Cross-border Insolvency Problems - Is the UNCITRAL Model Law the Answer.pdf
12-Maths-CBSE-Exam-Papers-2015-Allahabad-Set-1-Answer.pdf
-
Upload
geeteshgupta -
Category
Documents
-
view
212 -
download
0
description
Transcript of 12-Maths-CBSE-Exam-Papers-2015-Allahabad-Set-1-Answer.pdf
-
2QUESTION PAPER CODE 65/1/RUEXPECTED ANSWERS/VALUE POINTS
SECTION - A
1.333
yxzzyxzyxzyx
m
= 0 m
2. order 2, degree 1 (any one correct) msum = 3 m
3. ycotx.y12y
dydx
2 m
Integrating factor = 2y1log y1ore 2 m
4. a2ccbba22cba2cba2 2222 m
6cba2 m
5. ji011111kji
b a m
unit vector is 2j
2i m
6.10110
3z
15115
7y
515
3x
m
Direction cosines are 73,
72,
76 or 7
3,72,
76 m
Marks
-
3SECTION - B
7.
390002300030000
402050
15040050075250300
1003004002 m
cost incurred respectively for three villages is Rs. 30,000, Rs. 23,000, Rs. 39,000 1 m
One value : Women welfare or Any other relevant value 1 m
8.
318tan1x1x1
1 x 1 xtan 11 2 m
318
x22x
2 4x2 + 31x 8 = 0 1 m
,41x 8 (Rejected) 1 m
OR
L.H.S. =
zx1
xztanyz1zytan
xy1yxtan 111 2 m
RHS0xtanztanztanytanytanxtan 111111
2 m
9.ccbbabba
cacaabc
cbcbabacbaba
cacbca
22
22
22
Taking a, b & c common from C1 , C2 and C3 1 m
ccbcbabba
caccaabc2
-
4 3211 C CCC and taking 2 common from C1 1 m
bcbcbbbba
0ccaabc2
133 CCC 1 m
bcbcb
0cca0cca
abc2
322 RRR m
Expand by C3, = 2 abc ( b) ( ac c2 ac + c2) = 4a2 b2 c2 m
10. Adj A = 27A;366636
663
2+1 m
A. Adj A = 3IA100010001
27366636
663
122212221
1 m
11. f (x) = 1x1x
L 2
1x1x2lim
1x21x1xlim(1)f
1)(x1)(x
1 m
R 01x0lim
1x21x1xlim(1)f
1)(x1)(x
1 m
(x)f02 is not differentiable at x = 1
L 01x
0lim1x
21x1xlim(1)f 1x1x
1 m
R 21x1x2lim
1x21x1xlim(1)f
1x1x
1 m
(x)f20 is not differentiable at x = 1
-
512. 2xsinm
xsinm
x1em
dxdyget wex, w.r.t.atedifferenti,ey
11
1 m
dxdyx1 2 = my, Differentiate again w.r.t. x
dxdym
dxdy
x1x
dxydx1
22
22 1 m
(my)mdxdyx1m
dxdyx
dxydx1 22
22
m
0ymdxdyx
dxydx1 22
22 m
13. f (x) = 32x(x)h,1x
1x(x)g,1x 22
Differentiating w.r.t. x, we get
2(x)h,
1xx2x1(x)g,
1xx(x)f 22
2
2
1+1+1 m
5
2(x)ghf m
14.
dxxx22x1dx
21x
232dxxx22x3 2
222 2 m
cxx232
23
21x
sin89xx2
221x
2 23212
2 m
or
cxx2
32
312xsin
49xx2
212x 23212
OR
-
6
dx
2x1
53dx
1x12x
51dx
2x1x1xx
22
2
2 m
dx2x1
53dx
1x1
51dx
1x2x
51
22 m
c2xlog53xtan
511xlog
51 12 1 m
15. 4
04
4
03 dxtan x2xcos
1dx x2sin 2xcos
11 m
=
4
0
22
dxxsecxtan2xtan1
1 m
= 1
0
2dt
tt1
21 Taking, tan x = t; 1 m
= 1
0
25
t52t2
21
m
= 56
522
21
m
16. dx1x1
x1
1x1xlogdx
1x1xlog 2 2 m
= dx1x1dx1
1xxlog
x 1 m
= c1)(xlogxlog1xxlog
1 m
or c1xxlog1x
xlog
-
717. k5j5ibc;kjiba 1 m
k4j4551
111kji
bcba
1 m
Unit vector perpendicular to both of the vectors = 2k
2j 1 m
18. let the equation of line passing through (1, 2, 4) be
kcjbiak4j2ir 1 mSince the line is perpendicular to the two given lines
3a 16 b + 7 c = 0 1 m 3a + 8 b 5 c = 0
Solving we get, 6c
3b
2aor
72c
36b
24a
1 m
Equation of line is : k6j3i2k4j2ir mOR
Equation of plane is : 011112212
z2y1x
3 m
Solving we get, x + 2y + 3z 3 = 0 1 m
19. Let x = No. of spades in three cards drawnx : 0 1 2 3 1 m
P(x) :
6427
433
3C0
6427
434132
C1
649
43
4132
C2
641
43
41303
C3
2 m
x . P(x) : 0 6427
6418
643 m
Mean = 43
6448P(x)x m
-
8OR
let p = probability of success ; q = Probability of failure
then, 9 P(x = 4) = P(x = 2)
42262446 qpCqpC9 2 m
3pqq9p 22 1 m
Also, p + q = 1 p + 3p = 1 p = 41 1 m
SECTION - C
20. f : 5
35y54(y)f;96x5xf(x);)9,R 12
y95
35y5465
35y545(y)fof2
1
3 m
x5
396x5x554(x)f of2
1
2 m
Hence f is invertible with 5
35y54(y)f 1 m
OR
(i) commutative : let x, y then1R
x * y = x + y + xy = y + x + yx = y * x * is commutative 1 m
(ii) Associative : let x, y, z then1R
x * (y * z) = x * (y + z + yz) = x + (y + z + yz) + x (y + z + yz) = x + y + z + xy + yz + zx + xyz 1 m
(x * y) * z = (x + y + xy) * z = (x + y + xy) + z + (x + y + xy) . z = x + y + z + xy + yz + zx + xyz 1 m
x * (y * z) = (x * y) * z * is Associative
-
9(iii) Identity Element : let 1Raaa*ee*athatsuch1Re m
a + e + ae = a e = 0 m
(iv) Inverse : let a * b = b * a = e = 0 ; a, b 1R m
a + b + ab = 0 a1aaor
a1ab 1
m
21. Solving the two curves to get the points of intersection 8,p3 1 m
m1 = slope of tangent to first curve = 9p2x 1 m
m2 = slope of tangent to second curve = p2x
1 m
curves cut at right angle iff 1p2x
9p2x
m
)p3x(Put4x9p 22
p)(949p2
p = 0 ; p = 4 1 m
22. correct figure 1 m
ar (ABDOA) 316
12dy
41
4
0
34
0
2
yy ......(i) 1 m
ar (OEBDO) 4
0
323
4
0
24
0 12xx
34dx
4xdxx2
316
316
332
.........(ii) 1 m
ar (OEBCO) 3
1612xdxx
41
4
0
34
0
2
................(iii) 1 m
From (i), (ii) and (iii) we get ar (ABDOA) = ar (OEBDO) = ar (OEBCO)
-
10
23.1
xy
xy
dxdy
xxyy
dxdy
2
2
2
, Hence the differential equation is homogeneous 1 m
Put y = 1dxdxgetwe,
dxdx
dxdyandx
2
vvvvvvv 1+1 m
1
1dxdx
2
vvv
vvv
1 m
cxloglogdxx1d1 vvvv
v 1 m
cylog
xyor,cxlog
xylog
xy
1 m
OR
Given differential equation can be written as 21
2 y1ytanx
y11
dydx
1 m
Integrating factor =
dyy1eytanex:issolutionande 2
ytan1ytanytan
111 1+1 m
x t)ytan(wherec1)y(tanecetedttee 11ytantttytan 11 1 m
x = 1, y = 0 c = 2 21)y(taneex 1ytanytan 11 1 m
ytan1 1e21ytanxor
24. Equation of line through A and B is (say)6
5z1
4y13x
2 m
General point on the line is 564,3, 1 m
If this is the point of intersection with plane 2x + y + z = 7
then, 2756432 1 m
-
11
Point of intersection is (1, 2, 7) 1 m
Required distance = 7742413 222 1 m
25. Let the two factories I and II be in operation for x and y
days respectively to produce the order with the minimum cost
then, the LPP is :
Minimise cost : z = 12000 x + 15000 y 1 m
Subject to :
50x + 40y > 6400 or 5x + 4y > 640
50x + 20y > 4000 or 5x + 2y > 400 2 m
30x + 40y > 4800 or 3x + 4y > 480
x, y > 0
correct graph 2 m
Vertices are A (0, 200) ; B (32, 120)
C (80, 60) ; D (160, 0) m
z (A) = Rs. 30,00,000; z (B) = Rs. 21,84,000;
z (C) = Rs. 18,60,000 (Min.); z (D) = Rs. 19,20,000;
On plotting z < 1860000
or 12x + 15y < 1860, we get no
point common to the feasible region
Factory I operates for 80 days m
Factory II operates for 60 days
-
12
26. E1 : Bolt is manufactured by machine AE2 : Bolt is manufactured by machine BE3 : Bolt is manufactured by machine CA : Bolt is defective
;10020)P(E;
10050)P(E;
10030)P(E 321
1001)P(A/E;
1004)P(A/E;
1003)P(A/E 321 3 m
P(E2/A) = 3120
2020090200
1001
10020
1004
10050
1003
10030
1004
10005
2 m
P( E2/A) = 1 P(E2/A) = 3111 1 m