ANSWER KEY MARKING SCHEME MATHEMATICS HALF YEARLY EXAMINATION

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Preston Ferguson
5 years ago
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1 GENERAL INSTRUCTION Marking Sceme General Instructions :. Te marking sceme provides general guidelines to reduce subjectivity in te marking. Te answers given in te marking sceme are Text Book, Solution Book and COME BOOK bound. Te contends of e text book, Solution Book and Come Book are tus indicative.if a student as given any answer wic is different form one given in te marking sceme, but conveys te prescribed content meaning, Suc answers sould be given full credit.. Evaluation sould to be done strictly adering te instruction provided in te marking sceme in order to ensure uniformity in te assessment of candidate s performance. 3. In an answer to an question, between any two particular stages of marks (greater tan ) if a student start from a stage wit correct step but reac te next stage wit a wrong result ten suitable credits sould be given to te correct steps instead of denying te entire marks meant of tis page. 4. Only a roug sketc of te diagram is expected and full credit must be given for suc diagrams tat form part and parcel of te solution of a problem. 5. Marks sould not be deducted in questions on determinants and matrices, if arbitrary constants are assigned to variables oter tan tose in te marketing sceme.(te same will be applicable for sequence of elementary transformations applying to get ecelon form of a matrix). 6. Marks sould be awarded for correct answers to questions on te vector equations of lines and planes irrespective of teir equivalent vector form. Eg.r = a + t(b a ) or r = ( t)a + tb 7. Marks Sould not be deducted in questions on applications of integration if te correct dimensions suc as square units for area, cubic units for volume are not specified. 8. In questions on variable separable, omogeneous, linear differential equations of first order, full marks sould be given for an equivalent answer and student sould not be penalized for not getting te final answer mentioned in te marking sceme. 9. Tere is no separate mark allotted for formulae. If a particular stage is wrong ad if te candidate writes te appropriate formula ten award suitable mark for te formula, Tis mark is attaced wit tat stage.tis is done wit te aim tat a student wo did problem correctly witout writing formula is not penalized. Tat is mark sould not be deducted foe not writing te formula. SECTION A () mark to write te correct option or te corresponding answer or bot. If one writes bot option and answer wit one of tem is wrong, ten award ZERO mark only. Instead of,,3,4 if one writes a,b,c,d ten marks sould be awarded 3)3 5 4)y = 9 ) (x x + ) 3 3) k 4 6 4) 5 3 ) 3 3 4) no solutions 7 ) (4,4), ( 4, 4) 3 ) cos x 4 ) A as atleast one minar of r wic does not vanis and all iger order minor vanis. 8 ) focus 3 4) y -y +y= 5 3) a = 4, b = 4, c = 5 9 3) x 33 ) y /y b 6 3) c is parallel to a ).5 cm / sec 34 4) (ii), (iii), (iv) 7 ) 4 35 ) an increasing function in )5, 8 ) (,, 4) ) vertical tangent x x 36 3) Only Monoid 9 3 ),, and 3 ) x 6 37 ) 4 ) x + 9y + z = 4 ) 38 ) 5 4) e 9, 3π 5 ) 39 ) 4 )Cos -isin 6 d) b:a 4 ) -P(X<a) 3 ) x + 7 = 7 ) 4 )(-π, π) 8 4) 7/m 8

NO ANSWER MA R 4 4 43 Given Matix Equation [ 7 3 ] [x y ] = [ ] AX = B A = 7 3 = 7 6 = A is te non zero determinant so we ave invers adja = [ 3 7 ] = A = adja = A 3 3 [ ] = [ 7 7 ] X = A B X = ( x y

2 NO ANSWER MA R Given Matix Equation [ 7 3 ] [x y ] = [ ] AX = B A = 7 3 = 7 6 = A is te non zero determinant so we ave invers adja = [ 3 7 ] = A = adja = A 3 3 [ ] = [ 7 7 ] X = A B X = ( x y ) = [ 3 7 ] [ ] = [ + + ] = [ ] Solutions are x =, y = Δ = 3 = ( ) ( 3) + ( + 3) = = 5 Δ x = 4 = 5( ) ( 4) + ( + 4) = = 6 Since Δ = and Δx (atleast one of te values of Δx, Δy, Δz non zero) te system is inconsistent. i.e. it as no solution. i j k a b Requird Vectors 5nˆ i j k 8 i 8 j 9 k 9( i j k ) a b a b 5 a b 9( i j k ) ( i j k ) 3 () () () () () () () Q.NO ANSWER MARK S 43(ii) r r (4 i j 6 k ) r r (4 i j 6 k ) x y z 4x y 6z u 4;v ;w 6; c ( u, v, w) (,, 3) a u v w c [ a b b c c a ] ( a b ) ( b c ) ( c a ) [ a b c ] b [ a b b ] c ( c a ) [ a b c ] b ( c a ) [ a b b ] [ a b b ][ b c a ] [ a b c ] + i 3 = r (cos + i sin ) Þ r = and = π 3 + i 3 = (cos π 3 + i sin π 3 ) ( + i 3) n = n (cos nπ 3 + i sin nπ 3 ) () i 3 = (cos π 3 i sin π 3 ) ( i 3) n = n (cos n nπ 3 i sin nπ 3 ) () + Þ ( + i 3) n + ( i 3) n = n+ cos nπ 3. Respective intercept form c Were OA ct; OB t P ct, c t is any point on te rectangular yperbola xy c. Tangent at P ct, c t is x yt ct x y ct c t () () () () () () () () () () () () () () ()

3 (i) f(x)=x+cosx f (x)=-sinx limits -sinx f (x) Status [, π ) + > Increasing ( π, π) + > Increasing SOLUTION: T k l Þ logt log k log l T dt Þ dt dl Þ dl T l T T l T dl Þ T l..56 % 3. Substitute Q.NO ANSWER MARKS Q.NO ANSWER Te coordinates of AB, are 3x e 3cos x sin x / and. 3 / Mid point of AB is e 3cos sin c ct, t c ct, t t e 3cos sin 3 / wic is te point P. Tis sows tat te 3 e tangent is bisected at te point of contact.. 3 / 3 e 47 π/4 5(ii) I= x Sin x dx π/4 Let f(x)= x Sin x f(-x)=( x)sin ( x) = x Sin x = f(x) () f(x) is an odd function. I= () c, t AB ct, ax ax e e cos bx dx a cos bx bsin bx C a b / 3x a3, b 3x e e cos x dx 3cos xsin x 9 / () () () () 5 5 Caracteristic equation p 5 p p 4 p p p( p) ( p) ( p)( p) Þ p, C.F. Ae x x Be x P.I. e D 5D x x e f 4D 5 x 5 x e p q ~p ~q ~p v ~q (~p v ~q)vp T T F F F T T F F T T T F T T F T T F F T T T T x x x x e e 3 3 x x x x y C.F.+P.I. Ae Be e 3 MARKS () () () () () ()

4 Q.NO ANSWER MARKS Q.NO ANSWER MARKS a) d.f f(x) = d x, (F(x)) = {x dx elsewere (i)p(.5< x <.75) = F(.75)-F(.5) =.75.5 =.35 (ii) P(x.5) = F(.5) =.5 =.5 (iii) P(x>.75)=-P(X.75) = F(.75) =.75 =.4375 P(X < 4) =.;P(X > 9) =. P(X < 4) + P(4 < X < 9) + P(X > 9) =. + P(4 < X < 9) +. = P(4 < X < 9) =.. =.8 Te number of students scored between 4 and 9. = 8.8 = 64 if i is a root ten + i also is a root sum of te roots ; + i + i = 4. Product of te roots ( + i )( i ) = 4 i = 5 6x 4 5x x 3x (x + 4 x + 5) (6x + p x + ) Equalting coefficient of x 5 p + 8 = 3Þ p =. 6x x Is an anoter factor x = 3, 56 Te matrix equation corresponding to te given system is x [ ] [ y] = [ ] λ 4 z AX = B [AB] = [ ] λ 4 ~ [ 4 ] R R R λ 4 λ λ R 3 R 3 λr ~ [ 4 ] λ λ λ R 3 R 3 + R ~ [ 4 ] () 3λ R 3 R 3 λr Case (i) : λ = ρ(a) = ; ρ[a, B] = ρ(a) = ρ[a, B] = < 3 () (Number of unknowns.) Te given system is consistent but as an infinite number of solutions. Case (ii) : λ ρ(a) = 3; ρ[a, B] = 3 () ρ(a) = ρ[a, B] = 3 ( Number of unknowns.) Te given system is consistent and as a unique solution. () 55b) Required roots are ± i, 3, p q ~p ~q ~p v ~q T T F F F T F F T T F T T F T F F T T T p q p q ~ (p q) T T T F T F F T F T F T F F F T From & ~ (p q) ~p v ~q 57 Solution: Draw a circle of radius unit wit centre at O. Take te points P and Q on te circle suc tat XOP = A, XOQ = B POQ = POX QOX = A B Clearly te coordinates of P and Q are (cos A, sin A ) and (cos B, sin B ). Take te unit vectors i and j along x and y axes respectively. OP = cos A i + sin A,OQ j = cos B i + sin B j

5 58 By te value, we ave Q.NO ANSWER MARKS Q.NO ANSWER MARKS By definition of cross product, we ave 59 x 4 x 3 + x x + = () OQ OP = OQ OP sin(a B) k Multiply bot sides by x + we get, (x + )(x 4 x 3 + x x + ) = = ()() sin(a B) k x 5 + = x 5 = = sin(a B) k () i j k OQ OP = cosb sin B cosa sin A = (sin k A cosb cos A sin B) From () and sin(a B) = sin A cos B cos A sin B Solution: a = i j 3, k u = 3 i + j 4, k v = i 3 j + k Te required equation is r = a + s u + t v r = i j 3 k + s(3 i + j 4 ) k +t( i 3 j + ) k Cartesian form: (x, y, z ) = (,, 3); (l, m, n ) = (3,, 4) (l, m, n ) = (, 3,) Te equation of te plane is x x y y z z l m n l m n = x y + z = 3 (x )(4 ) (y + )(6 + 8) + (z + 3)( 9 4) = (x )( 8) (y + )(4) + (z + 3)( 3) = 8x + 6 4y 4 3z 39 = 8x 4y 3z 37 = 8x + 4y + 3z + 37 = () () 6 x = ( ) 5 = (cosπ + isinπ) 5 = (cos(kπ + π) + isin(kπ + π)) 5 kπ + π kπ + π = cos ( ) + isin ( ), k 5 5 =,,,3,4 = cos(k + ) π 5 + isin(k + ) π 5, k =,,,3,4 Te roots of x 5 + = are cis π 5, cis 3π 5, cis 5π 5 = cisπ =, cis 7π 5, cis 9π 5 Removing te root, we get, cis π 5, cis 3π 5, cis 7π 5, cis 9π 5 are te required roots. Equation of te cable is x = 4ay. Let AB and CD be te towers. Since te distance between te two towers is 4 m. VA = 5 m ; AB = 55 m A B = 55 5 = 5 ft Tus te point B is (, 5) () = 4a(5) 4a = = 8 5 Equation of te cable is x = 8y Let PQ be te vertical distance to te cable from te pole RQ. RQ = 3, RR = 5 R Q = 5 Let VR be x Q is (x, 5) Q is a point on parabola x = 8 5 = 4 5 x = 5 = PQ = x = m. () () ()

6 Q.NO ANSWER MARKS Q.NO ANSWER MARKS 6 Created 9x 6y 8x 64y + 89 = (9x 8x) + ( 6y 64y) = 89 9(x x) 6(y + 4y) = 89 9(x x + ) 6(y + 4y + 4 4) = 89 9(x x + ) 9 6(y + 4y + 4) + 64 = 8 9 9(x ) 6(y + ) = (x ) 6(y + ) = 44 (y + ) (x ) = 9 6 Y 9 X 6 = Here X = x ; Y = y +. () a = 6, b = 9 e = + b a e = ae = = 5 = 5 6 = ax by ; a x b y By Cramers Rule Referred to X, Y Referred x=x+ () y=y- () () center C(,) C(, ) Were mm=- () 66 Foci F (, ae) = F (,5) F (, ae) = F (, 5) F (,3) F (, 7) () () Vertices eccent ricity A(, a) = A(,4) A (, a) = A (, 4) e = 5 4 A(,) A (, 6) e = 5 4 () 63

y = x ( u y ) = ()y 3 ( x 3 ) = y 3 + x 3 () u y x = y ( u x ) = y 3 + ()x 3 = y 3 + x 3 u (), x y = u y x To find te volume of a cone wit abse radius r and eigt, revolve te area of a triangle wose

7 Q.N O ANSWER MARK S Q.NO ANSWER MARK S x and x axis, about x axis Volume of te cone V V r x dx y dx 3 3 r r x r V x dx r x y Solution: u xy yx y x u x = ()y y( x 3 ) = y + yx 3 u y = x( y 3 ) ()x = xy 3 x u x y = x ( u y ) = ()y 3 ( x 3 ) = y 3 + x 3 () u y x = y ( u x ) = y 3 + ()x 3 = y 3 + x 3 u (), x y = u y x To find te volume of a cone wit abse radius r and eigt, revolve te area of a triangle wose vertices are (,), (,) and (,r) Equation of te line joining (,) and r (,r) is y x Te volume of te cone is obtained by revolving te area bounded by r y x, x (4) 66 b y = sin x dy = cos x dx S.A. = πy + ( dy dx ) a π dx = π sin x + (cos x) dx Let t = cos x dt = sin x dx sin x = dt Cange of limits: x t S. A. = π + t ( dt) b = π + t dt [ f(t)dt = f(t)dt] = a b = π + t dt a π + t dt [ + t is an even ] = 4π [ t + t + log (t + + t )] = π[ + log( + ]

8 68 69 Q.N ANSWER MAR Q.NO ANSWER MAR 67 (x 3 + 3xy )dx + (y 3 + 3x y)dy = x 3 dx + 3xy dx + y 3 dy + 3x ydy = x 3 dx + y 3 dy = 3xy dx 3x ydy x 3 dx + y 3 dy = 3xy(ydx + xdy) x 3 dx + y 3 dy = 3 xy(ydx + xdy) x 3 dx + y 3 dy = 3 (xy)d(xy) x y4 (xy) = 3 + c 4 Multiplying by 4 on bot sides, we get x 4 + y 4 + 6x y = 4c x 4 + y 4 + 6x y = k were k = 4c Solution: Let x be te principal at time t. dx dx x dt dt = 4x x(t) = ce 4t t = x() = = ce c = x(t) = e 4t x = t =? = e 4t e 4t = = 4t = log e t = log e 4 = = = 7.3 After 7 years will te amount be twice te original principal. CLOSURE AXIOM : z z = z z =.=z z Closure Axiom is True. ASSOCIATIVE AXIOM : Te multiplication of complex numbers is always associative. IDENTITY AXIOM : =.Z=Z.=Z So tat is te identy element. INVERSE AXIOM : (/z).z=z.(/z)= /z is inverse of z. So M is a group under multiplication of Complex Numbers. (4) 7 Solution Semi-major axis CA is a = 9.9 million miles. Given e =.7. Te closest distance of te eart from te sun = F A and fartest distance of te eart from te sun = F A ae = 36 6 = 7 46 F A = CA CF = a ae = = F A = CA + CF = a + ae = = Let X be te number of incoming buses. Mean = λ =.9 (i) Mean number of buses during a period of 5 minutes is 5λ i.e., Mean = 5. 9 = 4. 5 P(X = 9) = e 5λ (5λ) x x! = e 4.5 (4.5) 9 9! (ii) Mean number of buses during a period of 8 minutes is 8λ i.e., Mean = 8. 9 = 7. P(X < ) = e 7. (7.) x x! x= (ii) Mean number of buses during a period of minutes is λ i.e., Mean =. 9 = 9. 9 P(X > 4) = e 9.9 (9.9) x Created By : 9 x=4 3 x! = e 9.9 (9.9) x x! x= S.Manikandan., Viceprincipal, Joti Vidyalaya matric Hr. sec. scool., Elmapillai., Salem ()

f a h f a h h lim lim

f a h f a h h lim lim Te Derivative Te derivative of a function f at a (denoted f a) is f a if tis it exists. An alternative way of defining f a is f a x a fa fa fx fa x a Note tat te tangent line to te grap of f at te point