Marking Scheme (Sample Paper II) Section A: FF (N) F (F (N) ) F (n1) N 1 1 N
Marking Scheme (Sample Paper II) Section A: FF (N) F (F (N) ) F (n1) N 1 1 N
Marking Scheme (Sample Paper II) Section A: FF (N) F (F (N) ) F (n1) N 1 1 N
Section A
Q1.
24
25
(1)
0 1 3
Q3. 1 0 2 (1)
3 2 0
Q4. 4%
(1)
Q5. 5
(1)
Q6. -14
(1)
Section B
(1+1/2)
(1)
OR
Let (a, b) N N. Then
(1)
(1)
Let (a, b), (c, d), (e, f ) N N be such that (a, b)R(c, d), (c, d)R(e, f ).
a 2 d 2 b 2 c 2 and c 2 f 2 d 2 e 2
a 2 d 2 c2 f 2 b2 c2 d 2 e2
a 2 f 2 b2 e2
(a, b)R(e, f )
Hence, R is transitive.
(1+1/2)
tan 1
x
2x
2sin 2
2 cos
1 cos x 1 cos x
2
2
1
tan
1 cos x 1 cos x
x
2x
2sin 2
2 cos
2
2
(1)
x
x
2 cos 2 sin
3
x 3
x
x
2
2
tan 1
( x
cos 0,sin 0) (1+1/2)
x
x
2
2 2 4
2
2
2 cos 2 sin
2
2
x
1 tan
2 tan 1 (tan( x ))
tan 1
x
4 2
1 tan
2
=
4 2
(1)
)
4 4 2
2
(1/2)
4 3 4 1
1 2 3 2
Q9. adjA
(2)
6
11 0
1 0
(adjA)A
11
0 11
0 1
(1/2)
11 0
1 0
A(adjA)
11
0 11
0 1
(1/2)
2 1
11
3 4
(1/2)
(1/2)
1 1 p
1 p q
1 1 p 1 p q
1
1 p (R 2 R 2 3R1, R 3 R 3 4R1)
Q(10) LHS = 3 4 3p 2 4p 3q 0
4 7 4p 2 7p 4q 0
3
2 3p
(2)
1 1 p 1 p q
1
1 p (R 3 R 3 3R 2 )
= 0
0
0
1
(1)
(1)
OR
0 2 3
0 2 3
4 2 0 4 (interchanging rows and columns)
Let 2 0
3 4 0
3 4 0
(1 + 1/2)
0 2 3
(1)(1)(1) 2 0 4
3 4 0
(1 +1/2))
(1/2)
2 0 0
(1/2)
7
2 0
2I
0 2
Q11. AB
(1/2)
2 3
1
1
A( B) I A 1 B
2
2
1 2
(1)
X (A)1C (A1)C
(1)
2 1 4 7
x 7, y 10
3 2 1 10
(1)
x 2
x 2
x 2
(1/2)
(1+1/2)
Since, f is differentiable at x = 2,
f (2 h) f (2)
f (2 h) f (2)
lim
(h 0)
h
h
h 0
h 0
(2 h)2 4
a(2 h) b 4
lim
lim
h
h
h 0
h 0
4 ah 4
lim (h 4) lim
4a
h
h 0
h 0
(1+1/2)
b = -4
(1/2)
Q13.
du
(log x) x
log(log x)
dx
log x
1 du
1
log(log x)
u dx log x
(1+1/2)
1 dv x cos x
(1+1/2)
dy du dv
(log x) x
log(log x)
dx dx dx
log x
x cos x
x
cos
x
cos
x(log
x)
x
sin
x
log
x
yuv
(1)
OR
dy
dx
dy
dy dt
b
ap cos pt, bpsin pt,
tan pt
dt
dt
dx dx
a
dt
(1+1/2)
d2 y
b
dt
psec2 pt
a
dx
dx
(1+1/2)
d2 y
dx
d2 y
dx
(1/2)
a cos3 pt
)
2 t 0
(1/2)
a2
sin x
sin x sin 2x dx sin x(1 2cos x) dx (1 cos x)(1 cos x)(1 2cos x) dx
dt
(cos x t sin xdx dt)
(1 t)(1 t)(1 2t)
9
(1)
1
A
B
C
(1+1/2)
1
1
4
log 1 t log 1 t log 1 2t c
6
2
6
1
1
2
log 1 cos x log 1 cos x log 1 2 cos x c
6
2
3
(1+1/2)
OR
sin
sin 2cos 3
2
sin
cos 2cos 4
2
( 5)2 (t 1)2
sin 1
sin
1 cos 2cos 3
2
1
t 2t 4
2
(1/2)
(1)
(1+1/2)
dt
t 1
cos 1
c sin 1
c
5
5
(1)
Q15. Let
2x(1 sin x)
1 cos x
2
0 2
2x sin x
2
0 1 cos x
dx
2x
2x sin x
1 cos2 x dx 1 cos2 x dx
dx
10
(as
2x
1 cos 2 x
is odd and
2x sin x
1 cos 2 x
is even)
x sin x
2
0 1 cos x
(1)
dx .
Let
x sin x
I1
2
0 1 cos x
I1
dx
0
( x) sin x
( x) sin( x)
1 cos 2 x
Adding, 2I1
1 cos 2 ( x)
dx
sin x
2
0 1 cos x
dt
dx
(1)
dx
1 t2
(1)
tan 1 t
I1
(1/2)
2
. Hence, I 2
4
(1/2)
x 2 y2
dy y
dy y
y
y
, x 0 or,
1 ( ) 2 f ( ), hence, homogeneous.
dx x
x
dx x
x
x
Put y = v x
(1/2)
dy
dv
dv
vx
v 1 v 2 (1)
. The differential equation becomes v x
dx
dx
dx
11
or,
dv
1 v
dx
x
(1/2).
(1)
log v 1 v 2 log x k v 1 v 2 x k
v 1 v 2 kx
y
y
1 ( ) 2 cx
x
x
y x 2 y 2 cx 2 ,
which gives the general solution.
(1)
dx (tan 1 y x) dx
x
tan 1 y
,
0r,
dy
dy 1 y2 1 y2
1 y2
which is linear in x
I.F. e
1
1 y2
dy
(1/2)
1
e tan y
(1)
xe tan
xe tan
1 y
1 y
e t tdt (tan 1 y t
te t e t c xe tan
1 y
1
1 y2
1 y
e tan
1 y
tan 1 y
1 y2
dy (1/2)
dy dt)
tan 1 ye tan
1 y
e tan
1 y
c
(2)
r 2i 3j (i 2j 3k)
r i 2j k (2i 3j 4k),
a1 i 2j k, b1 2i 3j 4k, a 2 2i 3j, b 2 i 2j 3k
(1)
12
i
j
b b 2 3
a 2 a1 3i j k,
1
2
1 2
k
4 17i 10j k
3
(a 2 a1 ).(b1 b 2 )
b1 b 2
42
units
390
(2)
(1/2)
(1/2)
Q19. Let us define the following events: E = A solves the problem, F = B solves the problem, G =
C solves the problem, H = D solves the problem
(i) The required probability = P(E
= 1 - P(E
F G
F G H)
(1/2)
H)
(1/2)
(1)
2 3 4 1 13
3 4 5 3 15
(1/2)
P(E) P(F) P(G) P(H) P(E) P(F) P(G) P(H) P(E) P(F) P(G) P(H)
P(E) P(F) P(G) P(H) P(E) P(F) P(G) P(H)
(1)
2 3 4 1 1 3 4 1 2 1 4 1
3 4 5 3 3 4 5 3 3 4 5 3
2 3 1 1 2 3 4 2 5
3 4 5 3 3 4 5 3 18
(1/2)
Section C
Q20. f (x) (x 1)2 (x 2)(5x 4)
(1/2)
13
f (x) 0 x 1, 2,
4
5
(1/2)
In the interval
Sign of f(x)
(, 2)
(+ve)(-ve)(-ve)= +ve
f is strictly increasing in
, 2
4
( 2, )
5
(+ve)(+ve)(-ve)= -ve
4
( ,1)
5
(+ve)(+ve)(+ve)= +ve
(1, )
(+ve)(+ve)(+ve)= +ve
f is strictly decreasing in
2, 5
f is strictly increasing
4
5
in ,1
f is strictly increasing in
1,
(2+1/2)
(1/2)
In the left nhd of -2, f(x)>0, in the right nhd of -2, f(x)<0 and f(-2) = 0, therefore, by the
first derivative test, -2 is a point of local maximum.
(1)
In the left nhd of -4/5, f(x)<0, in the right nhd of -4/5, f(x)>0 and f(-4/5) = 0, therefore, by
the first derivative test, -4/5 is a point of local minimum.
(1)
Q21. We have
(1)
Let a (b c)
(1)
Then
a (b c)
a
(b c)
a 2(b c)
sin
6
2 2
(2)
Now a b
b c c a a b b c . c a a b .c b c .a ( As the scalar
(1+1/2)
1
2a.( a) 1
2
(1/2)
4y x 2 4
dy
dy x
dy
2x
1
dx
dx 2 dx x 2
(1)
(1)
Graph sketch
15
(1)
dx
dx
1
dx
3
dx
xdx
3 2
2
3
2
2
3
2
3
2
3
(1)
x2
1 x2
1
6
x 3x x 2
2
2 3 2
2
3 4
(1+1/2)
= 1 square units
(1/2)
Q23. The equation of the line passing through the point(3, -2, 1) and parallel to the given line is
x 3 y 2 z 1
2
3
1
(1)
(1/2)
(1)
(1/2)
(1)
The equation of the line passing through (3, -2, 1) and perpendicular to the plane is
x 3 y 2 z 1
3
1
1
(1/2)
Any point on it is (3 3, 2, 1)
(1/2)
3 22 15
,
, )
7 7 7
OR
16
8
7
(1/2)
(1/2)
1 ( r.(i j 2k))
0
r.(2i 3j k)
1
or, r.((2 )i (3 )j (1 2)k)
(2)
If it contains the point (3, -2, -1), we have
2
3
(1)
2
2
4
3
r.((2 )i (3 )j ( 1 )k)
1or, r.(4i 7 j k)
3
3
3
(1)
If be the angle between the normals to the two given planes, then is the angle between
the planes and cos
n1.n 2
23 2
7
n1 n 2
14 6 2 21
(2)
Q24. Let us define the following events: E1 = Two white balls are transferred, E2 = Two red balls
are transferred, E3 = One red and one white balls are transferred, A = The ball drawn from
the Bag II is red
P(E1 )
P(E 2 )
P(E3 )
C2
C2
(1/2)
43
98
(1)
45
98
(1)
C2
C2
C1 4 C1
9
C2
4 5 2
98
(1)
17
P(A / E1 )
3
5
, P(A / E 2 ) ,
8
8
P(A / E 3 )
4
8
1 1 1
( )
2 2 2
P(E3 ) P(A / E3 )
P(E1) P(A / E1) P(E 2 ) P(A / E 2 ) P(E3 ) P(A / E3 )
(1/2)
= 20/37
(1/2)
OR
(1/2)
1 5
125
P(X 0) P(r 0) 3 C0 ( ) 0 ( )3
6 6
216
(1/2)
1 5
75
P(X 1) P(r 1) 3 C1 ( )1 ( ) 2
6 6
216
(1/2)
1 5
15
P(X 2) P(r 2) 3 C 2 ( ) 2 ( )1
6 6
216
(1/2)
1 5
1
P(x 3) p(r 3) 3 C3 ( )3 ( ) 0
6 6
216
(1/2)
xi
pi
xipi
(xi)2pi
125/216
75/216
75/216
75/216
15/216
30/216
60/216
1/216
3/216
9/216
1/2
2/3
Total
(2)
18
Mean =
xi pi 2 , var(X) xi2pi ( xi pi )2 3 4 12
(1)
15
6
(1/2)
Q25. Let the radius of the circular garden be r m and the side of the square garden be x m. Then
600 = 2r 4x x
600 2r
4
(1)
600 2r
dA
2
dA
300
2r (600 2r)(2) (4r 300 r),
0r
dr
16
2
dr
4
d 2A
dr 2
d 2A
(4 ), ( 2 ) 300 0
2
dr r
(1)
(1)
(1)
300
For this value of r, x 2r
4
(1)
To achieve any goal, there is every possibility that energy, time and money are required to
be invested. One must plan in such a manner that least energy, time and money are spent.
A good planning and execution, therefore, is essentially required.
(1)
Q26. Let the number of pieces of model A to be manufactured be = x and the number of pieces
of model B to be manufactured be = y.
(1/2)
(1/2)
(2)
19
Profit
(0,0)
Rs 0
(20,0)
Rs 160000
(12,6)
Rs 168000 (maximum)
(0.10)
Rs 120000
(1+1/2)
(1)
The number of pieces of model A =12, the number of pieces of model B =6 and the
maximum profit = Rs 168000.
(1/2)
20