1285053190 CBSE SamplePaper-1 XII Maths 2010 QuesSols0

8/3/2019 1285053190 CBSE SamplePaper-1 XII Maths 2010 QuesSols0

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=> I+(y-b)----.r+ =0' (d)'dx' dx => (y-b) = -

[1+( dy ) ' 1from (ii),

dx dy(x-a) = + . -d'y dx

putting in (i) to get

orI+ (X)'

d2ydx'

2

dx'

1+( , ( : ~ ) +d2ydx '

[1+( : ~ ) ' 1d' ydx '

1 + ( : ~ ) 2 ,

d'ydx '

19. Given differential equation can be written as~ 1 +y' ) +xy dy = 0dx

.JI +x' Y=> dx + dy = 0X ~ 1 + y..f y - f ~ xdx~ x

----------------(111)

= 9

Putting I+y2 = u' and I+x2= v' to get ydy = udu and xdx = vdv

I IV-I I r:-:-T Iu = -v--Iog - +c or vi+y = -,,\+x - --log2 v+1 2

10

I

\I,

Y,

\I,

\I,

\I,


11/16

20. Let ii, band cbe unit vector such that ii+b = c

21.

22.

23 .

:. lii+bl = 1 1 = lii+bl ' = ii '+ b'+2ii b = 2+2ii b=> 2(a b) = 1-2 = -1 ------------------------------------(i)

1_' ,- , -Nowii-b =a +b -2iib= 1+1-(-1) =3

II,

Given lines are r = (1+2A.); + (-I +).)J - kand

If lines are intersecting, then for some value of A and1+2A. = 2+).t, --0) -I +A = -l+).t --(ii) -1 = - ~ l --(iii)Solving (ii) and (iii) to get). = I, ).t = I ,which satisfY (i) hence the line are intersectingand point of intersection is (3, 0, -1)Let X denotes the random variable, 'number of green balls,X: 0

P(X) : SC 3 Sc , 4c,9c3 9c]5 10

42 21

IAI = 2(-1) - 1(4) + 3(1) = -3The cofactors are

2 3Sc , 4c, 4c39c 1 9c3S14 21

SECTIONC

OA" 1 d''f. = -aJAIAIAll = -I , AI2 = -4, AI3 = IA2I = S, An = 23, A2l = -I IA" = 3, A" = 12 , A" = -6

.. A" = - ~ ( ~ I ~ l 1 -II -6

II

I V,

I

2

2


12/16

Given equations can be WTitten as

:. X =A ' B

.. [ ~ ] = :2] [ ~ ] = ( - ; ~ ] z 1 -11 -6 2 14:. x = -6 , y=-27, z= 14

OR

-1 2] [ 1 - 12 -3 then 0 2-2 4 3 - 2

2] [1 0 ]3 = 0 1 0 A4 0 0 1

(1 1 2] [1 00]R, -7 R, -3R, 0 2 -3 = 0 lOAo +1 -2 -3 0 1

(1 1 2] [1 0 ]R, H R, 0 1 -2 = -3 0 1 Ao 2 -3 0 1 0

o 0]o 1 A1 -2

1 1 2] [1R, -7 R, -2R, 0 1 -2 = -3o 0 1 6[1 0 0] [-2R, -7 R, +R, 0 1 -2 = -3o 0 1 6

o 1]o 1 AI -2

[1 0 0] [-2 01]R, -7 R, + 2R, 0 1 0 = 9 2 -3 Ao 0 1 6 1 -212

1

y,

y,

1


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24.-2

D c AD =DC = BC = IOem.L1 ADM =L1BCN :. AM = BN = x (say)

Area (A) = .!. (IO+IO+2x) -JIOO -x ' = .!. (I0+x) -JIOO-x'2 2Let S =A' = (lO+x)' (100-x' )

= 0 => (I0+X)2 (-2x) + (I00-X 2) 2(10+x) = 0d'((I0+X)2 (-2x+20-2x) = 0 => x = 5d' s- , = (lO+X) 2 (-4) + (20-4x) 2(10+x) < 0 at x = 5dx-:. fo r Maximum area, x = 5Max imum area = 15m = 7513 em'

25. Correct figureSolving x2+y2 =4 and (X_2)2+y2 = 1

7we get x = -4

[, 7/ 4 ]: . Required area =2 JJJ4-x' dx + J - (x-2)' dx

7/4 ,

x , . ., x x-2 2 1 . . , 4[, 2]= 2 h -J4-x- +2sm + ["2 + '2 SII1 '(X-2)1

5" .Jls . . (1) , 7)2 -2 -Sin '4 -4 sin' '8 sq.u13

y,

y,

y,

x=7f4


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26. b-a IHere f(x) = (x 2+x+2) h = - = -, n n

f Lim If(x)dx = - [QI ) + f(l+h)+f(l +2h) + ............ .. + fll+(n-l)hJ]I n= Lim I [ " , 'J. 4+(4+3h+h ) + (4+6h+4h ) + .............. + (4+(n-I)3h+(n-l) hn ~ o o n 2= - 4n+- +im I [ 3 n(n-I) I n(n-I)(2n-I)]n --7 00 n n 2 n' 6= Lim [ 4 + ~ ( 1 - . ! . ) + .!.(1-.!.)(2-.!.)]11--700 2 n 6 n n y,

3 1= 4+-+ 24+9+2 - -52 3 6 6OR

put x = sina and .[;, = s i n ~ :. i n ' [ x ~ -.[;"h-x ' = sin" [sina c o s ~ - cosa s i n ~ l

. "[ . (A) A .,' . ,I r= Sin Sin a-I-' = a-I-' = Si n X - Sin "X y,

, ":. Given integral =f sin"x - sin"'[;')dx =f in"x dx - f in".[;, dxo 0 0, ,[ ., J' I f 2x [, ,r]' f 1 I= xsm' x -- r:-? dx - x'sm' "X + r'x dx

, 0 2 0 " I -x ' 0 0 " I -x 2"x

1

= -l+.!. J-M2tdt2 , t [1- x = t ' , dx = -2t dtJ= -1+ fMdt =1+[!..M + .!. sin"t]' = (_1 + 11:)a 2 2 a 4

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27. Let Q be the foot of perpendicular from P to the plane and P'(x, y, z) be the image ofP in theplane... The equations of line through P and Q is

x-I y-3 z-42 -1 1

The coordinates of Q (for some value of U are(21,,+I , -H3 , H4)Since Q lies on the plane, :,2(21,,+1) - 1 ( H 3 ) + ( H 4 ) + 3 ~ 0 Solving to get ), -1:. coordinates of foot of perpendicular (Q) are (-I, 4, 3)

2x-y+z+3=O

Perpendicular distance (PQ) = ~ ( 2 ? + ( - I ? + ( l Y = 16 unitsSince Q is mid point of PP'

x+1 y+3 z+4.. --:2 = -1 , --:2 =4 , --:2 =3 =!> x= -3 ,y=5 , z=2.. Image ofP is (-3 ,5,2)

II

P(1,3.4)

Q

: P'(X,y,z)

28 . Let, number of executive class tickets to be sold, be x and that of economy class be y.:. LPP becomes: Maximise Profit (P) 1000x + 600ySubject to : x 0, 0

x=20

y,y,y,

II

1

x+ y $ 200 200 nl,

For COlTect graph

y 4x or 4x-y $ 0x ~ 2 0

Getting vertices of feasible region asA( 20, 180 ), B(40, 160), C(20, 80)y,Profit at A Rs. 128000Profit at B Rs. 136000Profit at C Rs. 68000

A(20, 180)y=4x160 8(40, 160)

120

80

40 80 120 160 200 xx+y=200

:. Max profit = Rs. 136000 for 40 executive and 160 economy tickets

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29. Let the events be defined as :E, : Bag A is selectedE, : Bag B is selectedE, :Bag C is selectedA : A red ball is selected

I 2 1 3 1:. PCE,) = 6' PCE,) = 6 = "3 and PCE,) = 6 = "2

= 90293

I

2

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1285053190 CBSE SamplePaper-1 XII Maths 2010 QuesSols0