MODEL PAPER - I MATHEMATICS. Time allowed : 3 hours Maximum marks : 100
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1 MODEL PAPER - I MATHEMATICS Time allowed : 3 hours Maimum marks : General Instructions. All questions are compulsy.. The question paper consists of 9 questions divided into three sections A, B and C. Section A comprises of questions of one mark each, Section B comprises of questions of four marks each and Section C comprises of 7 questions of si marks each. 3. All questions in Section A are to be answered in one wd, one sentence as per the eact requirement of the question. 4. There is no overall choice. However, internal choice has been provided in 4 questions of four marks each and questions of si marks each. You have to attempt only one of the alternatives in all such questions. 5. Use of calculats is not permitted. SECTION A Question number to carry one mark each.. Find the value of, if 5 y y 4 y Let * be a binary operation on N given by a * b = HCF (a, b), a, b N. Write the value of 6 * Evaluate : 36 XII Maths
2 4. Evaluate : sec log 5. Write the principal value of cos 7 cos Write the value of the determinant : a b b c c a b c c a a b c a a b b c 7. F ind the value of from the following : 4 8. Find the value of i j k j k i k i j 9. Write the direction cosines of the line equally inclined to the three codinate aes.. If p is a unit vect and p. p 8, then find. SECTION B Question numbers to carry 4 marks each.. The length of a rectangle is decreasing at the rate of 5 cm/minute and the width y is increasing at the rate of 4 cm/minute. When = 8 cm and y = 6 cm, find the rate of change of (a) the perimeter, (b) the area of the rectangle. OR Find the intervals in which the function f given by f() = sin + cos, is strictly increasing strictly decreasing.. If (cos ) y = (sin y), find. dy OR XII Maths 37
3 4 4 4 f : R R defined as f Consider Show that f is invertible. Hence find f. 4. Evaluate : 5 4. OR Evaluate : sin. 5. If y = sin, show that d y dy 3 y. 6. In a multiple choice eamination with three possible answers (out of which only one is crect) f each of the five questions, what is the probability that a candidate would get four me crect answers just by guessing? 7. Using properties of determinants, prove the following : a b c c b c a b c a a b b c c a b a a b c 8. Solve the following differential equation : dy y y tan. 9. Solve the following differential equation : dy cos y tan.. Find the shtest distance between the lines r i j k r i j k i j k. 38 XII Maths
4 . Prove the following : sin sin cot,,. sin sin 4 OR Solve f : tan (cos ) = tan ( cosec ). The scalar product of the vect i j k with a unit vect along the sum of vects i 4 j 5k and i j 3k is equal to one. Find the value of. OR, and b c and c a are also coplanar. a b c are three coplanar vects. Show that a b, SECTION C Question number 3 to 9 carry 6 marks each. 3. Find the equation of the plane determined by the points A (3,, ), B (5,, 4) and C (,, 6). Also find the distance of the point P(6, 5, 9) from the plane. 4. Find the area of the region included between the parabola y = and the line + y =. 5. Evaluate :. a cos b sin 6. Using matrices, solve the following system of equation : + y + z = 6 + z = y + z = OR XII Maths 39
5 Obtain the inverse of the following matri using elementary operations: A Coloured balls are distributed in three bags as shown in the following table : Colour of the Ball Bag Red White Black I 3 II 4 III A bag is selected at random and then two balls are randomly drawn from the selected bag. They happen to be black and red. What is the probability that they came from bag I? 8. A dealer wishes to purchase a number of fans and sewing machines. He has only Rs. 576 to invest and has space f at most items. A fan costs him Rs. 36 and a sewing machine Rs. 4. His epectation is that he can sell a fan at a profit of Rs. and a sewing machine at a profit of Rs. 8. Assuming that be can sell all the items that he can buy, how should he invest his money in der to maimise the profit? Fmulate this as a linear programming problem and solve it graphically. 9. If the sum of the lengths of the hypotenuse and a side of a right-angled triangle is given, show that the area of the triangle is maimum when the angle between them is. 3 OR A tank with rectangular base and rectangular sides open at the top is to be constructed so that it s depts is m and volume is 8m 3. If building of tank cost Rs. 7 per sq meter f the base and Rs. 45 per sq meter f the sides. What is the cost of least epensive tank. 4 XII Maths
6 MODEL PAPER - I SECTION A Note : F mark questions in Section A, full marks are given if answer is crect (i.e. the last step of the solution). Here, solution is given f your help. Marks. We are given 5 y y 4 y y = 4 and y = y = and 5 = 4 5 = 5 =. 6 * 4 = HCF of 6 and 4 =. 3. sin sin sin 4. Let I 4 4 sec log Let then log = t dt XII Maths 4
7 = dt Marks I sec = tan t + c t dt = tan (log ) + c cos cos cos cos cos cos a b b c c a a b b c c a b c c a b c c a a b b c c a a b c a a b c a a b b c c a a b b c a b b c b c c a c a a b a b b c = 7. Here 4 8 = 4 = 8. = ± i j k j k i k i j k k i i j j = + + = 3 4 XII Maths
8 Marks 9. The d.c. of a line equally inclined to the codinate aes are,, p p p 8 As p is a unit vect, p 8 8 = 9 SECTION B. Let P be the perimeter and A be the area of the rectangle at any time t, then P = ( + y) and A = y It is given that and dt dy dt 5 cm/minute 4 cm/minute (i) We have P = ( + y) dp dy dt dt dt XII Maths 43
9 Marks = ( 5 + 4) = cm/minute...(½) (ii) We have A = y da dy y dt dt dt = [ ( 5)]...( = 8 and y = 6) = (3 3) = cm /minute...(½) OR The given function is f() = sin + cos, f () = cos sin sin cos sin cos cos sin 4 4 sin 4 F strictly decreasing function, f () < sin 4 sin XII Maths
10 4 4 Marks Thus f() is a strictly decreasing function on 5, () As sin and cos are well defined in [, ], f () = sin + cos is an increasing function in the complement of interval 5, 4 4 i.e., in. We are given 5,, 4 4 (cos ) y = (sin y) Taking log of both sides, we get y log cos = log sin y...(½) Differentiating w.r.t., we get dy y sin log cos cos dy cos y log sin y sin y...() dy dy y tan log cos cot y log sin y dy log cos cot y log sin y y tan dy log sin y y tan log cos cot y...(½) XII Maths 45
11 Marks 3. F showing f is one-one...(½) F showing f is onto As f is one-one and onto f is invertible F finding f...(½) 4. Let I (½) (½) 7 sin 7 c sin 7 c...() 46 XII Maths
12 OR Marks Let I sin sin. sin sin sin sin sin sin sin c sin sin sin c 4 4 sin sin c We have y sin XII Maths 47
13 Marks y sin Differentiating w.r.t., we get dy y. y Differentiating again, dy...(½) dy d y dy y d y dy 3 y...(½) which is the required result. 6. Here p, q, and n Let denote the number of crect answers. Probability of r successes is given by P(X = r) = n C r p r q n r, r =,, 3... P(X = 4 5) = P(X = 4) + P (X = 5)...(½) C 4 C (½) XII Maths
14 Marks 7. LHS a b c c b c a b c a b a a b c R R + R R R + R 3 a b a b a b b c b c b c b a a b c...() a b b c b a a b c...(½) C C + C 3 a b b c c a a a b c = (a + b) (b + c) (c + a)...(½) 8. The given differential equation is dy y y tan dy y y tan Let y z dy dz z XII Maths 49
15 dz z z tan z Marks dz tan z cot z dz...(½) log sin z + log = log c log ( sin z) = log c sin y c...(½) which is the required solution. 9. The given differential equation is cos dy y tan dy sec. y tan. sec It is a linear differential equation Integrating fact = e sec tan e Solution of the differential equation is tan tan y. e e. tan sec c...(½) Now, we find Let tan I e tan sec tan = t, sec = dt 5 XII Maths
16 I te dt t Marks t. e t t e dt = t. e t e t = (t )e t = (tan ) e tan...() From (i), solution is y. e tan = (tan ) e tan + c y = (tan ) + ce tan...(½). Equations of the two lines are : r i j k and r i j k i j k...(i) r i j k i j k...(ii) Here a i j k and a i j k and b i j k and b i j k a a i j k i j k i 3 j k...(½) and i i k b b i 3 j k 3 3i 3k XII Maths 5
17 b b Marks S.D. between the lines a a b b...(½) b b i 3j k 3i 3k units 3. sin sin cot sin sin cot sin cos cos sin sin cos cos sin..., 4 sin cos cos sin cot sin cos cos sin cot cos sin cot cot 5 XII Maths
18 OR Marks The given equation is tan (cos ) = tan ( cosec ) cos cos tan tan cosec...(½) cos sin cosec cos = cosec. sin cos = sin 4...(½). Unit vect along the sum of vects a i 4j 5k and b i j 3k is a b i 6j k a b 6 i 6j k...(½) 4 44 We are given that dot product of above unit vect with the vect i j k is ( + 6) = = XII Maths 53
19 Marks 8 = 8 =...(½) OR a, b and c are coplanar a b c...(½) a b, b c and c a are coplanar if a b b c c a (½) F showing a b b c c a a b c (3) SECTION C 3. Equation of the plane through the points A (3,, ), B (5,, 4) and C (,, 6) is 3 y z (½) i.e. 3 4y + 3z = 9...(½) Distance of point (6, 5, 9) from plane 3 4y + 3z = units The given parabola is y =...(i) It represents a parabola with verte at O (, ) 54 XII Maths
20 The given line is Marks + y = = y...(ii) Y P(, ) y = X' 3 4 X + y = Q(4, ) Y' Solving (i) and (ii), we get the point of intersection P (, ) and Q (4, ) Required area = Area of the shaded region ( y ) y dy...() 3 y y y sq. units XII Maths 55
21 Marks 5. Let I a cos b sin I = ( ) a cos ( ) b sin ( ) I = ( ) a cos b sin...(ii) Adding (i) and (ii), we get I a cos b sin...(iii) I. a cos b sin a a Using property f f, If f a f I sec a b tan Let tan = t then sec = dt When =, t = and when, t a b t dt I b dt a b t 56 XII Maths
22 t. tan a b a b b Marks bt tan ab a ab ab 6. The given system of equations is + y + z = 6 + z = y = z = AX = B, where 6 A, X y and B 7 3 z X = A B...(i) Now A 3 = ( ) ( 5) + () = Now cofacts of elements of Matri A are : = 4 A eists. A = ( ) ( ) =, A = ( ) 3 ( 5) = 5, A 3 = ( ) 4 (I) = XII Maths 57
23 A = ( ) 3 () =, A = () 4 ( ) =, A 3 = ( ) 5 ( ) = Marks A 3 = ( ) 4 () =, A 3 = ( ) 5 () =, A 33 = ( ) 6 ( ) = adj A 5...() adj A A = A 5 A 4 From (i), X = A B...(½) 6 y z = 3, y = and z =...(½) OR 6. By using elementary row transfmations, we can write A = IA 58 XII Maths
24 Marks 3 i.e., 3 A 4 Applying R R R, we get 3 3 A 4 Applying R R R, we get A 4 Applying R R + R 3, we get 9 3 A 4...(½) Applying R R R 3, we get 3 A 4...(½) Applying R R, we get A 4...(½) XII Maths 59
25 Applying R 3 R 3 4R, we get A 8 9 Marks...(½) A Let the events be E : Bag I is selected E : Bag II is selected E 3 : Bag III is selected and A : a black and a red ball are drawn P (E ) = P (E ) = P (E 3 ) = P(A E ) 6 C 5 5 P(A E ) C P(A E 3) C 66...(½) P(E A) P(A E ). P(E ) P(A E ) P/(E ) P(A E ) P E P(A E ). P(E ) (½) 6 XII Maths
26 Marks Let us suppose that the dealer buys fans and y sewing machines, Thus L.P. problem is Maimise Z = + 8y subject to constraints, 3 + y = 48 5 Y + y y y 48, y...() B(,) P(8,) + y = A(,) D (6,) X XII Maths 6
27 Marks F crect graph...() The feasible region ODPB of the L.P.P. is the shaded region which has the cners O (, ), D (6, ), P (8, ) and B (, ) The values of the objective function Z at O, D, P and B are : At O, Z = + 8 = At D, Z = = 35 At P, Z = = 39 Maimum and At B, Z = + 8 = 36 Thus Z is maimum at = 8 and y = and the maimum value of z = Rs 39. Hence the dealer should purchase 8 fans and sewing machines to obtain maimum profit. 9. Let ABC be a right angled triangle with base BC = and hypotenuse AB = y such that + y = k where k is a constant...(½) Let be the angle between the base and the hypotenuse. The area of triangle, A BC AC A y y y...(½) B C 6 XII Maths
28 Marks A ( y ) 4 ( k ) 4 3 k k A k k..(i) 4 4 Differentiating w.r.t. we get da k 6k A 4...(ii) da k 3k 4A F maimum minimum, da k 3k 4 k 3 Differentiating (ii) w.r.t.. we get da d A k k A 4 Putting, da k and, we get 3 d A k 4A XII Maths 63
29 A is maimum when Marks k 3 Now k k k y k cos k cos y k Let the length of the tank be metres and breadth by y metres OR Depth of the tank = metre Volume = y = 8 y = 4 y 4 Area of base = y sq m Area of 4 walls = [ + y] = 4 ( + y) Cost C (,y) = 7 (y) + 45 (4 + 4y) C (, y) = ( + y) 4 C( ) (½) dc 4 Now 8 F maimum minimum, dc XII Maths
30 = 4 Marks =...(½) and d C d C C is minimum at = Least Cost = Rs [(8 + 8 ( + )] = Rs [(8 + 7] = Rs XII Maths 65
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