SAMPLE QUESTION PAPER
MARKING SCHEME
CLASS XII
MATHEMATICS (CODE-041)
SECTION: A (Solution of MCQs of 1 Mark each)
Q no.
ANS
1
(d)
 0 1  2  1 0
A 
 , A   0 1 .
1 0


2
(d)
 A  B
3
(b)
3 0 1
1
Area 
3 0 1 , given that the area  9 sq unit .
2
0 k 1
HINTS/SOLUTION
1
 B1  A1 .
3 0 1
1
  9  3 0 1 ; expanding along C 2 , we get  k  3.
2
0 k 1
4
(a)
Since, f is continuous at x  0 ,
therefore, L. H . L  R. H . L  f  0   a finite quantity .
lim f  x   lim f  x   f  0
x0
x0
 lim
x 0
5
(d)
 kx
 lim 3  3  k  3.
x 0
x
Vectors 2i  3 j  6k &6i  9 j  18k are parallel and the fixed point i  j  k on the



line r  i  j  k   2i  3 j  6k does not satisfy the other line

r  2i  j  k   6i  9 j  18k ; where  &  are scalars.


6
(c)
  dy  2   d 2 y 
The degree of the differential equation 1       2  is 2
  dx    dx 
7
(b)
Z  px  qy     i 
3
At
 3,0  ,
2
Z  3 p     ii  and at  1,1 , Z  p  q      iii 
From  ii  &  iii  , 3 p  p  q  2 p  q .
Page 1 of 19
8
(a)
 
Given, ABCD is a rhombus whose diagonals bisect each other. EA  EC and
 
EB  ED but since they are opposite to each other so they are of opposite signs




 EA   EC and EB   ED .
  
  
 EA  EC  O .....  i  and EB  ED  O ....  ii 





Adding (i) and (ii), we get EA  EB  EC  ED  O .
9
(b)
f  x   e cos x sin 3  2n  1 x
2
f   x  e
cos2   x 
sin3  2n  1  x 
2
f (  x )   e cos x sin 3 (2n  1) x
 f ( x )   f ( x )

So,  ecos x sin3 (2n  1) x dx  0
2

10
(b)
Matrix A is a skew symmetric matrix of odd order.  A  0.
11
(c)
We observe,  0,0  does not satisfy the inequality x  y  1
So, the half plane represented by the above inequality will not contain origin
therefore, it will not contain the shaded feasible region.
12
(b)


  a .b   18
3 j  4k .
Vector component of a along b   2 b 


25
 b 


13
(d)
adj  2 A    2 A  23 A
14
(d)

2


2

 26 A  26   2   28 .
2
2
Method 1:
1
1
Let A, B , C be the respective events of solving the problem. Then, P  A  , P  B  
2
3
1
and P  C   . Here, A, B , C are independent events.
4
Problem is solved if at least one of them solves the problem.
     
Required probability is  P  A  B  C   1  P A P B P C
Page 2 of 19
1 
1 
1
1 3

 1   1    1   1    1   .
2 
3 
4
4 4

Method 2:
The problem will be solved if one or more of them can solve the problem. The probability is

 
 
 
 
 

P ABC  P ABC  P ABC  P ABC  P ABC  P ABC  P  ABC 
1 2 3 1 1 3 1 2 1 1 1 3 1 2 1 1 1 1 1 1 1 3
 . .  . .  . .  . .  . .  . .  . .  .
2 3 4 2 3 4 2 3 4 2 3 4 2 3 4 2 3 4 2 3 4 4
Method 3:
Let us think quantitively. Let us assume that there are 100 questions given to A. A
1
 100  50 questions then remaining 50 questions is given to B and B solves
2
1
2
50   16.67 questions . Remaining 50  questions is given to C and C solves
3
3
solves
2 1
50    8.33 questions.
3 4
Therefore, number of questions solved is 50  16.67  8.33  75 .
So, required probability is
15
(c)
75 3
 .
100 4
Method 1:
ydx  xdy  0 
 x
ydx  xdy
1
 0  d    0  x  y  y  cx.
2
y
c
 y
Method 2:
ydx  xdy  0  ydx  xdy 
dy dx
; on integrating

y
x

dy

y

dx
x
log e y  log e x  log e c
since x , y , c  0 , we write log e y  log e x  log e c  y  cx .
16
(d)
Dot product of two mutually perpendicular vectors is zero.
 2  3   1   2  1  0    8.
17
(c)
Method 1:
2 x, x  0
f  x  x  x  
 0 ,x  0
There is a sharp corner at x  0 , so f  x  is not differentiable at x  0 .
Method 2:
Page 3 of 19
Lf '  0   0 & Rf '  0   2 ; so, the function is not differentiable at x  0
For x  0, f  x   2 x (linear function) & when x  0, f  x   0 (constant function)
Hence f  x  is differentiable when x    ,0    0,   .
18
(d)
 1  1  1
1
We know, l  m  n  1           1  3    1  c   3 .
c c c
c
19
(a)
d
3
2
f  x     x  1  x  3 

dx
2
2
2
2
2
2
2
Assertion : f  x  has a minimum at x  1 is true as
d
d
f  x    0,  x   1  h,1 and

 f  x    0,  x  1,1  h ; where,
dx
dx
' h ' is an infinitesimally small positive quantity , which is in accordance with
20
the Reason statement.
Assertion is false. As element 4 has no image under f , so relation f is not a function.
(d)
Reason is true. The given function f : 1, 2, 3   x , y , z , p is one – one, as for each
a  1, 2,3 , there is different image in  x , y, z , p under f .
Section –B
[This section comprises of solution of very short answer type questions (VSA) of 2 marks each]
21
  33  
3 

 3
1
1
sin1  cos 
   sin cos  6 
  sin cos 
5 

 5
  5 

21 OR
22
23

2


1

  3 
1
  sin sin  


2 5 
3

 .
5
10
1

1  x 2  4  1  3  x 2  5  3  x  5
1
 x    5,  3    3, 5  . So, required domain is   5,  3    3, 5  .
1
1
f  x   x e x  f '  x   e x  x  1
When x   1,   ,  x  1  0 & e x  0  f '  x   0  f  x  increases in this interval.
1
or, we can write f  x   x e x  f '  x   e x  x  1
1
2
For f  x  to be increasing, we have f '  x   e x  x  1  0  x  1 as e x  0,  x  
1
Hence, the required interval where f  x  increases is  1,   .
1
2
Method 1 : f  x  
1
4x  2x  1 ,
2
Page 4 of 19
2
1 1 3
1 3 3
 2

Let g  x   4 x  2 x  1  4  x  2 x     4  x    
4 16  4
4 4 4


2

1
, let g  x   4 x 2  2 x  1
4x  2x  1
2
d
1
d2
g  x    g '  x   8 x  2 and g '  x   0 at x   also 2  g  x    g"  x   8  0

dx
4
dx
1
1
so , f  x  is maximum at x  
4
4
1
4
 1
maximum value of f  x   f    
 .
2
3
 4
 1
 1
4     2    1
 4
 4
 g  x  is minimum when x  
Method 3 : f  x  
1
2
1
2
4
maximum value of f  x   .
3
Method 2 : f  x  
1
1
1
2
1
2
1
4x  2x  1
2
On differentiating w.r.t x ,we get f '  x  
  8 x  2
4 x
2
 2x  1

2
....  i 
1
For maxima or minima , we put f '  x   0  8 x  2  0  x   .
4
Again, differentiating equation (i) w.r.t. x ,we get

 4 x2  2 x  1

f " x    

  8   8 x  2 2   4 x
 4 x  2 x  1
2
2
4
2
1
2
1
2

 2 x  1  8 x  2  


1
2
1
 1
At x   , f "     0
4
 4
1
4
f  x  is maximum at x   .
1
 maximum value of f  x  is f    
 4
Method 4: f  x  
1
2
 1
 1
4    2    1
 4
 4
1
2
4
 .
3
1
4x  2x  1
2
On differentiating w.r.t x ,we get f '  x  
  8 x  2
4 x
2
 2x  1

2
....  i 
1
For maxima or minima , we put f '  x   0  8 x  2  0  x   .
4


1
1
2
1
2
1
When x    h  ,   , where ' h ' is infinitesimally small positive quantity.
4 4

4 x  1  8 x  2  8 x  2  0    8 x  2   0 and  4 x 2  2 x  1  0  f '  x   0
2
Page 5 of 19
 1 1

and when x    ,   h  , 4 x  1  8 x  2  8 x  2  0    8 x  2   0
 4 4


and 4 x 2  2 x  1

2
1
 0  f '  x   0 . This shows that x   is the point of local maxima.
4
maximum value of f  x  is f    
2
 4
 1
1
1
 1
4    2    1
4


 4
23 OR
4
 .
3
For maxima and minima, P '  x   0  42  2 x  0
1
2
So, P  x  is maximum at x  21 .
The maximum value of P  x   72   42  21    21   513 i.e., the maximum profit is 513.
2
1
2 x

2 x
Let f  x   log 
2 x
2 x 
We have, f   x   log 
   log  2  x    f  x 
2 x 


1
1
2 x
So, f  x  is an odd function.   log 
 dx  0.
1
2 x
25
1
2
1
2
 x  21 and P "  x   2  0
24
1
2
1
f  x   x 3  x , for all x   .
d
f  x    f '  x   3 x 2  1; for all x   , x 2  0  f '  x   0

dx
1
1
2
1
2
Hence, no critical point exists.
Section –C
[This section comprises of solution short answer type questions (SA) of 3 marks each]
26
We have,
So,
2x2  3
 x x
2
27

1
5
2t  3
A
B
, we get A  & B 
 
3
3
t  t  9 t t  9
2x2  3


x2 x2  9
. Now , let x 2  t
2
9

dx 
1 dx 5

3 x2 3

x
dx
2
9
1 5
 x
 tan 1    c , where ' c ' is an arbitrary constant of integration.
3x 9
3
We have, (i)
1
 P  X   1  k  2 k  3k  1  k  6 .
1
2
1
1
2
1
1
i
1
Page 6 of 19
(ii) P  X  2   P  X  0   P  X  1  k  2k  3k  3 
1 1
 .
6 2
1
(iii) P  X  2   0.
28
3 12
x dx
2
1
2
x
2
dt
dx  
3
1 x
3 1 t2
1
2
3
Let x 2  t  dt 


1
2 1
sin  t   c
3
 3
2
 sin1  x 2   c , where ' c ' is an arbitrary constant of integration.
3
 
28 OR
1

Let I   4 log e  1  tan x  dx ------(i)
0

a
a



 I   4 loge  1  tan   x   dx , Using,  f ( x )dx   f (a  x )dx
0
0
0
4




1  tan x 
2



4
 I   4 log e  1 
dx

log e 

 dx 

0
0
1

tan
x
1

tan
x






4
0
1
log e 2 dx  I ( Using ------(i)
1
1
 2I 
29

4
log e 2  I 

8
log e 2.
x
x
 xy

 ydx  xdy 
2
2
y
Method 1: ye dx   xe  y  dy  e  ydx  xdy   y dy  e y 
  dy


y2




x
y
x
y
 x
 e d    dy
 y
x
y
1
1
x
 x
  e d     dy  e y  y  c , where ' c ' is an arbitrary constant of integration.
 y
1
x
dx xe y  y 2
Method 2: We have ,

x
dy
y .e y

dx x
y
  x ……………. (i)
dy y
ey
Let x  vy 
dx
dv
 v  y. ;
dy
dy
1
2
1
2
Page 7 of 19
So equation (i) becomes v  y
y
dv
y
v v
dy
e
1
2
1
2
1
2
dv
y
 v
dy e
 e v dv  dy
On integrating we get,
e
v
dv   dy  e v  y  c  e x / y  y  c
1
2
where ' c ' is an arbitrary constant of integration.
29 OR
The given Differential equation is
dy

 cos x  dx
2
y  tan x
Dividing both the sides by cos 2 x , we get
dy
y
tan x


2
dx cos x cos 2 x
dy
 y sec2 x  tan x sec2 x ........  i 
dx


Comparing with


1
2
dy
 Py  Q
dx
P  sec 2 x , Q  tan x .sec 2 x
The Integrating factor is, IF  e 
P dx
 e
sec2 x dx
1
2
 e tan x
On multiplying the equation  i  by e tan x , we get
d
y .e tan x  e tan x tan x sec2 x  d y .e tan x  e tan x tan x sec2 x dx
dx








1
On integrating we get , y .e tan x   t .e t dt  c1 ; where, t  tan x so that dt  sec2 x dx
 te t  e t  c   tan x  e tan x  e tan x  c


 y  tan x  1  c . e  tan x , where ' c1 '& ' c ' are arbitrary constants of integration.
30
1
The feasible region determined by the
constraints, x  2 y  100, 2 x  y  0, 2 x  y  200, x , y  0 , is given below.
Page 8 of 19
1
1
2
A  0, 50  , B  20, 40  , C  50, 100  and D  0, 200  are the corner points of the feasible
region.
1
The values of Z at these corner points are given below.
Corner point
Corresponding value of
Z  x  2y
30 OR
A  0, 50 
100
Minimum
B  20, 40 
100
Minimum
C  50, 100 
250
D  0, 200 
400
1
2
The minimum value of Z is 100 at all the points on the line segment joining the points  0,50 
and  20,40  .
The feasible region determined by the constraints, x  3, x  y  5, x  2 y  6, y  0.
is given below.
1
1
2
Page 9 of 19
Here, it
be seen
the
can
that
1
feasible region is unbounded.
The values of Z at corner points A  3, 2  , B  4, 1 and C  6, 0  are given below.
Corner point
Corresponding value of Z   x  2 y
A  3, 2 
1 ( may or may not be the maximum value)
B  4, 1 
-2
C  6, 0 
-6
1
2
Since the feasible region is unbounded, Z  1 may or may not be the maximum value.
Now, we draw the graph of the inequality, – x  2 y  1 , and we check whether the resulting
open half-plane has any point/s, in common with the feasible region or not.
Here, the resulting open half plane has points in common with the feasible region.
Hence, Z  1 is not the maximum value. We conclude, Z has no maximum value.
31
y
 x 
 log e 
  log e x  log e  a  bx 
x
 a  bx 
1
2
On differentiating with respect to x , we get

x
x
dy
y
1
1 d
1
b
dx
 
 a  bx   
2
x
x a  bx dx
x a  bx
1
dy
b 
ax
1
 y  x2  


dx
 x a  bx  a  bx
1
2
On differentiating again with respect to x , we get
x
d 2 y dy dy  a  bx  a  ax  b 



2
dx 2 dx dx
 a  bx 
1
2
Page 10 of 19
2
d2 y  a 
 x 2 
 .
dx
 a  bx 
1
2
Section –D
[This section comprises of solution of long answer type questions (LA) of 5 marks each]
32
1
To find the point of intersections of the curve y  x 2  1 and the line y  x  1 ,
we write x 2  1  x  1  x  x  1  0  x  0,1.
So, the point of intersections P  0,1  and Q  1,2  .
1
Area of the shaded region OPQRTSO = (Area of the region OSQPO + Area of the region
STRQS )
1


2
  x 2  1 dx    x  1 dx
0
1
2
 x3
  x2

   x    x
3
0  2
1
33
 1   
 1 
   1   0   2  2     1 
 2 
 3   
23
23

sq units.
Hence the required area is
6
6
Let  a , b  be an arbitrary element of    . Then,  a , b      and a, b 
We have, ab  ba ;
1
1
1
2
1
2
1
(As a, b   and multiplication is commutative on  )
  a , b  R  a , b  , according to the definition of the relation R on   
Thus  a , b  R  a , b  ,   a , b      .
So, R is reflexive relation on    .
1
Let  a , b  ,  c , d  be arbitrary elements of    such that  a , b  R  c , d  .
Page 11 of 19
Then,  a , b  R  c , d   ad  bc  bc  ad ;
 cb  da;
(changing LHS and RHS)
(As a, b, c, d   and multiplication is commutative on  )
  c , d  R  a , b  ; according to the definition of the relation R on   
Thus  a , b  R  c , d    c , d  R  a , b 
So, R is symmetric relation on    .
1
Let  a , b  ,  c , d  ,  e , f  be arbitrary elements of    such that
 a , b  R  c , d  and  c , d  R  e , f  .
Then
 a , b  R  c , d   ad  bc 
   ad  cf    bc  de   af
 c , d  R  e , f   cf  de 
  a, b  R  e, f ;
Thus
 be
(according to the definition of the relation R on    )
 a , b  R  c , d  and  c , d  R  e , f 
  a, b  R  e, f

So, R is transitive relation on    .
As the relation R is reflexive, symmetric and transitive so, it is equivalence relation on    .
 2, 6     x , y      :  x , y  R  2, 6 
  x , y      : 3 x  y
  x ,3 x  : x     1, 3  ,  2,6 ,  3,9  ,.........
33 OR
1
1
2
1
2
1
 x
 1  x , if x  0
We have, f  x   
 x , if x  0
 1  x
Now, we consider the following cases
Case 1: when x  0 , we have f  x  
x
1 x
Injectivity: let x, y   0 such that f  x   f  y  , then


x
y

 x  xy  y  xy  x  y
1 x 1 y
1
So, f is injective function.
Surjectivity : when x  0 , we have f  x  
x
1
 0 and f  x   1 
 1,as x  0
1 x
1 x
y
y
1 y
 0 such that f  x  
Let y   0,1  , thus for each y  0,1 there exists x 
 y.
y
1 y
1
1 y

1
Page 12 of 19
So, f is onto function on  0,   to  0,1 .
Case 2: when x  0 , we have f  x  
x
1 x

Injectivity: Let x, y   i.e., x , y  0 , such that f  x  f  y  , then

x
y

 x  xy  y  xy  x  y
1 x 1 y
So, f is injective function.
x
x
1
Surjectivity : x  0 , we have f  x  
 0 also, f  x  
 1 
 1
1 x
1 x
1 x
1  f  x   0 .
Let y    1, 0  be an arbitrary real number and there exists x 
1
y
 0 such that,
1 y
y
 y 
1 y
f  x  f 
 y.

 1 y  1 y
1 y
So, for y    1, 0  , there exists x 
y
 0 such that f  x   y .
1 y
1
Hence, f is onto function on   , 0  to   1, 0  .
Case 3:
(Injectivity): Let x  0 & y  0 such that f  x   f  y  
x
y

1 x 1 y
 x  xy  y  xy  x  y  2 xy , here LHS  0 but RHS  0 , which is inadmissible.
Hence , f  x   f  y  when x  y.
Hence f is one-one and onto function.
34
1
The given system of equations can be written in the form AX  B,
 2 3 10 
1 / x 
 4




 
Where, A   4 6 5  , X  1 / y  and B  1 
 6 9 20 
 1 / z 
 2
2 3 10
Now, A  4 6 5  2  120  45  3  80  30   10  36  36 
6 9 20
1
2
 2  75   3   110   10  72   150  330  720  1200  0  A1 exists.
1
2
T
72 
75 
 75 110
 75 150



 adj A  150 100 0   110 100 30 
 75
 72
30 24 
0
24 
1
1
2
Page 13 of 19
 75 150 75 
1
1 
110 100 30 
Hence, A 
 adjA 
A
1200 
 72
0
24
1
1
 
x
 75 150 75   4 
1 
1
1
As, AX  B  X  A B    
110 100 30  1 
y
1200 
 
 72
0
24   2
1
z
 
1
2
1
1
 
 
x
 600   2 
 300  150  150   
1 
1 
1
1

440  100  60     
400    



1200
y
1200
3
 288  0  48   
 240   
1
1
z
 5 
 
1
2
1 1 1 1 1 1
 ,  , 
Hence, x  2, y  3, z  5 .
x 2 y 3 z 5
1
Thus,
35
1
2
Let P 1,6,3  be the given point, and let ' L ' be the foot of the perpendicular from ' P ' to the
given line AB (as shown in the figure below). The coordinates of a general point on the
given line are given by
x 0 y 1 z  2


  ;  is a scalar, i.e., x   , y  2  1 and z  3  2
1
2
3
Let the coordinates of L be   , 2  1,3  2  .
So, direction ratios of PL are   1, 2  1  6 and 3  2  3, i .e .   1, 2  5 and 3  1.
Direction ratios of the given line are 1, 2 and 3, which is perpendicular to PL .
1
2
1
2
Therefore,    1 1   2  5  2   3  1  3  0  14  14  0    1
So, coordinates of L are 1,3,5  .
1
Page 14 of 19
Let Q  x1 , y1 , z1  be the image of P 1,6,3  in the given line. Then, L is the mid-point of
PQ .
Therefore,
 x1  1  1,  y1  6   3 and  z1  3  5  x
1
2
2
2
Hence, the image of P  1,6,3  in the given line is 1,0,7  .
1
 1, y1  0 and z1  7
Now, the distance of the point  1,0,7 from the y  axis is
12  72  50 units.
1
1
35 OR
Method 1:
Given that equation of lines are


r   i  j  k ..............  i  and r  i  j   2 j  k ..............  ii 




The given lines are non-parallel lines as vectors i  j  k and 2 j  k are not parallel. There is a
unique line segment PQ ( P lying on line  i  and Q on the other line  ii  ), which is at right
angles to both the lines. PQ is the shortest distance between the lines. Hence, the shortest possible
distance between the aeroplanes  PQ .

Let the position vector of the point P lying on the line r   i  j  k where '  ' is a scalar, is



1
2

 i  j  k , for some  and the position vector of the point Q lying on the line

r  i  j   2 j  k ; where '  ' is a scalar, is i   1  2  j     k , for some  .


  
Now, the vector PQ  OQ  OP   1    i   1  2    j       k ; (where ' O ' is the

origin), is perpendicular to both the lines, so the vector PQ is perpendicular to both the vectors
1
2
i  j  k and 2 j  k .
  1    .1   1  2    .  1       .1  0 &
  1    .0   1  2    .  2        .1  0
 2  3  3  0 & 2  5   3  0
1
2
1
2
Page 15 of 19
On solving the above equations , we get  
2
and   0
3
So, the position vector of the points, at which they should be so that the distance between them is
the shortest, are


2   
i  j  k and i  j .
3
1
2
2
2

   1
1
2
1  1  2
PQ  OQ  OP  i  j  k and PQ            
3
3
3
 3  3  3
The shortest distance 
1
2
3
1
2
units.
3
Method 2:
The equation of two given straight lines in the Cartesian form are
x y z
  ........ i  and
1 1 1
x1
y1 z

 .........  ii 
0
2
1
The lines are not parallel as direction ratios are not proportional. Let P be a point on straight line
i
and Q be a point on straight line  ii  such that line PQ is perpendicular to both of the lines.
Let the coordinates of P be   ,   ,   and that of Q be  1, 2  1,   ; where '  ' and '  ' are
scalars.
Then the direction ratios of the line PQ are    1,    2  1,    
Since PQ is perpendicular to straight line  i  , we have,
(  1).1  (    2  1).( 1)  (   ).1  0
 3  3  2......  iii 
1
2
1
2
1
2
Since , PQ is perpendicular to straight line  ii  , we have
1
2
0.    1      2  1 .( 2)       .1  0  3  5  2........  iv 
1
Solving  iii  and  iv  , we get   0,  
2
3
 2 2 2
Therfore , the Coordinates of P are  ,  ,  and that of Q are  1, 1, 0
 3 3 3
1
1
Page 16 of 19
2
So, the required shortest distance is
2
2
2 
2 
2

 1  3    1  3    0  3  

 
 

2
units.
3
Section –E
[This section comprises solution of 3 case- study/passage based questions of 4 marks each with two sub
parts. Solution of the first two case study questions have three sub parts (i),(ii),(iii) of marks 1,1,2
respectively. Solution of the third case study question has two sub parts of 2 marks each.)
36
Let E1 , E2 , E3 be the events that Jayant, Sonia and Oliver processed the form, which are clearly
pairwise mutually exclusive and exhaustive set of events.
Then P  E1  
50
5
20 1
30
3
 , P  E2  
 and P  E3  
 .
100 10
100 5
100 10
Also, let E be the event of committing an error.
We have, P  E | E1   0.06 , P  E | E 2   0.04 , P  E | E 3   0.03.
(i)
The probability that Sonia processed the form and committed an error is given by
P  E  E2   P  E2  . P  E | E2  
(ii)
1
 0.04  0.008.
5
1
The total probability of committing an error in processing the form is given by
P  E   P  E1  . P  E | E1   P  E2  . P  E | E2   P  E3  . P  E | E3 
P E 
(iii)
50
20
30
 0.06 
 0.04 
 0.03  0.047.
100
100
100
1
The probability that the form is processed by Jayant given that form has an error is given by
P  E1 | E  
P  E | E1   P  E1 
P  E | E1  . P  E1   P  E | E2  . P  E2   P  E | E3  . P  E3 
50
30
100


50
20
30 47
0.06 
 0.04 
 0.03 
100
100
100
0.06 
1
Therefore, the required probability that the form is not processed by Jayant given that form has an


30 17
error  P E1 | E  1  P  E1 | E   1 
 .
47 47
(iii) OR
3
 PE E  P E
i 1
i
1
| E   P  E 2 | E   P  E3 | E   1
1
1
1
Since, sum of the posterior probabilities is 1.
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3
( We have ,  P  E i E   P  E1 | E   P  E 2 | E   P  E 3 | E 
i 1



37
P  E  E1   P  E  E2   P  E  E3 
P E
P   E  E1    E  E2    E  E3  
P E
P  E  ( E1  E2  E3 
PE

PE  S
PE
as Ei & E j ; i  j , are mutually exclusive events

PE
PE
 1; ' S ' being the sample space )
We have ,


F1  62  02  6 kN , F2 
 4 
2

 42  32  4 2 kN , F3 
 3 
2
  3   18  3 2 kN .
2
1
(i) Magnitude of force of Team A  6 kN .
 

(ii) Sin ce a  c  3(i  j )and b  4 (i  j )

 
So, b and a  c are unlike vectors having same intial point

 
and b  4 2 & a  c  3 2
  

 
Thus F 2  F 1  F 3 also F 2 and F 1  F 3 are unlike
1
1
1
Hence B will win the game
   
(iii) F  F1  F2  F3  6iˆ  0 ˆj  4iˆ  4 ˆj  3iˆ  3 ˆj   iˆ  ˆj

F 
1
 1    1 
2
2
 2 kN .
1
OR

F   iˆ  ˆj
 3
1
     tan 1      
; where' ' is the angle made by the resultant force with the
4
4
1
 ve direction of the x  axis.
38
1
y  4x  x2
2
(i) The rate of growth of the plant with respect to the number of days exposed to sunlight
is given by
dy
 4  x.
dx
(ii) Let rate of growth be represented by the function g  x  
2
dy
.
dx
Page 18 of 19
Now, g '  x  
d  dy 
 1  0
dx  dx 
 g  x  decreases.
1
So the rate of growth of the plant decreases for the first three days.
1
Height of the plant after 2 days is y  4  2 
1
2
 2   6 cm .
2
Page 19 of 19