 = 0 A = 6 r2 So, 2 r2 + 2 rh 2 2 rh = 4 r h = 2r 2 Again, And, ...(ii) h r = 2 = 6 r dV dr d 2 V w 2 = A 2 3 r 2 = 0 6 r for any value of r i.e., r > 0 d 2 V 2 dr < 0 Hence, volume is maximum when r = Hence Proved h 2 OR Let ABCD be a rectangle in a given circle of radius 'a' with centre at O. Let AB = 2x and AD = 2y be the sides of the rectangle. Applying Pythagoras theorem in OAM, we get AM2 + OM2 = OA2 x2 + y2 = a2 y = a2 2 x Let P be the perimeter of the rectangle ABCD. Then,we have, P = 4x + 4y 2 x P = 4x + 4 a2 dP dx =4 4x 2 2 a x [Using (i)] [differentiating both sides w.r.t x] The critical points of P are given by dx = 0. dP dx 4- 4x a2 x2 4= = 0 2 2 a x 4x a2 x2 =x a2 - x2 = x2 2x2 = a2 a x= 2 Now, dP dx =4- 4x 2 2 a x 2 x( x) 2 4{ a x = 0 dP d 2 = 2 dx ( d 2 } a2 x2 P 2 2 { a x } P 2 dx ) x=a/ 2 2 2 (a 2 a 2 2 a 2 = 3/2 3/2 8 2 a < 0 ) Thus, P is maximum when x = Putting x = 2 2 ( a x ) 4a = 4a = 2 a 2 . in (i), we obtain y = a 2 a 2 . x=y= 2x = 2y AB = BC ABCD is a square. Hence, P is maximum when the rectangle is square of side 2x = 2a 2 = 2a . 14. Read the text carefully and answer the questions: Family photography is all about capturing groups of people that have family ties. These range from the small group, such as parents and their children. New-born photography also falls under this umbrella. Mr Ramesh, His wife Mrs Saroj, their daughter Sonu and son Ashish line up at random for a family photograph, as shown in figure. (a) Sample space is given by {MFSD, MFDS, MSFD, MSDF, MDFS, MDSF, FMSD, FMDS, FSMD, FSDM, FDMS, FDSM, SFMD, SFDM, SMFD, SMDF, SDMF, SDFM DFMS, DFSM, DMSF, DMFS, DSMF, DSFM}, where F, M, D and S represent father, mother, daughter and son respectively. n(S) = 24 Let A denotes the event that daughter is at one end n(A) = 12 and B denotes the event that father, and mother are in the middle n(B) = 4 Also, n(A B) = 4 4 P (A/B) = P (A B) P (B) = 24 4 24 = 1 (b) Sample space is given by {MFSD, MFDS, MSFD, MSDF, MDFS, MDSF, FMSD, FMDS, FSMD, FSDM, FDMS, FDSM, SFMD, SFDM, SMFD, SMDF, SDMF, SDFM DFMS, DFSM, DMSF, DMFS, DSMF, DSFM}, where F, M, D and S represent father, mother, daughter and son respectively. n(S) = 24 Let A denotes the event that mother is at right end. n(A) = 6 and B denotes the event that son and daughter are together. n(B) = 12 Also, n (A B) = 4 4 P (A/B) = P (A B) P (B) 24 = 12 = 1 3 24 (c) Sample space is given by {MFSD, MFDS, MSFD, MSDF, MDFS, MDSF, FMSD, FMDS, FSMD, FSDM, FDMS, FDSM, SFMD, SFDM, SMFD, SMDF, SDMF, SDFM DFMS, DFSM, DMSF, DMFS, DSMF, DSFM}, where F, M, D and S represent father, mother, daughter and son respectively. n(S) = 24 Let A denotes the event that father, and mother are in the middle. n(A) = 4 and B denote the event that son is at right end. n(B) = 6 Also, n (A B) = 2 2 P (A/B) = P (A B) P (B) 24 = 6 = 1 3 24 (d) Sample space is given by {MFSD, MFDS, MSFD, MSDF, MDFS, MDSF, FMSD, FMDS, FSMD, FSDM, FDMS, FDSM, SFMD, SFDM, SMFD, SMDF, SDMF, SDFM DFMS, DFSM, DMSF, DMFS, DSMF, DSFM}, where F, M, D and S represent father, mother, daughter and son respectively. n(S) = 24 Let A denotes the event that father and son are standing together. n(A) = 12 and B denote the event that mother and daughter are standing together. n(B) = 12 Also, n (A B) = 8 8 P (A/B) = P (A B) P (B) 24 = 12 = 2 3 24 SECTION B - 15 MARKS 15. In subparts (i) and (ii) choose the correct options and in subparts (iii) to (v), answer the questions as instructed. (a) (b) 3 2 (i + j) Explanation: ^ ^ ^ ^ ^ Let a = a1 ^ i + 2 j + a3 k , = 1 i + 2 j + 3 k ^ ^ b = 3 i + 4k + = 3 i + 4k ^ ^ ( 1 + 1 ) i + ( 2 + 2 ) j + ( 3 + 3 ) k = 3 + 4k 1 + 1 = 3 2 + 2 = 0 a3 + 3 = 4 Given that a is parllel to a = 0 i 1 ^ k ^ j 1 a 2 3 = 0 1 0 {Given a = i + j } ^ ^ ^ 3 i + 3 j + ( 1 2 ) k = 0 3 = 0, 1 2 = 0 3 = 0, 1 = 2 Given is perpendicular to a = 0 a ~ ~ ~ ~ ~ ( 1 i + 2 j + 3 k) ( i + j ) = 0 1 + 2 = 0 1 = 2 Solving 3 = 0, 1 = 2 , 1 + 1 = 3 2 + 2 = 0, 3 + 3 = 4, 1 = 2 3 1 = 2 = 2 , 3 = 0 ^ = 1 ^ i + 2 ^ j + 3 k 3 = 2 (i + j ) (b) (1, 0, 0), (0, 1, 0), (0, 0, 1) respectively. ^ ^ ^ (c) i j k a b = 2 3 1 3 2 5 ^ = ^ i ( 2 15) ^ j ( 4 9) + k (10 3) ^ ^ ^ = 17 i + 13 j + 7k (d) (b) 1 6 Explanation: Multiplying the first equation of the plane by 2 ; 4x + 4y - 2z + 4 = 0 4x + 4y - 2z = -4 ...(i) The second equation of the plane is 4x + 4y - 2z + 5 = 0 4x + 4y - 2z = -5 ...(ii) We know that the distance between two planes ax + by + cz = d1 and ax + by + cz = d2 is |d2 d1 | 2 a2 +b +c2 So, the required distance | 5+4| = 2 2 2 4 +4 +( 2) = | 1| 16+16+4 1 = 36 = 1 6 units (e) The required distance = the length of the perpendicular from P(2, 1, -1) to the plane x - 2y + 4z - 9 = 0 |2 2 1+4 ( 4) 9| = 2 2 2 | 1 +( 2) +4 | 16. |a b| 8 2 5 13 21 units. sin = 4 5 1 sin2 cos = 1 = |a||b| sin = = 16 = 25 3 5 a.b = |a ||b| cos =2x5x a b 3 5 =6 OR a b = 0 & |a | = 5 Now, |a + b| = 13 2 | a + b| 2 | a | = 169 2 + | b| + 2( a b) 2 25 + |b| | b| = 12 + 2 0 = 169 = 169 17. We are given that ^ ^ and r = (1 ) ^ i + (2 1) j + ( + 2) k The vector equations of the given lines can be re-written as ^ ^ ^ r = ( 1) i + ( + 1) j (1 + ) k ^ ^ and r = ^ i ^ j + 2k + ( ^ i + 2^ j + k) Comparing the given equations with the equations ^ ^ r = ^ i +^ j k + ( ^ i +^ j k) r = a1 + b1 a2 + b2 and r = We get, ^ a1 = ^ i +^ j k ^ a2 = ^ i ^ j + 2k ^ ^ ^ b1 = i + j k ^ ^ ^ b2 = i + 2 j + k ^ ^ ^ a2 a1 = 2 i 2 j + 3k and | b1 = b2 | ^ i ^ j 1 1 1 = | b1 b2 | 9 + 9 3 2 = = ^ k 1 2 1 ^ ^ 3 i + 3k 2 2 +3 = 3 ( a2 a1 ) ( b1 b2 ) ^ ^ ^ ^ = (2^ i 2 j + 3k) (3 i + 3k) = 6 + 9 = 15 Hence the shortest distance between the lines, r = a1 + b1 d=| and r = ( a2 a1 ) ( b1 b2 ) | a2 + b2 is given by | b1 b2 | = = 15 3 2 5 2 OR Given the equation of the plane is, 2x - 3y - 6z = 14 = 14 ^ ^ ^ ^ ^ ^ (x i + y j + zk) (2 i 3 j 6k) (2)2 + ( 3)2 + ( 6)2 Dividing the equation by r ^ ^ ^ (2 i 3 j 6k) 4+9+36 r ( r ( 2 7 2 7 ^ i 3 ^ i 3 7 7 = 14 4+9+36 ^ j 6 ^ j 6 7 7 ^ k) = ^ k) 14 7 = 2 ...(i) ^ We know that the vector equation of a plane with distance d from the origin and normal to unit vector n is given by ^ = d ...(ii) r n Comparing (i) and (ii), d = 2 and ^ = n 2 7 ^ i 3 7 ^ j 6 7 ^ k So, the distance of the plane from origin = 2 unit 2 3 6 Direction cosine of normal to plane = , , . 7 7 7 18. The given equations are y = x ... (1) and x2 + y2 = 32 ... (2) Solving (1) and (2), we find that the line and the circle meet at B(4, 4) in the first quadrant. Draw perpendicular BM to the x-axis. Therefore, the required area = area of the region OBMO + area of the region BMAB. Now, the area of the region OBMO = 4 ydx = 0 = 1 2 4 2 4 [x ] 0 ...(3) xdx 0 = 8 Again, the area of the region BMAB = 4 2 4 = [ = ( 1 2 1 2 ydx = 4 2 4 32 x2 dx 4 2 2 x 32 x + 4 2 0 + 1 1 2 1 32 sin 1 32 sin 2 x 4 2 1) ( ] 4 4 2 32 16 + 1 2 1 32 sin 1 2 ) ...(4) Adding (3) and (4), we get, the required area = 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 instructed. = 8 (8 + 4 ) = 4 8 (a) (d) Both MP and AP rise Explanation: Both MP & AP rise (b) (a) R Explanation: Corner points Value of Z = 2x + 5y P(0, 5) Z = 2(0) + 5(5) = 25 Q(1, 5) Z = 2(1) + 5(5) = 27 R(4, 2) Z = 2(4) + 5(2) = 18 Minimum S(12, 0) Z = 2(12) + 5(0) = 24 Thus, minimum value of Z occurs ar R(4, 2) (c) Given x = 18, y = 100, x = 14, y = 20, r = 0.8 bxy = r y x = 0.8 20 14 = Regression equation y on x ) y - y = byx (x x y - 100 = 8 7 8 7 (x - 18) 7y - 700 = 8x - 144 8x - 7y + 556 = 0 (d) Given x = 24 2p 3 and price per unit = p. So, R = p x = p (8 2 3 p) = 8p 2 3 2 p (e) Given P = So, R = p q = d(R) MR = 100q q+2 - 4q (q+2) 100 100q 1 = dq 2 - 4 (q+2) MR = - 4 100 q+2 -4 200 2 (q+2) 20. Given C(x) = i. MC = 1 3 x + x2 - 8x + 5 3 C(x) = d dx ( 1 3 x 3 + x2 - 8x + 5) MC = x2 + 2x - 8 1 C(x) ii. AC = d dx = x AC = 1 x 3 3 2 x +x 8x+5 3 x + x - 8x + 2 5 x iii. Slope of average cost function = d dx (AC) = 2 3 x +1- 5 2 x OR Let x be the number of units produced and sold. Revenue function = R(x) = 6x Given that variable cost is estimated to 35% of R(x) V(x) = 35% of R(x) 35 = 6x 100 21x V (x) = 10 Hence, cost function = C(x) = Fixed cost + V(x) = 20000 + [Given, fixed cost = 20000] 21x 10 Now, profit function P(x) = R(x) - C(x) 21x = 6x - 20000 - 10 39x P (x) = 21. x y xy x2 1 7 7 1 2 6 12 4 3 5 15 9 4 4 16 16 5 3 15 25 = 15 x x = x = n 5 = n y = 3 15 y and y = 20000 10 25 5 =5 x y xy n byx = 2 ( x) x = 65 = 15.25 5 (15) 55 2 n 2 5 65 75 55 45 = 10 10 = -1 Regression line y on x is given by ) (y - y ) = byx (x - x y - 5 = -1(x - 3) y - 5 = -x + 3 x + y = 8 ...(i) When x = 6, From (i), we get = 25 xy = 65 x2 = 55 y = 8 - 6 y = 2 22. Let number of Souvenirs of type A be x, and that of type B be y. L.P.P is maximise P = 50x + 60y such that 5x + 8y 200 10x + 8y 240 x, y 0 P(at A) = 1500 P(at B) = (400 + 1200) = 1600 P(at C) = (1200) Max Profit = 1600, when number of Souvenirs of type A = 8 and number of Souvenirs of type B = 20. OR Maximise Z = 3x + 4y. Subject to the constraints x + y 1, x 0, y 0 The Shaded region shown in the figure as OAB is bounded and the coordinates of corner points O, A and B are (0, 0), (1, 0) and (0, 1), respectively. Corner Points Corresponding value of Z (0, 0) 0 (1, 0) 3 (0, 1) 4 (Maximum) Hence, the maximum value of Z is 4 at (0, 1).

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