Class : XIII (Sterling)
Time : 3 hour Max. Marks : 60
INSTRUCTIONS
1. The question paper contain pages and 3-parts. Part-A contains 6 objective question , Part-B contains 2
questions of"Match the Column" type and Part-C contains 4 subjective type questions.Allquestions are
compulsory. Please ensure that the Question Paper you have received contains all the QUESTIONS
and Pages. If you found some mistake like missing questions or pages then contact immediately to
the Invigilator.
PART-A
(i) Q.1 to Q.6 have One or More than one is / are correct alternative(s) and carry 4 marks each.
There is NEGATIVE marking. 1 markwillbe deducted for eachwrong answer.
PART-B
(iii) Q.1 to Q.2 are "Match the Column" type whichmayhave oneor more than one matching options and
carry8 marks for each question. 2 marks willbe awarded for each correct matchwithin a question.
ThereisNONEGATIVE marking. Markswillbeawardedonlyifallthecorrectalternativesareselected.
PART-C
(iv) Q.1 to Q.4 are "Subjective" questions. There is NO NEGATIVE marking. Marks will be awarded
onlyifallthe correct bubbles are filled in your OMR sheet.
2. Indicate the correct answer for eachquestionbyfilling appropriate bubble(s) inyour answer sheet.
3. Use onlyHB pencilfor darkening the bubble(s).
4. Use ofCalculator, Log Table, Slide Rule and Mobile is not allowed.
5. The answer(s) ofthe questionsmust bemarked byshading the circlesagainst the questionbydark HBpencil
only.
PART TEST-2
PART-B
For example if Correct
match for (A) is P, Q;for
(B) is P, R; for (C) is P
and for (D) is S then the
correct method for filling
the bubble is
P Q R S
(A)
(B)
(C)
(D)
PART-C
Ensure that all columns
(4 before decimal and 2
after decimal) are filled.
Answer having blank
column willbe treated as
incorrect. Insert leading
zero(s) if required after
rounding the result to 2
decimalplaces.
e.g. 86 should be filled as
0086.00
.
.
.
.
.
.
.
.
.
.
PART-A
For example if only 'B'
choice is correct then,
the correct method for
filling thebubble is
A B C D
For example ifonly'B &
D' choices are correct
then, thecorrect method
for fillingthe bubbles is
A B C D
The answer of the
question in any other
manner (such as putting
, cross , or partial
shading etc.) will be
treated as wrong.

Class - XIII MATHS Part Test - 2
PARTA
Select the correct alternative(s) (choose one or more than one)
Q.1 You are given an unlimited supply ofeach ofthe digits 1, 2, 3 or 4. Using only these four digits, you
construct n digit numbers. Suchn digit numbers willbe calledLE G IT I MAT E ifit contains the digit
1 either aneven number times or not at all. Number of n digit legitimate numbers are
(A) 2n + 1 (B) 2n + 1 + 2 (C) 2n + 2 + 4 (D*) 2n – 1(2n + 1)
[Sol. × × ×............× (n places)
Totalnumbers = 3n + nC2 · 3n – 2 + nC4 · 3n – 4 + .......... (nC2 indicates selection of places)
=
2
)
1
3
(
)
1
3
( n
n



=
2
1
[4n + 2n] = 22n – 1 + 2n – 1 = 2n – 1(2n + 1) ]
Q.2 Whichofthefollowing statements are correct?
(A*) Number ofwords that can beformed with 6 onlyofthe letters ofthe word "CENTRIFUGAL" if
each word must contain allthe vowels is 3 · 7!
(B*) There are 15 balls ofwhichsome are white and the rest black. Ifthe number ofways in which the
balls can be arranged in a row, is maximumthan the number ofwhite balls must be equalto 7 or 8.
Assume balls ofthe same colour to be alike.
(C) There are12 things, 4 alikeofone kind, 5alike ofanother kind andtherest are alldifferent. The total
number ofcombinations is 240.
(D*) Number ofselections that canbe made of6 letters fromthe word "COMMITTEE" is 35.
[Sol. (A) 11
c
7
v
4
/
 7C2 · 6! = 3 · 7 · 6! = 3 · 7!
(B) No. of ways =
)!
r
15
(
!
r
!
15

= 15cr





r
W
.....
W
W





r
15
B
......
B
B
B

This is maximumif r = 7 or 8
(C)  Totalno. ofcombinations = 5 · 6 · 23 – 1 = 240 –1 = 239
(D) 2 alike + 2 other alike + 2 other alike = 1
2 alike + 2 other alike + 2 different = 3C2 · 4C2 = 18
2 alike + 4 different =3C1 · 5C4 = 15
All6 different = 1
——
= 35 Ans. ]
Q.3 Given three points P1, P2 and P3 with position vectors k̂
j
ˆ
î
2
r1 



; ĵ
3
î
r2 


and k̂
3
ĵ
î
r3 



respectively. Equationofa line orthogonalto the plane containing these points and passing through the
centroid P0 ofthe triangle P1P2P3, is
(A)  
k̂
2
j
ˆ
3
î
4
3
1
r 



+ )
k̂
2
j
ˆ
3
î
14
( 

 (B*) k̂
3
2
j
ˆ
î
3
4
r 



+ )
k̂
2
j
ˆ
3
î
14
( 


(C) k̂
2
j
ˆ
3
î
4
r 



+ )
k̂
2
j
ˆ
3
î
14
( 

 (D)  
k̂
2
j
ˆ
3
î
4
3
1
r 



+ )
k̂
2
j
ˆ
3
î
10
( 



[Hint: P0 = 




 
3
2
,
1
,
3
4
vector normalto the plane
b
a
n




 =
1
4
1
3
2
0
k̂
ĵ
î


 = )
2
0
(
k̂
)
3
0
(
j
ˆ
)
12
2
(
î 





= – )
k̂
2
ĵ
3
î
14
( 

equationofthe line is k̂
3
2
j
ˆ
î
3
4
r 



+ )
k̂
2
j
ˆ
3
î
14
( 

 Ans. ]
Q.4 Evaluate  








3
0
2
1
2
dx
x
1
x
2
·sin
x
1
1
=
(A*)
72
7 2

(B)
16
2

(C)
9
2

(D) none
[Sol: I =  








3
0
2
1
2
dx
x
1
x
2
·sin
x
1
1
sin–1 





 2
x
1
x
2
=













1
x
if
,
x
tan
2
1
x
1
if
,
x
tan
2
1
1
I =   


















1
0
3
1
2
1
2
2
1
2
dx
x
1
x
2
·sin
x
1
1
dx
x
1
x
2
sin
x
1
1
=   






1
0
3
1
2
1
2
1
dx
x
1
x
tan
2
dx
x
1
x
tan
2
= 
  






 3
1
2
1
1
0
3
1
2
2
1
dx
x
1
x
tan
2
dx
x
1
1
dx
x
1
x
tan
2
=   








3
/
4
/
3
1
1
4
/
0
,
dt
t
2
x
tan
dt
t
2 Put, tan–1x = t
=  
3
/
4
/
2
1
1
4
/
0
2
2
t
2
1
tan
3
tan
2
t
2

























=





 








 





16
9
4
3
16
2
2
2
=
2
72
7
 ]

Q.5 Which ofthe following is always true about a function f(x) on the interval[a, b] = {x |a < x <b|}?
(A) If f (x) > 0 on [a, b], then 
 
b
a
b
a
dx
)
x
(
dx
)
x
( 2
f
f
(B) Iff (x) is increasing on[a, b] then f 2(x) is increasing on [a, b]
(C) If f (x) attains a minimumat c where a < c < b, then f ' (c) = 0
(D*) none
[Sol: (A) may be false if 0 < f (x) < 1, f 2 (x) < f (x) and 
 
b
a
b
a
dx
)
x
(
dx
)
x
( 2
f
f
(B) may be false since if f (x) < 0,
dx
d
f 2 (x) = 2 f (x) f  (x) < 0 when f  (x) > 0 and so f  (x) is
decreasing while f (x) is increasing.
(C) maybefalse since a functionmaynot be differentiableat x= cfor whichit attainsits minimum. Since
none of the statement are always true, the correct answer is (D) ]
Q6. Let 




dx
e
2
x
=  and f (t) = 




dx
e
2
x
t
, (t > 0)
(A) f (x) is equal to
x

(B*) f(x)is monotonic decreasing
(C*) if area bounded by y = f (x), x = 2, x = 16 and y = c is minimum then c =
3

(D*) equation of normal to curve y = f (x) at x =  is y – 1 = 2(x – )
[Sol. 




dx
e
2
x
=  ; put x = t u  dx = t du
 = t 




du
e
2
tu
= t 




dx
e
2
tx
 f (t) =
t

 f (x) =
x

f ' (x) = – 2
3
x
2

 f (x) is monotonic decreasing.
if area bounded by y = f (x), x = 2, x = 16 and y = c is minimum then
c = f 




 
2
16
2
= f (9) =
3



x
N
m = 


2
= 2
equation of normal is (y– 1) = 2(x – ) ]

PART B
Match the column
Q1. ColumnI ColumnII
(A) P(m, n) (where m, n  N) is anypoint inthe interior of (P) 20
the quadrilateral formed by the pair of lines xy= 0 &
8x2 + 5y2 + 14xy – 48x – 30y + 40 = 0 the possible
number ofpositions of the point P is
(B) The graph of y= x + k intersect the graph of (x–3)2 + (y + 2)2 (Q) 4
= 50 in one or more points if and only if a < k < b then b–a is
(C) If a

, b

and c

are three mutuallyperpendicular unit vectors and (R) 12
d

is unit vector whichmakes equalangle with c
and
b
,
a



then
2
d
c
b
a






 = p + q r where q, r and p are three consecutive
natural numbers then find the value ofp + q + r
(D)Abscissa and ordinates ofngivenpoints are inAPwithfirst terma (S) 9
and common difference 1 and 2 respectively. Ifalgebraic sumof
perpendiculars fromthese givenpoints on a variable line which
always passes through the point 





11
,
2
13
is zero, then a + n =
[Ans: A  Q; B  P; C  S; D  R]
[Hint: OP < OM
2
2
n
m  <
25
16
20

m2 + n2 <
41
400
m2 + n2 < 9 ..............(1)
( m, n  N)
& OP > OL
2
2
n
m  >
5
2
m2 + n2 >
5
4
m2 + n2 > 2 ..............(2)
From (1) & (2)
2 < m2 + n2 < 9
Possibilities
If m2 + n2 = 2  m = n = 1  P(1, 1)
If m2 + n2 = 5  m = 2, n = 1 or m = 1, n = 2  P(2, 1) & P(1, 2)
If m2 + n2 = 8  m = n = 2  P(2, 2)
Totalno. ofpossibilities are 4 ]

 The value ofk can be obtained ifperpendicular distance from(3, – 2) onthe line x– y+ k =0 is  50 ,
which gives | 5 + k|  10  – 10  5 + k  10  – 15  k  5
a = –15; b = 5
(C)
2
d
c
b
a






 = 
  b
·
a
2
a
2 


= 4 +  
c
b
a
·
d
2






Let c
b
a
d









 , then c
·
d
b
·
d
a
·
d







 = cos
  =  =  = cos
also 2 + 2 + 2 = 1  3cos2 = 1  cos =
3
1


2
d
c
b
a






 =
3
3
.
2
4 = 4 + 3
2
q = 2, r = 3, p = 4
(D) a = 2, n = 10 ]
Q.2 ColumnI ColumnII
(A) The value of
 






n
4
r
1
r
2
n
n
4
r
3
r
n
Lim =
q
p
(P) 2
in its lowest form where p + q =
(B) No. ofintegralvalues ofa for which the cubic (Q) 0
f(x) = x3 + ax + 2 is non monotonic and has
exactly one real root.
(C) The radicalcentre ofthree circles is at the origin. (R) 1
The equations of two of the circles are x2 + y2 = 1
and x2 + y2 + 4x + 4y – 1 = 0. If the equation of
third circle passes through the point (1, 1) and (–2, 1)
is x2 + y2 + 2gx + 2fy + c = 0 then f – c =
(D) If x2
+ y2
+ z2
– 2xyz = 1 , then the value of (S) 11
2
2
2
z
1
dz
y
1
dy
x
1
dx





[Ans: A  S; B  P; C  Q; D  Q]
[Sol (A) Tr = 2
4
n
r
3
n
·
n
r
1









S = 









n
4
1
2
n
r
·
4
n
r
3
1
n
1
=  
4
0
2
)
4
x
3
(
x
dx
put 4
x
3  = t  dt
dx
x
1
2
3


= 
10
4
2
t
dt
3
2
= 













10
1
4
1
3
2
t
1
3
2
4
10
=
10
1
40
6
·
3
2

(C) Ans: x2 + y2 + x – 2y – 1 = 0
Let the equation ofthe 3rd circle be
x2 + y2 + 2gx + 2fy + c = 0
radicalaxis between 1st and 3rd is
2gx + 2fy + c + 1 = 0
It passes through (0, 0)  c = – 1
Hence equation of the 3rd circle is x2 + y2 + 2gx + 2fy – 1 = 0
since the circle
x2 + y2 + 2gx + 2fy – 1 = 0
passes through (1, 1) and (–2, 1), hence
2 + 2g + 2f – 1 = 0  2g + 2f + 1 = 0 ....(1)
or 4 + 1 – 4g + 2f – 1 = 0 2g – f = 2 ....(2)
Solving (1) and (2) f = – 1 ; g =
2
1
Hence the equation ofthe third circle is
x2 + y2 + x – 2y – 1 = 0
Alternatively : Equation of the third circle canbe written as
(x – 1) (x + 2) + (y – 1)2 +  (y – 1) = 0 [ familyofcircle through (1, 1) and (– 2, 1) ]
x2 + y2 + x + ( – 2) y – (1 + ) = 0
Radical axis of S2 and S3 is
S2 – S3 = 0
i.e. 3x + (6 – )y +  = 0
point (0, 0) willsatisfythis equation
  = 0
hence S3 = x2 + y2 + x – 2y – 1 = 0
(D) Diff. thegivenrelation
2x dx + 2y dy + 2z dz – 2 (xy dz + yz dx + zx dy ) = 0
( x – yz ) dx + ( y – zx ) dy + ( z – xy ) dz = 0
dx
y zx z xy
dy
x yz z xy
dz
x yz y zx
( )( ) ( )( ) ( )( )
 

 

 
 0
dx
P
dy
P
dz
P
1 2 3
0
  
Now consider
P y zx z xy
1
2 2 2
  
( ) ( )
    
( ) ( )
y xyz z x z xyz x y
2 2 2 2 2 2
2 2
 
 
  
 

)
y
x
y
x
1
(
)
x
z
z
x
1
( 2
2
2
2
2
2
2
2







= ( ) ( ) ( ) ( )
1 1 1 1
2 2 2 2
   
z x x y
)
x
1
(
·
)
z
1
)(
y
1
(
)
x
1
(
P 2
2
2
2
1 





|||ar
expressions for P and P
2 3 Hence the result . ]

PART C
Q.1 Number of eight digit numbers which can be formed using the digits 1, 3, 4, 5, 6, 7, 8, 9 (without
repetition) ifeachnumber has to be divisible by275 is
[Hint: Anumber willbe divisible by275 ifit is divisible by 25 and 11 .
For divisible by x + y = 1 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 43
For divisible by25 last two digits must be 7 5
Hence
Now we have 1, 3, 4, 6, 8, 9 to be filled in six places, such that
sum of even places = 27  5 = 22 or 16  5 = 11
sum of odd places = 27 7 = 20 or 16  7 = 9
36 is possible when we have 1, 4, 6 at even places and 3, 8, 9 at odd places ]
Q.2 The least integral value of a for which the function f(x) =
3
x
3
a
+ (a + 2)x2 + (a – 1)x + 2 possess a
negativepointofminimum. [Ans: a = 2]
Q.3 Let c
and
b
,
a



be threenon-coplanar unit vectors equallyinclinedto one another at anacute angle. If
c
b
b
a






 = c
r
b
q
a
p




 , then p2 + q2/cos + r2 is equal to. [Ans: 2]
[Sol: 1
c
·
c
b
·
b
a
·
a 







 and 


 cos
c
·
b
c
·
a
b
·
a






consider c
r
b
q
a
p
c
b
b
a












 . Taking dot product with b

we get
0 = q + (p + r) cos .....(1)
Also c
r
b
q
a
p
c
b
b
a













    )
c
·
a
pr
c
·
b
qr
b
·
a
pq
(
2
c
·
c
r
b
·
b
q
a
·
a
p
c
b
·
b
a
2
c
b
b
a 2
2
2
2
2 































      
   
pr
qr
pq
cos
2
r
q
p
c
·
a
b
·
b
c
·
b
b
·
a
2
sin
2 2
2
2
2


















    
pr
qr
pq
cos
2
r
q
p
cos
cos
sin
2 2
2
2
2
2












    








 cos
1
2
pr
qr
pq
cos
2
r
q
p 2
2
2
 p2 + r2 + q2 + 2q(p + r)cos + 2prcos = 2(1 – cos)
From (1) we get 2q (p + r) cos = –2q2 and p2 + r2 + 2pr =

2
2
cos
q
 p2 + r2 – q2 + 2pr cos = 2(1 – cos)
 p2 + r2 – q2 + 










 2
2
2
2
r
p
cos
q
cos = 2(1 – cos)
 (p2 + r2) ( 1 –cos) +

cos
q2
(1 – cos) = 2(1 – cos)
 p2 + r2 +

cos
q2
= 2 ]

Q.4 Equation ofstraight linemeeting the circle x2 + y2 = 9intwo points at equaldistances of5 froma point
(x1, y1) on the circumference ofthe circle is given byxx1 + yy1 +
q
p
= 0. Where p + q = (no.
q
p
isinits
lowest form) [Ans:
q
p
=
2
7
; p + q = 9]