Class : XII (NUCLEUS)
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-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
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 - XII Part Test - 3
PART -A
Select the correct alternative. Choose one or more than one.
Q1. Let P & Q are two points denoting the complex number  &  respectively on the complex plane.
Which ofthe following equations canrepresent the equationsofthe circle passing through P & Q with
least possible area?
(A) 











Z
Z
Arg =
2

(B*) Re (Z – )  


Z = 0
(C*)
2
Z 
 +
2
Z 
 =
2



(D) Z
Z + 






 


2 Z + 




 


2 Z +   +   = 0
[Hint: (A) is not correct as it should be +
2

(B) equation ofcircle is




Z
Z
is PI
 Re 











Z
Z
= 0
or Re (Z – A)  


Z = 0
 B is correct
(C) is obviouslycorrect frompythagoras theorem
(D) incorrect because centre ofrequired circle is 




 


2
but centre ofgiven circle is –
– coefficient of Z = – 




 


2
Q.2 The equation to the orthogonaltrajectories ofthe systemofparabolas y= ax2 is
(A*)
2
2
y
2
x
 = c (B)
2
y
x
2
2
 = c (C)
2
2
y
2
x
 = c (D)
2
y
x
2
2
 = c
[Sol: ax
2
dx
dy

a =
dx
dy
x
2
1
dx
dy
x
y
2


Orthogonaltrajectoryis
dy
dx
x
y
2


2y dy + x dx = 0
y2 +
2
x2
= C ]
Q.3 Consider thefollowing regions intheplane :
R1 = {(x, y) : 0  x  1 and 0  y  1}
R2 = {(x, y) : x2 + y2  4/3}
The area ofthe region R1  R2 can be expressed as
9
b
3
a 

, where a and b are integers. Then the
value of (a + b) equals
(A) 2 (B) 3 (C*) 4 (D) 5
[Sol. A = 1
A
1
3
1


A1 =  
 




 
1
3
1
2
dx
x
3
4
put x =
3
2
sin
= 





d
cos
3
2
·
cos
3
2
3
6
= 





d
)
1
2
(cos
3
2
3
6
=
3
6
2
sin
2
1
3
2











= 




 




6
2
3
·
2
1
3
2
3
·
2
1
3
2
=
6
·
3
2 
=
9

A =
9
3
1 
 =
9
3
3 

a = 3, b = 1
a + b = 4 Ans. ]
Q.4 Imagine that you have two thumbtacks placed at two points,Aand B. Ifthe ends ofa fixed length of
string are fastened to the thumbtacks and the string is drawn taut with a pencil, the path traced by the
pencilwillbe anellipse. Thebest wayto maximisethe area surrounded bythe ellipse with afixedlength
ofstringoccurs when
I the two pointsAand Bhave the maximumdistance between them.
II two pointsAand B coincide.
III Aand B are placed vertically.
IV The area is always same regardless ofthe location ofAand B.
(A) I (B*) II (C) III (D) IV
[Sol. A = ab 'a' is constant and b varies
A2 = 2a2(a2 – c2)
forAto be maximumc must be minimum; A& B  centre
as A  B  c  0
ellipse becomes circle ]

Q.5 Consider the expansion , (a1 + a2 + a3 + ....... + ap)n where nN and
n  p. The correct statements are
(A*) num
ber of di f f erent t er m
s i n t he expansi on i s , n+p 1Cn
(B)co-efficientofanyterminwhichnoneofthevariables a
1,a2,......,a
p occurmorethanonce
i
s'
n
'
(C*)co-efficien
tofanyterminwhichnoneoft
hevariables a
1,a2,....
..,a
p occurmor
ethanonce
i
s n
!
(D*)Numberoftermsinwhichnoneofthevariables a
1,a2,......,a
p occurmorethanonceis 





n
p
.
[
S
o
l
. (a1 +a
2 +a
3 +.....+a
p)n; p n
wehavepbeggarsandn identicalcoinstobedistributed. 







n
n
Ø
Ø
......
Ø
Ø
0
0
......
0
0
Total ways = n + p – 1Cn  (
A
)i
sc
o
r
r
e
c
t
a
g
a
i
n  degreeofeverytermwillbenandp n,hencenumberoftermswillbeasmanyas
thenumberofwaysinwhichntermscanbeselectedoutofp
 number of terms = pCn (
D
)i
sc
o
r
r
e
c
t
a
g
a
i
n coe
ffic
ientofthet
ermsay a
1,a2,a3.
...a
n=



terms
n
!
1
!.....
1
!
1
!
n
(usingthetheoryofmultinomialse.g.coefficientofa
2b3c4 in(a+b+c)
9is
!
4
!
3
!
2
!
9
]
Q.6 The probabilitythat a positive two digit number selected at randomhas its tens digit at least three more
thenitsunit digit is
(A*) 14/45 (B) 7/45 (C) 36/45 (D) 1/6
[Sol. n (S) = 9 · 10 = 90
6
,
5
,
4
,
3
,
2
,
1
,
0
9
2
,
1
,
0
5
1
,
0
4
0
3
possible
not
is
2
or
1
)
units
(
y
)
Tens
(
x

n (A) = 1 + 2 + 3 + ..... + 7 =
2
8
·
7
= 28
p =
90
28
=
45
14
Ans. ]
PART - B
MATCH THE COLUMN
INSTRUCTIONS:
Column-Iand column-IIcontains fourentrieseach. Entries ofcolumn-Iare to be matchedwith some
entriesofcolumn-II.Oneormorethanoneentriesofcolumn-Imayhavethematchingwiththesameentries
ofcolumn-IIandoneentryofcolumn-Imayhaveoneormorethanonematchingwithentriesofcolumn-II.
Q1. Column–I Column–II
(A) The differentialequationofallparabolas having (P) 1
their axisofsymmetrycoinciding withthe axis of
x has its degree
(B) The differentialequation ofallparabolaseach of (Q) 2
which has a latus rectum'4a' &whose axes are
parallelto x-axis is ofdegree
(C) Degree ofthe differentialequationy= a  
a
x
e
1 
 , (R) 3
a being the parameter is

(D) The polynomialf (x) satisfies the condition f (x+ 1) = (S) does not exist
x2 + 4x. The area enclosed by y = f (x – 1) and the
curve x2 + y = 0, is
k
2
16
where k is [Ans: (A)  P;(B)  P;(C)  S; (D)  R]
[Sol: (A) equation (x  a)2 + y2 = (x  b)2 [ S = (a , 0) ; D : x = b ]
y2 = (b2  a2) + 2x (a  b)
differentiate twice to get y
d y
dx
2
2 +
dy
dx






2
= 0 ; y 2
2
dx
y
d
+
2
dx
dy






= 0
(B) Equation to the familyof parabolas is (y– k)2 = 4a(x  h)
2(y – k)
dx
dy
= 4a  (y – k)
dx
dy
= 2a
(y – k) 2
2
dx
y
d
+
2
dx
dy






= 0
2a 2
2
dx
y
d
+
3
dx
dy






= 0. Hence the degree is 1.
(C) DE =
dx
dy
x – y ln
dx
dy
– x = 0
(D) f (x + 1) = x2 + 4x
put x = x – 2
f (x – 1) = (x – 2)2 + 4(x – 2)
f (x – 1) = x2 – 4
A = 2  


2
0
2
2
dx
)]
4
x
(
)
x
[(
= 2  
2
0
2
dx
)
x
2
4
( =
3
2
16
Ans. ]
Q2. Column–I Column–II
(A) The name ofthe conic represented bythe (P) Ellipse
equation px + qy = 1, where p, q  R,
p, q > 0 is
(B) Two parabola y2 = 4a (x – 1), and x2 = 4a(y – 2) (Q) Circle
always touch each other. 1 and 2 being variable
parameters. Then, their points ofcontact lie on a
(C) Ifparabola oflatus-rectuml, touches a fixed equal (R) parabola
parabola, the axes ofthe two curves being parallel,
then the locusofthe vertexofthemoving curve is
(D) Froma point P tangents are drawnto the parabola (S) hyperbola
y2 = 4ax. Ifthe chord ofcontact ofthese tangents
touches the rectangular hyperbola x2 – y2 = a2, then
the locus ofP is
[Ans: (A)  R;(B)  S;(C)  R; (D)  P]
[Sol: (A) We have, px + qy = 1

 ( px + qy )2 = 1
 px + qy + xy
pq
2 = 1
 (px + qy – 1)2 = 4pq xy
 p2x2 – 2pq xy + q2y2 – 2px – 2qy + 1 = 0
Oncomparing thisequationwiththe equation
a = p2, b = q2, c = 1, g = –p, f = –q and h = –pq
  = abc + 2fgh –af2 – bg2 – ch2
= p2q2 – 2p2q2 – p2q2 – p2q2 – p2q2
= –4p2q2  0
and, h2 – ab = p2q2 – p2q2 = 0
Thus we have
D  0 and h2 = ab So the givenrepresents a parabola
(B) Equate
dx
dy
to get xy= 4a2 which is a hyperbola
(C) Let the vertexofthe moving parabola is (h, k)
the the equationof the moving parabola (y– k)2 = –4a(x – h) .........(1)
Equation offixed parabola is y2 = 4ax .........(2)
where 4a = l
since these two parabolas touch each other
Solving eq. (1) and (2)
2y2 – 2ky +k2 – 4ah = 0
D = 0
k2 = 8ah
k2 = 2lh
y2 = 2lx
(D) Let P  (h, k). Chord ofcontact to parabola is
yk = 2a (x + h)  2ax – yk + 2ah = 0
Equation oftangent to hyperbola at anypoint
(asec, atan ) is
xsec – ytan – a = 0
On comparing equations (1) and (2)

sec
2a
=

tan
k
=
a
ah
–
2
 sec =
h
a
 , tan =
2h
k

So, 2
2
4h
k
h
a
2
2
 = 1
 k2 + 4h2 = 4a2
So the locus of(h, k) is 2
2
4a
y
a
x
2
2
 = 1
Therefore ans is Ellipse]

PART - C
Q1. Let Z = 18 + 26i where Z0 = x0 + iy0 (x0, y0  R) is the cube root of Z having least positive
argument. Find the value of x0y0(x0 + y0). [Ans. 12]
[Sol. Let Z = 18 + 26i
let r cos  = 18 and r sin  = 26
 r2 = 324 + 676 = 1000  r = 10
10
tan  =
18
26
=
9
13
; hence   




 
2
,
0 ;
3

 




 
6
,
0
 Z1/3 = [ 10
10 (cos  + i sin )]1/3 = 10 




 


3
sin
3
cos i
now tan  =
)
3
(
tan
3
1
)
3
(
tan
)
3
tan(
3
2
3





= 2
3
t
3
1
t
t
3


where t = tan
3

13(1 – 3t2) = 9(3t – t3)  13 – 39t2 = 27t – 9t3  9t3 – 39t2 – 27t + 13 = 0
3t2(3t – 1) – 12t(3t – 1) – 13(3t – 1) = 0  (3t – 1)(3t2 – 12t – 13) = 0
 tan
3

=
3
1

3

 




 
6
,
0
sin
3

=
10
1
and cos
3

=
10
3
 Z1/3 = 3 + i  x0 = 3 and y0 = 1
 x0y0 (x0 + y0) = 12 Ans. ]
Q2. A die isweighted such that the probabilityofrolling an n isproportionalto n2 (n= 1, 2, 3, 4, 5, 6). The
die is rolled twice, yielding the numbers a and b. The probability that a < b in lowest form is
q
p
where p + q =
[Sol. Given P(1) = K ; P(2) = 22K ; P(3) = 32K ; P(4) = 42 K ; P(5) = 52K ; P(6) = 62K
 Total = 91 K = 1  K =
91
1
;  P(1) =
91
1
; P(2) =
91
4
and so on
Let three eventsA, B, C are defined as
A : a < b
B : a = b
C : a > b
By symmetry, P(A) = P(C). Also P(A) + P(B) + P(C) = 1 ....(1)
since P(B) =
91
25
)]
i
(
P
[
6
1
i
2



2P(A) + P(B) = 1 ( from(i) )
P(A) =
2
1
[1 – P(B)] =
91
33
Ans. 33 + 91 = 124 ]
Q.3 If the expression z5 – 32 can be factorised into linear and quadratic factors over realcoefficients as
(z5 – 32) = (z – 2)(z2 – pz + 4)(z2 – qz + 4) then find the value of (p2 + 2p). [10]
[Ans. 4]

[Sol. z5 – 32 = (z – z0)(z – z1)(z – z2)(z – z3)(z – z4) [T/S, Q.24, Ex-1, Complex]
Where zi's , i = 0, 1, 2, 3, 4 are given by
zi = 2 




 


5
m
2
sin
5
m
2
cos i (using Demoivre'sTheorem)
with m = 0, 1, 2, 3, 4, we get z0 = 2,
amd z1 = 




 


5
2
sin
5
2
cos
2 i ; z2 = 




 


5
4
sin
5
4
cos
2 i ; z3 = 




 


5
6
sin
5
6
cos
2 i
z4 = 




 


5
8
sin
5
8
cos
2 i
hence, z5 – 32 = (z – 2) 







 4
z
5
2
cos
4
z2








 4
z
5
4
cos
4
z2

















z
z
)
(
z
)
z
)(
z
(
using
2
 p =
5
2
cos
4

= 4 cos 72° = 4 sin 18°
 p2 + 2p = [16sin218° + 8 sin18°] = 8[1 – cos 36° + sin 18°]
= 8 




 



4
1
5
4
1
5
1 = 4 Ans.]
Q.4 A movable parabola touches the x and the y-axes at (1, 0) and (0, 1). The locus of the focus of the
parabola is ax2–ax + ay2 – ay + 1 = 0 where a is
[Ans: 2]
[Sol: Since the x-axis and y-axis are two perpendicular tangents to the parabola and both meet at the origin,
the directrixpassesthroughthe origin.
Let y= mx be the directrix and (h, k) be the focus.
FA = AM
   2
2
k
1
h 
 = 2
m
1
m

.............(1)
and FB = BN

  2
2
1
k
h 
 = 2
m
1
m

.............(2)
Fromequations (1) and (2), we get
(h – 1)2 + h2 + k2 + (k – 1)2 = 1
 2x2 – 2x + 2y2 – 2y + 1 = 0 is the required locus. ]