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CBSE XII Sample / Mock 2017 : MATHEMATICS (with Marking Scheme / Solutions)

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SAMPLE QUESTION PAPER CLASS-XII (2016-17) MATHEMATICS (041) Time allowed: 3 hours Maximum Marks: 100 General Instructions: (i) All questions are compulsory. (ii) This question paper contains 29 questions. (iii) Question 1- 4 in Section A are very short-answer type questions carrying 1 mark each. (iv) Question 5-12 in Section B are short-answer type questions carrying 2 marks each. (v) Question 13-23 in Section C are long-answer-I type questions carrying 4 marks each. (vi) Question 24-29 in Section D are long-answer-II type questions carrying 6 marks each. SECTION-A Questions from 1 to 4 are of 1 mark each. 1. What is the principal value of 2. A and B are square matrices of order 3 each, | | = 2 and | | = 3. Find | 3. What is the distance of the point (p, q, r) from the x-axis? 4. Let f : R R be defined by f(x) = 3x2 5 and g : R R be defined by g(x) = . /? | . Find gof SECTION-B Questions from 5 to 12 are of 2 marks each. 5. How many equivalence relations on the set {1,2,3} containing (1,2) and (2,1) are there in all ? Justify your answer. 6. Let li,mi,ni ; i = 1, 2, 3 be the direction cosines of three mutually perpendicular vectors in space. Show that AA = I3 , where A = [ ]. 7. If ey (x + 1) = 1, show that 8. Find the sum of the order and the degree of the following differential equations: + +( 1 +x) =0 9. Find the Cartesian and Vector equations of the line which passes through the point ( 2 4 5) and pa a e to the ine gi en by 10. Solve the following Linear Programming Problem graphically: Maximize Z = 3x + 4y subject to x+y 11. A couple has 2 children. Find the probability that both are boys, if it is known that (i) one of them is a boy (ii) the older child is a boy. 12. The sides of an equilateral triangle are increasing at the rate of 2 cm/sec. Find the rate at which its area increases, when side is 10 cm long. SECTION-C Questions from 13 to 23 are of 4 marks each. 13. If A + B + C = then find the value of ( | ( ) | ) OR Using properties of determinant, prove that | 14. | = 3abc It is given that for the function f(x) = x3 6x2 + ax + b Ro e s theo em ho ds in * 1 3+ with c = 2 + . Find the a ues of a and b 15. Determine for what values of x, the function f(x) = x3 + strictly decreasing OR Find the point on the curve y = ( x 0) is st ict y inc easing o at which the tangent is y = x 11 16. Evaluate ( 17. Find the area of the region bounded by the y-axis, y = cos x and y = sinx, 0 18. Can y = ax + be a solution of the following differential equation? ) dx as limit of sums. y=x + ...............(*) If no, find the solution of the D.E.(*). OR Check whether the following differential equation is homogeneous or not xy = 1 + cos . / x 0 Find the general solution of the differential equation using substitution y=vx. 19. If the vectors c 1 show that a + + , +b + + + + + are coplanar, then for a, b, =1 20. A p ane meets the coo dinate axes in A B and C such that the cent oid of ABC is the point ( , , ) . Show that the equation of the plane is 21. If a 20 year old girl drives her car at 25 km/h, she has to spend Rs 4/km on petrol. If she drives her car at 40 km/h, the petrol cost increases to Rs 5/km. She has Rs 200 to spend on petrol and wishes to find the maximum distance she can travel within one hour. Express the above problem as a Linear Programming Problem. Write any one value reflected in the problem. 22. The random variable X has a probability distribution P(X) of the following form, where k is some number: k , if x =0 P(X) = 2k , if x = 1 3k , if x = 2 0 , otherwise (i) Find the value of k (ii) Find P(X <2) (iii) Find P(X ) ( ) P(X ) 23. A bag contains ( 2n +1) coins. It is known that n of these coins ha e a head on both its sides whereas the rest of the coins are fair. A coin is picked up at random from the bag and is tossed. If the probability that the toss results in a head is find the a ue of n . SECTION-D Questions from 24 to 29 are of 6 marks each 24. Using properties of integral, evaluate dx OR Find: 25. dx Does the following trigonometric equation have any solutions? If Yes, obtain the solution(s): . /+ . /= OR Determine whether the operation define below on is binary operation or not. a b = ab+1 If yes, check the commutative and the associative properties. Also check the existence of identity element and the inverse of all elements in . 26. Find the value of x, y and z, if A = [ ] satisfies A = A 1 OR Verify: A(adj A) = (adj A)A = | ||I for matrix A = [ 27. Find if y = { ] } 28. Find the sho test distance between the ine x y + 1 = 0 and the cu e y2 = x 29. Define skew lines. Using only vector approach, find the shortest distance between the following two skew lines: = (8 + 3 ) (9 + 16 ) +(10 + 7 ) = 15 + 29 +5 + (3 + 8 5 ) SAMPLE QUESTION PAPER CLASS-XII (2016-17) MATHEMATICS (041) Marking Scheme 1. 2. . | | /= . /= 1 3 | | | | = 27 3 2 1 3. Distance of the point (p, q, r) from the x-axis 1 = Distance of the point (p, q, r) from the point (p,0,0) = 4. 5. gof(x) = g{ f(x)} = g(3x2 5) = ( = ) 1 Equivalence relations could be the following: { (1,1), (2,2), (3,3), (1,2), (2,1)} and (1) { (1,1), (2,2), (3,3), (1,2), (1,3), (2,1), (2,3), (3,1), (3,2)} (1) 2 So, only two equivalence relations.(Ans.) 6. AA = [ ] [ ] =[ ] = I3 (1) 2 because = 1, for each i = 1, 2, 3 = 0 ( i j) fo each i j = 1 2 3 7. On differentiating ey (x + 1) = 1 w.r.t. x, we get ey + (x + 1) ey ey + =0 Here, { (1) +( ) +3 = Thus, order is 2 and degree is 3. So, the sum is 5 9. 2 =0 8. (1) (1) 2 (1) Here, Cartesian equation of the line is (1) Vector equation of the line is = ( 2 + 4 5 ) + (3 + 5 + 6 ) (1) 2 10. The feasible region is a triangle with vertices O(0,0), A(4,0) and B(0,4) Zo = 3 0 + 4 0 = 0 (1) ZA = 3 4 + 4 0 = 12 2 ZB = 3 0 + 4 4 = 16 Thus, maximum of Z is at B(0,4) and the maximum value is 16 11. Sample space = { B1B2, B1G2, G1B2, G1G2} , B1 and G1 are the older boy and girl respectively. Let E1 = both the children are boys; E2 = one of the children is a boy ; 2 E3 = the older child is a boy Then, 12. Here, Area(A) = So. 13. (i) P(E1/ E2) = P ( )= = (1) (ii) P(E1/ E3) = P ( )= = (1) x2 whe e x is the side of the equi ate a t iang e (1) (10) ( ) = 10 cm2/sec 2 As A + B + C = ( | ) ( ) =0 | | | = | | sin B | | (2) 4 | + cos C | = 0 sin B tan A cos C + cos C sin B tan A = 0 (Ans.) OR (2) Let = | | 4 | Applying C1 C1 + C3 we get = (a + b + c)| Applying R2 R2 (1) R1, and R3 R3 R1 , we get | = (a + b + c)| (1) Expanding a ong fi st co umn we ha e the esu t 14. (2) Since Ro e s theorem holds true, f(1) = f(3) (1)3 6(1)2 + a(1) + b = (3)3 6(3)2 + a(3) + b i.e., i.e., a + b + 22 = 3a + b a = 11 Also, f (x) = 3x 2 (2) 12x + a or 3x 2 4 12x + 11 As f (c) = 0 , we have 3( 2 + )2 12(2 + ) +11 = 0 As it is independent of b, b is arbitrary. 15. He e f (x) = 3x2 3x = ( 4 = ) ( ( (2) ) (1) )( ) C itica points a e 1 and 1 f (x) 0 if x 1 o x (1) 1 and f (x) 0 if ( { ) 1 x 1 + Hence, f(x) is strictly increasing for x > 1 o x (1) 1 and st ict y dec easing fo (-1,0)u(0,1) [1] (1) OR Here, 4 4 11 So, slope of the tangent is 11 Slope of the given tangent line is 1. Thus, 11 = 1 (1) that gives x = When x = 2, y = 2 hen x = 2 y = 2 ut of the two points (2 on y the point ( 2 ) and ( 2 13) ) ies on the cu e hus the equi ed point is ( 2 16. Here, f(x) = ( ) dx = , ( ) = , = , ( ) (1) ( )+..... ( ( * , * = , = , =6+ ( ( ) ( * ( * ) ) ( ) (1) ] ) +] 4 +] ( ) ( )] ) ) +] (1) +] ,i.e., (1) The rough sketch of the bounded region is shown on the right. (1) Required area = dx (1) )- (1) dx = ( = sin + cos sin y = ax + 4 cos = 18. ) a = 0 b = 2 and nh = b a = 2 = 17. (2) 1 , i.e, ( ) (1) ( ) =a Substituting this a ue of a in (1) we get . / 4 y= x Thus, y = ax + . + is a solution of the following differential equation y = x + . / 1 OR Given differential equation can be written as . / Let F(x,y) = . / +[ Then F( x, y) = +[ . / = +[ . / = +[ - ......(1) -. . ( ) / 4 ] - F(x y) Hence, the given D.E. is not a homogeneous equation. Putting y = vx and = . / (1) in (1), we get dx = dx (1) Integrating both sides, we get 2 tan = or 2 tan 19. Since the vector [ [ i.e., | +C = 1 +C and are coplanar -=0 - (1) |=0 (1) 4 or | |=0 a( b -1)(c-1) -1 (1-a)( c -1) - 1(1-a)( b 1) = 0 i.e., a( 1 - b)(1- c) + (1-a)( 1 - c) + (1- a)( 1 b) = 0 (1) Dividing both the sides by (1-a)( 1 - b)(1- c), we get . i.e., / i.e., 20. (1) We know that the equation of the plane having intercepts a, b and c on the three coordinate axes is (1) Here, the coordinates of A, B and C are (a,0,0), (0,b,0) and (0,0,c) respectively. he cent oid of ABC is ( ). ) to ( , , ) Equating ( 4 (1) b ( ) Thus, the equation of the plane is or 21. (1) Let the distance covered with speed of 25 km/h = x km and the distance covered with speed of 40 km/h = y km ( ) Total distance covered = z km The L.P.P. of the above problem, therefore, is (1) 4 Maximize z = x + y subject to constraints 4x + 5y (1) x (1) Any one value 22. ( ) Here, X P(X) 0 1 2 k 2k 3k (i) Since P(0) + P(1) + P(2)= 1, we have 4 k + 2k + 3k = 1 i.e., 6 k = 1, or k = (1) (ii) P(X <2)= P(0) + P(1)= k + 2k = 3k = ; 23. (1) (iii) P(X ) = P(0) + P(1) + P(2)= k + 2k + 3k = 6k = 1 (1) (iv) P(X ) = P(2) = 3k = (1) Let the events be described as follows: E1 : a coin having head on both sides is selected. E2 : a fair coin is selected. A : head comes up in tossing a selected coin P(E1 ) = ; P(E2 ) = ; P(A/E1 ) = 1; P(A/E2 ) = (2) 4 It is given that P(A) = P(E1 ) P(A/E1 ) + P(E2 ) P(A/E2 ) = 1+ [n+ -= ( (1) = ) ( ) 24. I= (1) dx = = ) dx . / . / (1) dx 6 dx ( (1) dx dx dx 2I = ( ) dx I = [-2tan 0. I = , ( (1) /1 (2) )- = (1) OR Let I = dx = dx ( ) On substituting tan x = t and I= dt = ( )( = 25. | . /+ | + = | | + = | | + . / /. / dt (1) dt + | dt | + | | + | | + ]= 0( ( ( , if . ( ) ( ) ( )( ) ) ( ) ( . / ( ( dt ( ) ) (2) ) +c ) ) = /( ) < 1 .....(*) (1) (2) 1= ) , 6 ] = (1) = ( ) /= / . . ( | + | + ) =0 =2 (1) Let us now verify whether x = 2 satisfies the condition (*) For x = 2, . /( ) = 3 = which is not less than 1 Hence, this value does not satisfy the condition (*) i.e., there is no solution to the given trigonometric equation. OR Given on Q , defined by a b = ab+1 Let, a Q, b Q then ab Q 6 dt | . [ ( ) = | (1) dt dt + = = ) , we get (1) (1) and (ab+1) Q 6 a b=ab+1 is defined on Q is a binary operation on Q (1) Commutative: a b = ab+1 b a = ba+1 ( ba = ab in Q) =ab+1 a b =b a So is commutative on Q (1) Associative: (a b) c= (ab+1) c =(ab+1)c+1 = abc+c+1 a (b c)=a (bc+1) = a(bc+1)+1 = abc+a+1 (a b) c a (b c) So is not associative on Identity Element : Let e (1) be the identity element, then for every a a e=a and e a=a ae+1=a and ea+1=a e= 26. and e= (1) e is not unique as it depend on `a hence identity e ement does not exist fo (1) Inverse: since there is no identity element hence, there is no inverse. (1) he e ation A = A 1 gi es A A = A 1A = I Thus, [ ] [ ] (1) [ . ] [ ] [ / ] 6 [ ] [ ] (2) x= ; y= OR . / Here, | | | | = 1(0+0) +1( +2) +2(0 0) =11 | |I = [ (1) ( ) ] ....................(1) 6 adj A = [ (2) ] Now, A(adj A) = [ ][ (adj A)A = [ and ] ] [ [ ] (1) ]= [ ] (1) Thus, it is verified that A(adj A) = (adj A)A = | |I 27. Putting x = cos 2 in { } , we get (1) i.e., ( ) =2 ( ) = 2 = (2) 6 Hence, y = log y = x + log ( ) 2 sin x cos x + 28. ,sin 2x - = sin 2x - (2) - (1) Let (t2, t) be any point on the curve y2 = x. Its distance (S) from the ine x y + 1 = 0 is gi en by S= | = | { = and = . ( ) (1) ( ) 6 = >0 = / + >0} ( (1) ) Thus, S is minimum at t = (1) So, the required shortest distance is 29. ( ) ( ) = , or (1) 1) the line which are neither intersecting nor parallel. (1) 2) The given equations are =8 9 + 10 + (3 16 + 7 ) ....................(1) ( ) = 15 + 29 +5 + (3 + 8 5 ) ....................(2) Here, = 8 9 + 10 = 15 + 29 +5 = 3 16 + 7 = 3 + 8 5 ) + (29 + 9) + (5 10) Now, = (15 ( ) and = | ( ).( ) = ( Shortest distance = | =| |= ( ).( ) ( ) | | ) (1) 1176 (1) (1) | = |= = (1) --0-0-0--

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