12 th Class Mathematics Paper

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1 th Class Mathematics Paper Maimum Time: hours Maimum Marks: 00 General Instructions: (i) All questions are compulsory. (ii) The question paper consists of 9 questions divided into four sections A, B, C and D. Section A comprises of 4 questions of one mark each, Section B comprises of 8 questions of two marks each, Section C comprises of questions of four marks each and Section D comprises of 6 questions of si marks each. (iii) All questions in Section A are to be answered in one word, one sentence or as per the eact requirement of the question. (iv) There is no overall choice. However, internal choice has been provided in questions of four marks each and questions of si marks each. You have to attempt only one of the alternatives in all such questions. (v) Use of calculators is not permitted. You may ask for logarithmic tables, if required. SECTION A Question numbers to 4 carry mark each.. Find the value of tan cot. Given: tan cot tan cot π 5π tan tan cot tan 6 π 5π 6 π 5π 6 π. If the matri A 0 a 0 is skew symmetric, find the values of a and b. b 0 0 a Given: A 0 b 0 We know that if A is skew symmetric matri then, A T A. Pioneer Education SCO 0, Sector 40 D, Chandigarh , Page of

2 0 b 0 a a b 0 0 b 0 a a b 0 a and b a and b. Find the magnitude of each of the two vectors a and b, having the same magnitude such that the angle between them is 60 0 and their scalar product is 9. Suppose a and b be the two vectors. Now, a. b a b cos θ We are given that 0 9 a b, θ 60 and a. b 9 a a cos a a 9 a b 4. If a * b denotes the larger of a and b and if a o b (a * b) +, then write the value of (5) o (0), where * and o are binary operations. Given: a b (a * b) + Here a * b denotes the larger of 'a' and 'b'. Substitute a 5 and b 0 in a b (a * b) + therefore 5 0 (5 * 0) [5 * 0 0]. SECTION-B sin sin 4,, Given: sin sin ( 4 ) LHS sin ( 4 ) Substitute sin θ 5. Prove that: sin sin θ 4 sin θ sin sin θ θ Pioneer Education SCO 0, Sector 40 D, Chandigarh , Page of

3 sin RHS Since, LHS RHS Hence proved. 6. Given A sin θ θ sin, 4 7 computer A and show that A 9I A. Given A Now, L.H.S A A RHS 9I A Since, LHS RHS Hence proved. cos 7. Differentiate tan with respect to. sin cos Given tan sin tan cos cos tan sin sin cos tan cot Pioneer Education SCO 0, Sector 40 D, Chandigarh , Page of

4 tan tan π π Now, y 8. The total cost C() associated with the production of units of an item is given by C() Find the marginal cost when units are produced, where by marginal cost we mean the instantaneous rate of change of total cost at any level of output. Given C() d Marginal Cost C () 0.04() Evaluate: Given: cos cos cos cos sin sin cos sin cos sin sin cos cos cos sin cos cos sec tan + C b 5 0. Find the differential equation representing the family of curves y ae, where a and b are arbitrary constants. Given: y ae y a e b 5 b5 b 5 Differentiate y ae with respect to ae b 5 b Pioneer Education SCO 0, Sector 40 D, Chandigarh , Page 4 of

5 aby a by d y y d y y y a e b5 b y. If θis the angle between two vectors i j k and i j k, find sinθ. Using cross product of two vectors i j k i j k i j k i j k cos θ i j k. i j k cos θ i j k i j k cos θ 0 cos θ 4 5 cos θ sin θ cos θ A black and a red die are rolled together. Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4. Let A be the event when the sum of the observations is 8 {(, 6), (, 5), (5, ), (4, 4), (6, )} na 5 Let B be the event when the sum of the observation on red die is less than 4 {(, ), (, ), (, ), (4, ), (5, ), (6, ), (5, ), (6, ) (6, )} n(b) 8 8 PB 6 Pioneer Education SCO 0, Sector 40 D, Chandigarh , Page 5 of

6 Now, na B na B PA / B 6 8 PA B 8 PB 9. SECTION-C Question numbers to carry 4 marks each. y z Given: y z 9 (yz + y + yz + z) Apply R R R and R R R y 0 y y z 0 z z Taking common from R and R, we get y 0 y 0 y z 0 y z 0 z z 9[( y) ( z) 0 + (y + yz + z) 9[yz + y + yz + z] 9 [yz + y + yz + z] Hence proved. Pioneer Education SCO 0, Sector 40 D, Chandigarh , Page 6 of

7 4. If ( y ) y, find. If a ( θ sin θ ) and y a cosθ Given: y y y y y 4 y 4y y 4 y y 4y y 4 y y 4y y 4 y y 4y y Given: a(θ sinθ ) and y a( cos θ) acos θ and a sin θ dθ dθ Now, dθ dθ a sin θ a cos θ sin θ cosθ π sin π π θ cos d y 5. If y sin (sin ), prove that y sin (sin ) Differentiate it with respect, find when tan y cos 0. π θ. Pioneer Education SCO 0, Sector 40 D, Chandigarh , Page 7 of

8 cos (sin )cos Again differentiate it w. r. t. sin (sin )cos sin cos (sin ) + sin(sin )cos + sin cos (sin ) 0 d y sin ycos 0 cos sin cos cos sin d y cos ycos tan 0 cos d y tan ycos 0 Hence proved. 6. Find the equations of the tangent and the normal, to the curve 6 + 9y 45 at the point (, y), where and y > 0. Find the intervals in which the function f() is (a) strictly increasing, (b) strictly decreasing. Given: 6 + 9y 45. Substitute in 6 + 9y 45. 6() + 9y y 45 9y 8. y. (Neglecting y as y > 0) Now, differentiate 6 + 9y 45 with respect to which will give us the slope of the tangent. + 8y 0 8y Substitute and y in The equation of tangent is given by y 7 7y y y Pioneer Education SCO 0, Sector 40 D, Chandigarh , Page 8 of

9 The slope of the normal is The equation of normal is given by y 7 y y Slope of tangent 4 Given: f() f () Find the critical points by equation f () to zero ( ) ( ) ( ) 0 ( ) ( ) 0 ( ) ( 4) ( + ) 0, and 4 Therefore, we have the intervals,,,,, 4 and 4, (a) For f() to be strictly increasing, we should have f () > 0 ( ) ( 4) ( + ) > 0 Therefore,, 4, (b) For f() to be strictly decreasing, we should have f () < 0 ( ) ( 4) ( + ) < 0 Therefore,, 4, 7. An open tank with a square base and vertical sides is to be constructed from a metal sheet so as to hold a given quantity of water. Show that the cost of material will be least when depth of the tank is half of its width. If the cost is to be borne by nearby settled lower income families, for whom water will be provided, what kind of value is hidden in this question? Let the dimensions of the tank be y. Therefore, the total surface area of the open tank would be ( +y+y) +4y Let V be the volume of the tank. V V y y Therefore, the surface area will be least when y. Pioneer Education SCO 0, Sector 40 D, Chandigarh , Page 9 of

10 8. Find: Given: cos sin sin cos sin sin Suppose sin t cos dt cos sin sin t t Using partial fraction, we get A Bt C t t t t A + At + Bt + C Bt Ct (A B) t + (B C) t + A + C A B 0, B C 0 and A + C Solving all these equations we get A B C Substituting the values, we get t dt dt dt t t t t dt dt dt t t t dt t dt dt t t t ln t ln t tan t C ln t ln t tan t C dt ln t ln t tan t C Substitute back the value of t in ln t ln t tan t C. ln sin ln sin tan sin C 9. Find the particular solution of the differential equation e tan y + ( e ) sec y 0, given that y π when 0. 4 Pioneer Education SCO 0, Sector 40 D, Chandigarh , Page 0 of

11 Find the particular solution of the differential equation + y tan sin, given that y 0 when π. Given: e tan y e sec y 0 e tan y e sec y e sec y e tan y Integrating both sides, we get e sec y e tan y dt ds Let e t and tan y s t s e dt and sec y ln t ln s + ln C t sc Substituting t e and s tan y, we get e C tan y π Now, y when 0 4 o π e C tan 4 C C e tan y tan y e y tan e Given: y tan sin Comparing y tan sin with general linear equation + Py Q, we get P tan and Q sin tan logsec log sec Now, integrating factor e e e sec The differential equation is given by y sec sin sec C Pioneer Education SCO 0, Sector 40 D, Chandigarh , Page of

12 sin y sec sec C cos ysec tan sec C y sec sec C Now, y 0 when π y sec sec C C - π π y sec sec Integrating both sides, we get e sec y e tan y dt ds Let e t and tan y s t s e dt and sec y ln t ln s + ln C t sc Substituting t e and s tan y, we get e C tan y Now, y π 4 when 0 o π e C tan 4 C C e tan y tan y e y tan e 0. Let a 4i 5j k, b i 4j 5k and c i j k. Find a vector d which is perpendicular to both c and d. a. We are given that d which is perpendicular to both c and b d.b 0 and d. c 0 Let us suppose d i yj zk Now, d. b 0 Pioneer Education SCO 0, Sector 40 D, Chandigarh , Page of

13 i yj zk. i 4j 5k 0 4y + 5z 0 () d. c 0 i yj zk. i j k 0 + y z 0 () d. a i yj zk. 4i 5j k 4 + 5y z () Solving (), () and (), we get 6, y and z 6 d i j k. Find the shortest distance between the lines r 4i j λi jk and r i jk μi 4j 5k. Here, a 4i j, b i j k and a i j k, b i 4j 5k a a i j k 4i j i k Now, Also, i j k b b 4 5 b b 0 i 5 6 j 4 4 k b b i j b b 4 5 Therefore, the shortest distance is given by d b b a a i j. i k b b units Pioneer Education SCO 0, Sector 40 D, Chandigarh , Page of

14 . Suppose a girl throws a die. If she gets or, she tosses a coin three times and notes the number of tails. If she gets, 4, 5 or 6, she tosses a coin once and notes whether a head or tail is obtained. If she obtained eactly one tail, what is the probability that she threw, 4, 5 or 6 with the die? Let A be the event when she get or and A be the event when she get, 4, 5 or 6 PA and PA Let B be the event of getting eactly one tail. PB/ A and PB/ A 8 Required probability P(A/B) PAPB/ A P A P B/ A P A P B/ A Two numbers are selected at random (without replacement) from the first five positive integers. Let X denote the larger of the two numbers obtained. Find the mean and variance of X. The first five positive integers are,,, 4 and 5. We can select two numbers from 5 numbers in 5 P Now, let X denote the larger of two numbers, then X can take values from,, 4 and 5. For, the opposite observations are (, ) and (, ) PX 0 For X, the possible observations are (, ), (, ), (, ) and (, ) P(X ) 4 0 For X 4, the possible observations are (, 4), (, 4), (, 4), (4, ), (4, ) and (4, ) 6 PX 4 0 For X 5, the possible observations are (, 5), (, 5), (, 5), (4, 5), (5, ), (5, ), (5, ) and (5, 4) Pioneer Education SCO 0, Sector 40 D, Chandigarh , Page 4 of

15 8 P(X 8) 0 Therefore, we have X 4 5 P(X) Therefore, Mean E X EX Variance E(X ) [E(X)] 7 (4) Therefore, mean and variance are 4 and respectively. SECTION-D Question numbers 4 to 9 carry 6 marks each. 4. Let A { Z : 0 }. Show that R {(a, b) : a, b A, a b is divisible by 4} is an equivalence relation. Find the set of all elements related to. Also write the equivalence class []. Show that the function f : defined by f(), if g: is defined as g(), find fog(). Given A {0,,,, 4, 5, 6, 7, 8, 9, 0,, } Also, R {(a, b) : a, b A, a b For refleive Substitute a b R bb 0which is divisible by 4 Therefore, R is refleive. For symmetric Substitute a b and b a Pioneer Education SCO 0, Sector 40 D, Chandigarh , Page 5 of 8 0 is neither one-one nor onto. Also,

16 R ba a b which is divisible by 4 Therefore, R is symmetric For Transitive If a b is divisible by 4 i.e., a b k Also, If b c is divisible4 by 4 i.e., b c 4k a c 4k 4k is also divisible by 4 Therefore, R is Transitive. Now, set of elements related to is {(, ), (, 5), (, 9), (5, )-, (9, )} Let (, ) R 4k where k,6, 0 Therefore, the equivalence class [] is {, 6, 0}. Solutions: Given: f() Let y y y y y 0 4y y Since for any value of we will get two values for y. Hence, f() is many-one not one-one. Since is real number. Hence, 4y 0 y y 0 y Therefore, we always get the value of y as Hence, f() is not onto. Now, g() fog , Pioneer Education SCO 0, Sector 40 D, Chandigarh , Page 6 of

17 5 5. If A 4, find A. Use it to solve the system of equations y + 5z + y 4z 5 + y z. Using elementary row transformations, find the inverse of the matri A 5 Given : A 4 A Now, adj. A A 9 5 T Pioneer Education SCO 0, Sector 40 D, Chandigarh , Page 7 of

18 0 9 5 We can write the given equation as AX B 5 4 y 5 z 0 y 9 5 z 5 y z, y and z Given : A We know that A IA A Apply R R R and R R R A Apply R R R A Apply R R R and R R R Pioneer Education SCO 0, Sector 40 D, Chandigarh , Page 8 of

19 A A Using integration, find the area of the region in the first quadrant enclosed by the -ais, the line y and the circle + y. Given: y and + y Substitute y in + y, we get Therefore, the required area is given by 4 y y sin 0 4π sq units 7. Evaluate: π/4 0 sin cos 6 9 sin 4 Pioneer Education SCO 0, Sector 40 D, Chandigarh , Page 9 of

20 Evaluate: Given: π/4 0 e, as the limit of the sum. sin cos 6 9 sin Let sin cos t (cos + sin ) dt Squaring both sides in sin cos t (sin cos ) t sin + cos sin t sin t sin cos dt 6 9sin 6 9 9t dt 5 9t dt 9 9 t 5 t log t 5 Substitute t sin cos Now, sin cos sin cos log sin 9 0 sin cos 5 π/4 0 sin cos sin cos log 5 6 9sin 9 0 sin cos 5 π/4 0 log 4 0 Given: Here, e a, b f() + + e b a h n n n Pioneer Education SCO 0, Sector 40 D, Chandigarh , Page 0 of

21 e lim f f h f h f h... f n h h0 4 e h h e h h e h h e lim h0 n h... n h n h h h h h hn n n n n n n n e e lim 4n e h h h e h0 h 6 e h hn n n n n n n n e e lim 4n e h h h e h0 h 6 e 6 e e 8. Find the distance of the point (, 5, 0) from the point of intersection of the line and the plane r i j k λ i 4j k Given: r i j k λi 4j k r λ i 4λ j λ k r. i j k 5. Substitute r λ i 4λj λ k in ri j k 5. λ i 4λj λ k. i j k 5 λ 4λj λ 5 λ 4λ λ 5 λ 5 5 λ 0 Substituting λ 0 in λ i 4λ j λ k we get r i j k Therefore, the coordinates of the points are (,, ) and (, 5, 0). The distance between the two points is given by units 9. A factory manufactures two types of screws A and B, each type requiring the use of two machines, an automatic and a hand-operated. It takes 4 minutes on the automatic and 6 minutes on the handoperated machines to manufacture a packet of screws A while it takes 6 minutes on the automatic and minutes on the hand-operated machine to manufacture a packet of screws B. Each machine is available for at most 4 hours on any day. The manufacturer can sell a packet of screws A at a profit of 70 paise and screws B at a profit of Rs. Assuming that he can sell all the screws he manufactures, how Pioneer Education SCO 0, Sector 40 D, Chandigarh , Page of

22 many packets of each type should the factory owner produce in a day in order to maimize his profit? Formulate the above LPP and solve it graphically and find the maimum profit. Let the factory manufacture screws of type A and y screws of type B on each day. Therefore, 0 and y 0 We can write the two equations as given below: 4 6y y 40 and 6 + y y 40 Now, the total profit for A and B are 70 paise and Rs. Therefore, we have to maimize y First we draw the graph of 0 y 0, 4 6y 40 and 6 y 40 The value of y at the corner points y Objective Function y A B C D E Thus, the factory should produce 0 packages of screws A and 0 packages of screws B to get the maimum profit of Rs 4. Pioneer Education SCO 0, Sector 40 D, Chandigarh , Page of

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