Class : XIII (Sterling)
Time : 1 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 theColumn" type and Part-C contains 4 subjective type questions.All questions 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 mark will be deducted for eachwrong answer.
PART-B
(iii) Q.1 to Q.2 are"Match the Column" typewhich mayhave oneor more than onematching options and
carry8 marks foreach question. 2 marks will be awarded for each correct match within a question.
ThereisNONEGATIVEmarking.Markswillbeawardedonlyifallthecorrectalternativesareselected.
PART-C
(iv) Q.1 to Q.4 are "Subjective"questions. There is NO NEGATIVE marking. 5 Marks will be awarded
onlyifall the correct bubbles are filled in yourOMR sheet.
2. Indicate the correctanswer for each question byfillingappropriate bubble(s) inyour answer sheet.
3. Use onlyHB pencil fordarkeningthe bubble(s).
4. Use ofCalculator, LogTable, SlideRule and Mobile is not allowed.
5. Theanswer(s)ofthequestionsmustbemarked byshadingthecirclesagainst thequestionbydarkHBpencil
only.
PART TEST-3
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
thebubbleis
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 befilled as
0086.00
.
.
.
.
.
.
.
.
.
.
PART-A
For example if only 'B'
choice is correct then,
the correct method for
fillingthebubbleis
A B C D
Forexampleifonly'B&
D' choices are correct
then, the correct method
forfillingthebubblesis
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.
MATHEMATICS

PARTA
Select the correct alternative(s) (choose one or more than one)
Direction for Q.1 & Q.2 (2 questions)
Cards aredrawnonebyonefromawell shuffledpack of52 playingcardswithoutreplacementuntilthe
queen of hearts appears. Let the chance that 8th card drawn the queen ofhearts is s
r where r and sare
relativelyprime.Acircle of radius r obtained as above is tangent to two congruent circles which are
tangent toeachother.Allthethreecircles aredisjoint andthesethreecircles havecommon tangent line.
Let xbetheradii ofthetwo congruent circles.Suppose threenumbers a2, b2 and c2 forman arithmetical
progression. If (a + b) = – 12 and (c – b) = x and
a
b
a
c


has the value equal to q.
If now the roots of the equation x2 + Bx + D = 0 are r1 and r2 where r1 + r2 = 2 and 2
2
2
1 r
r  = q then
Q.1 The value of 'x'is equal to
(A) 2 (B) 4 (C) 6 (D) 8
Q.2 The value of 'D' is equal to
(A)
2
1
(B)
4
1
(C)
6
1
(D)
8
1
Q.3 Refering to the parabola y2 = 4ax (a > 0) four statements are given
I Chord of contact of the pair of tangent from (–a, 0) passes through the focus of the parabola.
II All chords subtendinga right angle at the vertex passesthrough focus.
III Tangent totheparabolaat anypoint Pmeets thetangent at thevertexinY.Alinedrawn through
Y perpendicularto the tangent passes through the focus.
IV Acirclecircumscribinganytriangleformedby3co-normalpointsontheparabolapassesthrough
itsfocus.
(A) exactlyonestatement is correct (B) exactlytwo statements are correct.
(C) exactlythreestatements are correct (D) all thestatements are correct.
Q.4 The area of region log2(logyx) > 0  x  





2
,
2
1
is
(A)
8
5
sq. units (B)
8
7
sq. units (C)
8
13
sq. units (D)
8
15
sq. units
Q.5 The vertices of a triangle in the argand plane are 3 + 4i, 4 + 3i and 2 6 + i then distance between
orthocentreandcircumcentreofthetriangleis
(A) 6
28
137  (B) 6
28
137  (C) 6
28
137
2
1
 (D) 6
28
137
3
1

Q6. A purse contains 3 cards. Oneof them is red on both the sides, one is blue on both the sides and one is
red on oneside and blue onthe other side.Acard is randomlychosen andput on the table. It shows red
colour on theupper side. Then the probabilitythat the other side of the card is also red is
(A)
3
2
(B)
2
1
(C)
3
1
(D) none

PART B
MATCH THE COLUMN:
INSTRUCTIONS:
Column-Iand column-IIcontainsfour entries each. Entries ofcolumn-Iareto bematchedwith some
entriesofcolumn-II.Oneormorethanoneentriesofcolumn-Imayhavethematchingwiththesameentries
ofcolumn-IIandoneentryofcolumn-Imayhaveoneormorethanonematchingwithentriesofcolumn-II.
Q1. ColumnI ColumnII
(A) The fractional part of
15
2 n
4
is
15
k
where k is (P) 0
(B) If C0, C1, C2, ..........Cn denote the binonial (Q) 1
coefficients in the expansion of (1 + x)n, then
 

 


n
0
r
r
n
e
e
r
n
r
10
log
1
10
log
r
1
C
)
1
( is equal to
(C) The remainder obtained when 1! + 2! + 3! + .... (R) 2
.............. 95! is divided by 15 is
(D) The possible values of b > 0, so that the area of (S) 3
the bounded region enclosed between the parabola
y = x – bx2 and y =
b
x2
is maximum is
Q.2 ColumnI ColumnII
(A) The area bounded by y = x2 + 2 and y = 2| x | – cosx (P) 2
is of the form of
q
p
where p & q are relatively prime
then p – q =
(B) Tangent drawn to circle | z | = 2 atA(z1) and normal at (Q) 8
B(z2) meet at the point P(zp) then zp =
2
1
2
1
2
2
1
z
z
z
z
z
|
z
|
k

where
k =
(C)Acircle cuts two perpendicularlines so that each intercept is (R) 0
of given length a and b, the locus of the centre of the circle is
given by ),
b
a
(
m
y
x 2
2
2
2




where l, m, are relatively
prime numbers then the value of l + m is
(D)Theorderofdifferentialequationoffamilyof circleswithone (S) 5
diameter along y = x

PART C
Q.1 If a function f(x) is defined as f(x) = max {4 – x2, |x – 2|, (x – 2)1/3} for x  [–2, 4]. Then find the area
bounded bythe curve and x-axis is
q
p
where p & q are co-prime then p+q is equal to.
Q.2 P is anypoint, O being the origin. The circle on OP as diameter is drawn. Points Q and R are taken on
the cirlceto lies on thesame side of OPsuch that POQ= QOR = . If P,Q, R are z1, z2, z3 such that
3
2 z2
2 = (2 + 3 ) z1z3. Then find angle .
Q.3 A student can solve 2 out of 4 problems of mathematics, 3 out of 5 problems of physics and 4 out of 5
problemsofchemistry.Thereareequalnumberofbooksofmaths,physicsandchemistryinhisshelf.He
selects one book randomlyand attempts 10 problems from it. If he solves the 1st problem, find the
probabilitycorrectly upto 2 places ofdecimal that he will be able to solvethe second problem.
Q.4 From an external point P(, 2) a variable line is drawn to meet the ellipse
4
y
9
x 2
2
 = 1 at the pointA
A
and D. Same line meets thex-axis and y-axis at the points B and C respectively. Find theleast positive
value of '' such that PA·PD = PB·PC.

SOLUTIONS
PARTA
Q.1 B Q.2 A
[Q1 & Q2 Sol. Prove that queen of hearts is obtained and the 8th draws =
52
1
Hence r = 1
Now radius of congurrent circle is x as shown
(x + 1)2 = (x – 1)2 + x2
x2 + 2x + 1 = 2x2 – 2x + 1
x2 – 4x = 0
x = 0 (not possible)
 x = 4
now b2 – a2 = d
c2 – a2 = 2d
8
c
a
—
—
—
—
—
4
b
c
x
12
b
a
given









2
2
2
2
a
b
a
c


= 2
a
b
a
c


·
a
b
a
c


= 2 q =
c
a
)
b
a
(
2


=
8
)
12
(
2


= 3
now r1 + r2 = 2;
2
2
2
1 r
r  + 2r1r2 = 4  q + 2D = 4  D = 1/2]
Q.3 (B)
[Sol I & III correct ; II – passes through (4a, 0)  False ; IV – passes through vertex hence false]
Q.4 (A)
[Hint: logy(x) > 1
Case-I: y  (0, 1), x < y
Case-II: y  (1, ), x > y
Area =
8
5
2
1
8
1
)
1
)(
1
(
2
1
2
1
2
1
2
1
















Ans. ]
Q.5 (B)
[Sol: Clearly |Z1
| = |Z2
| = |Z3
| = 5,
 points would lie on the circle centred at origin O.
 Circumcenter C is origin
Centroid G = 









3
8
3
6
2
7 i
OC = 3CG
= 3 x
3
1
  64
6
2
7
2


= 6
28
137  Ans. ]
Q6. (A)
[Hint: Bayes theorem 
A : card randomly chosen shows up red
B1 : card with both side red
B2 : card which both side blue
B3 : one side red & one side blue
P(B1) = P(B2) = P(B3) =
3
1
A(A/B1) = 1 P(A/B2) =
2
1
P(A/B3) = 0 P(B1/A) =
3
2
]

PART B
MATCH THE COLUMN:
Q1. [Ans: A  Q; B  P; C  S; D  Q]
[Sol:(A)
15
2 n
4
=
15
16n
=
15
)
15
1
( n

=
15
15
C
.......
15
C
15
C
1 n
n
n
2
2
n
1
n




=
15
k
15
1
, where k  N, =
15
1
+ k
 







15
2n
= 





 k
15
1
=
15
1
Ans: Q
(B) Let loge10 = x, then
 

 


n
0
r
r
n
e
e
r
n
r
10
log
1
10
log
r
1
C
)
1
( =
 

 


n
0
r
r
r
n
r
nx
1
rx
1
C
)
1
(
=
 





 










n
0
r
r
1
r
1
n
r
n
0
r
r
r
n
r
nx
1
rx
C
r
n
)
1
(
nx
1
1
C
)
1
(
= 

























n
0
r
1
r
1
r
1
n
1
r
n
0
r
r
r
n
r
nx
1
1
C
)
1
(
·
nx
1
nx
nx
1
1
C
)
1
(
=
1
n
n
nx
1
1
1
nx
1
nx
nx
1
1
1


















 =
n
n
nx
1
nx
nx
1
nx















= 0 Ans: P
(C) 1! + 2! + 3!+ 4! = 33  remainder is 3 Ans: S
(D) The given parabolas are
y = x – bx2 ..........(1)
and y =
b
x2
..........(2)
The abscissae of their points of intersection are given by
b
x2
= x – bx2  x2 (1 + b2) – bx = 0  x = 0, x = 2
b
1
b

Thus, the two parabolas intersect at O (0, 0) and
A
  








 2
2
2
b
1
b
,
b
1
b
The region enclosed by the two parabolas is shaded in figure. To find the area of this region, let us slice it into vertical strips. The
approximating rectangle shown in figure has length = y2 – y1, width = x and it can move between x = 0 and x = 2
b
1
b

. So, the
area A enclosed by the two parabolas is given by
A =  



2
b
1
b
0
1
2 dx
y
y
= 











2
b
1
b
0
2
2
dx
b
x
bx
x =
2
b
1
b
0
3
3
2
b
3
x
3
bx
2
x 








=  
3
2
2
2
2
b
1
b
b
1
b
3
1
b
1
b
2
1
















=
 2
2
2
2
2
2
2
b
1
b
6
1
b
1
b
3
1
b
1
b
2
1

















Now, A =
 2
2
2
b
1
b
6
1


   
  











 4
2
2
2
2
2
b
1
b
2
·
b
1
2
·
b
b
2
·
b
1
6
1
db
dA

 
 
2
2
3
2
b
2
b
1
b
1
3
b
db
dA






  
 3
2
b
1
3
b
1
b
1
b
db
dA




For maximum value of A, we must have
db
dA
= 0  b(1– b) (1 + b) = 0  1 – b = 0  b = 1  
0
b 

Since b > 0. Therefore, 1 + b > 0. Thus,
db
dA
changes its sign from positive to negative in the neighbourhood of b = 1. Ans: Q]
Q.2 [Ans: A  S; B  P; C  S; D  P]
[Sol:(A) x2 + 2 = 2| x | – cosx
(|x| – 1)2 + 1 = –cosx  x = ± 1
Area =  






1
1
2
dx
x
cos
|
x
|
2
2
x =
3
8
(B) 2
z
z
z
0
arg
1
p
1 












p
1
1
p
1
1
z
z
z
z
z
z


 = 0 ..................(1)
Also B, O and P are collinear  arg
p
2
z
z
= 
or
p
2
z
z
=
p
2
z
z
..................(2)
from (1) & (2) zp =
2
1
2
1
2
2
1
z
z
z
z
z
|
z
|
2

;
p
2
z
z
=
p
2
z
z
,
2
2
p
p
z
z
z
z 
(C) Let the perpendicular lines are x-axis and y-axis and let the equation of circle is
x2 + y2 + 2gx + 2fy + c = 0
xint = a
c
g
2 2

 and b
c
f
2
y 2
int 









 



4
b
a
f
g
2
2
2
2
 locus of centre is x2 – y2 = )
b
a
(
4
1 2
2

5
m
,
4
m
,
1 




 

(D) (x – a)2 + (y – a)2 = r2 ]
PART C
Q.1 [Ans: 63]
[Sol: y = 4 – x2 is a parabola EMA.
y = |x – 2| is a pair of straight lines ABF and ACD
y = (x – 2)1/3 is the curve IHABG
Thus frm the graph
f(x) = max {4 – x2, |x – 2|, (x – 2)1/3} is
f(x) =  






















4
x
3
if
2
x
3
x
2
if
2
x
2
x
1
if
x
4
1
x
2
if
x
2
3
/
1
2
Now, area bounded by the curve and x-axis
=  

 









3
2
4
3
3
/
1
2
1
2
1
2
dx
)
2
x
(
dx
)
2
x
(
dx
)
x
4
(
dx
)
x
2
(
=  
4
3
2
3
2
3
/
4
2
1
3
1
2
2
x
2
2
x
2
x
4
3
3
x
x
4
2
x
x
2 


































=
4
59
2
3
4
3
9
2
7



 sq. units ]

Q.2 [Ans: 15o]
[Sol: By rotation formula we have
z2 =

i
1e
z
OP
OQ
= z1cos ei ...........(1)
(considering  OQP)
z3 =

i
2
1e
z
OP
OR
= z1cos2 e2i ...........(2)
(considering  ORP)
Now z2
2 = z1
2 cos2 e2i z2
2 =



cos
z
cos
z
z
1
2
3
2
1
(from 2)
 z2
2 cos2 z1z3 cos
 3
2 z2
2 cos2 3
2 z1z3 cos .............(3)
It is given that 3
2 z2
2 =  
3
2  z1z3
Using this in equation (3)  
3
2  z1z3cos2 = 3
2 z1z3cos2   
3
2  (2cos2 – 1) = 3
2 cos2
 cos2
 =
4
3
2 
=
2
2
2
1
3







 
 cos =
2
2
1
3 
(where  is acute and hence cos > 0)   = 15°.
Q.3 [Ans: 0.65]
[Sol: Let P(m), P(p), P(c) be the probability of selecting book of maths, physics and chemistry respectively.
Clearly P(m) = P(p) = P(c) =
3
1
Again let P(s1) and P(s2) be the probability that he solves the 1st and 1st as well as 2nd problem respectively.
Then P (s1) = P(m) × P 





m
s1
+ P(p) × P 







p
s1
+ P(c) × P 





c
s1
 P(s1) =
30
19
5
4
3
1
5
3
3
1
2
1
3
1






Similarly P(s2) =
300
125
5
4
3
1
5
3
3
1
2
1
3
1
2
2
2























 
)
s
(
P
)
s
(
P
1
2
=
30
19
300
125
=
38
25
= 0.65 ]
Q.4 [Ans:  = 6]
[Sol: We have been give PA·PD = PB·PC
Equation of any line through point 'P' is :
=






sin
2
y
cos
x
= r
or, x =  + r cos, y = 2 + r sin
Putting this point in the equation of given ellipse we get:
4 (r cos + )2 + 9(rsin + 2)2 = 36
or r2 (4cos2 + 9sin2) + 4r (9cos + 2cos) + 42 = 0
Since PA and PD are the roots of this quadratic in r, we get
PA · PD =
 




2
2
2
sin
9
cos
4
4
Similarly, putting x = r cos + , y = r sin + 2 in the equation of the coordinate axes i.e. xy = 0 we get; (r cos + ) (r sin + 2)
= 0 or r2 sin cos + r(2cos + sin) + 2 = 0
Since PB and PC are the roots of this quadratic in 'r', we get
PB · PC =



cos
sin
2
=


2
sin
4
Thus, we finally get:


2
sin
4
=




2
2
2
sin
9
cos
4
4
=
   






2
cos
1
9
2
cos
1
4
8 2


2
sin
1
=



2
cos
5
13
2
 13 = 5cos2 + 2sin2
for existence of  






 4
25
13
4
25 2
R

 is the solution of L.H.S. inequality ....(1)
and )
,
6
[
U
]
6
,
( 



 is the solution of R.H.S. inequality ....(2)
 (1)  (2) is  (–, –6] U [6, )