1. Subject : Mathematics Date : DPP No. : Class : XIII Course :
DPP No. – 01
Total Marks : 26 Max. Time : 22 min.
Single choice Objective ('–1' negative marking) Q.2, 4, 5, 6 (4 marks 3 min.) [16, 12]
Multiple choice objective ('–1' negative marking) Q.3 (6 marks 5 min.) [6, 5]
Subjective Questions ('–1' negative marking) Q.1 (4 marks 5 min.) [4, 5]
1. ABCD is a quadrilateral and E the point of intersection of the lines joining the middle points of opposite
sides . Show that the resultant of

OA ,

OB ,

OC and

OD is equal to 4

OE , where O is any point.
2. P, Q have position vectors
 
a b
& relative to the origin 'O' & X, Y divide PQ

internally and externally
respectively in the ratio 2 : 1 . Vector XY

=
(A)  
3
2
 
b a
 (B)  
4
3
 
a b
 (C)  
5
6
 
b a
 (D*)  
4
3
 
b a

3. P is a point on the line through the point A whose position vector is a

and the line is parallel to the
vector b

. If PA = 6 , the position vector of P is :
(A) a

+ 6 b

(B*) a

+
b
6
 b

(C*)
|
b
|
b
6
–
a 


(D) b

+
a
6
 a

4. Let c
,
b
,
a



be three unit vectors such that c
b
a




 = 1 and b
a


 . If c

makes angles
,  with b
,
a


respectively then cos + cos is equal to
(A) 3/2 (B) 1 (C*) – 1 (D) 1/2
5. )
r
k̂
(
k̂
.
r
)
ĵ
.
r
(
)
r
î
)
î
.
r
( )
(
)
r
ĵ
(
(




 

 

 =
(A*) 0 (B)

r (C) 2

r (D) 3

r
6. If k̂
–
ĵ
2
b
,
ĵ
î
a 




and b
a
b
r
,
a
b
a
r













 then
r
r


is equal to
(A*) )
k̂
ĵ
3
î
(
11
1

 (B)  
k̂
ĵ
3
–
î
11
1
 (C) )
k̂
ĵ
–
î
(
3
1
 (D) none of these
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2. Subject : Mathematics Date : DPP No. : Class : XIII Course :
DPP No. – 02
Total Marks : 28 Max. Time : 23 min.
Single choice Objective ('–1' negative marking) Q.1, 2, 3, 4, 5 (4 marks 3 min.) [20, 15]
Match the Following (no negative marking) (2 × 4) Q.6 (8 marks 8 min.) [8, 8]
1. If b
,
a


are nonzero and noncollinear vectors then ,      k
k
b
a
j
j
b
a
i
i
b
a













 =
(A) b
a


 (B*) b
a


 (C) b
a


 (D) a
b



2. A vector c

of magnitude 6
20 parallel to the bisector of the angle between k̂
4
ĵ
4
î
7
a 



and
k̂
2
ĵ
î
2
b 




is
(A) ±  
k̂
ĵ
7
î
2
3
20

 (B) ±  
k̂
2
ĵ
7
î
20
3

 (C) ±  
k̂
7
ĵ
2
î
3
20

 (D*) ±  
k̂
2
ĵ
7
î
3
20


3. Let the centre of the parallelopiped formed by PA i j k

  
  
2 2 ; PB i j k

  
4 3
   ;
PC i j k

  
3 5
   is given by the position vector (7, 6, 2) . Then the position vector of the point P is:
(A*) (3, 4, 1) (B) (6, 8, 2) (C) (1, 3, 4) (D) (2, 6, 8)
4. If 

x
Lim
2
)
rx
–
qx
px
( 2

 , then
(A*) q = 4r (B) q = 2r (C) q = r (D) q = 6r
5. The number 6
4 )
2000
(
log
2
+ 6
5 )
2000
(
log
3
can be written as
n
m
where m and n are relatively prime
positive integers, then (m + n) equals
(A) 4 (B) 5 (C) 6 (D*) 7
6. Match the column
Column - I Column - II
(A) If c
,
b
,
a



are three mutually perpendicular vectors where (p)
4
3
–
1
c
and
2
b
a 





, then ]
a
c
c
b
b
a
[
12
1 







 is
(B) If b
,
a


are two unit vectors inclined at
3

, then (q) 0
]
b
b
a
b
a
[






 is
(C) If c
,
b


are orthogonal unit vectors and a
c
b




 , then (r)
3
4
]
c
b
b
a
c
b
a
[










 is
(D) If ]
b
y
x
[
]
a
y
x
[






 = ]
c
b
a
[



= 0 each vector being a (s) 1
non-zero vector, and no two vectors are collinear then ]
c
y
x
[



=
Ans. (A)  (r) ; (B)  (p) ; (C)  (s) ; (D)  (q)
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3. Subject : Mathematics Date : DPP No. : Class : XIII Course :
DPP No. – 03
Total Marks : 24 Max. Time : 20 min.
Comprehension ('–1' negative marking) Q.1 to Q.3 (4 marks 3 min.) [12, 9]
Single choice Objective ('–1' negative marking) Q.4, 6 (4 marks 3 min.) [8, 6]
Subjective Questions ('–1' negative marking) Q.5 (4 marks 5 min.) [4, 5]
Comprehension (Q.no. 1 to 3)
Let a, b, c, d  R. Then the cubic equation of the type ax3 + bx2 + cx + d = 0 has either one root real
or all three roots are real. But in case of trigonometric equations of the type a sin3 x + b sin2 x + c sinx
+ d = 0 can possess several solutions depending upon the domain of x.
To solve an equation of the type a cos + b sin = c. The equation can be written as
cos ( – ) = c/ )
b
a
( 2
2
 .
The solution is  = 2n +  ± , where tan  = b/a, cos  = c/ )
b
a
( 2
2
 .
1. On the domain [–, ] the equation 4sin3 x + 2 sin2 x – 2sinx – 1 = 0 possess
(A) only one real root (B) three real roots
(C) four real roots (D*) six real roots
2. In the interval [–/4, /2], the equation, cos 4x +
x
tan
1
x
tan
10
2

= 3 has
(A) no solution (B) one solution (C*) two solutions (D) three solutions
3. |tan x| = tan x +
x
cos
1
(0  x  2) has
(A) no solution (B*) one solution (C) two solutions (D) three solutions
4. If 5
|
b
|
,
2
|
a
| 



and 0
b
.
a 


, then )))))
b
a
(
a
(
a
(
a
(
a
(
a












 is equal to
(A) a
64

(B) b
64

(C*) – b
64

(D) – a
64

5. Find the exact value of the expression
º
10
tan
º
40
tan
2
º
20
tan
º
70
tan 

.
Ans. 4
6. The matrix A has x rows and (x + 5) columns. The matrix B has y rows and (11 – y) columns. Both
AB and BA exist. The value of x and y is
(A) 8, 3 (B) 3, 4 (C*) 3, 8 (D) 8, 8
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