1. PAPER CODE : A
P A P E R - 1
Class : XII & XIII
Time : 3 hour Max. Marks : 297
INSTRUCTIONS
1. The question paper contains 99 questions and 00 pages. Each question carry 3 marks and all of them are
compulsory.There is NEGATIVE marking. 1 mark will be deducted for eachwrong answer.
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.
2. Indicate the correct answer(s) for each questionbyfillingappropriate bubble(s) in your OMR sheet.
3. Use onlyHB pencil fordarkeningthe bubble(s).
4. Use ofCalculator, LogTable, SlideRule and Mobile is not allowed.
5. For exampleifonly'B'choice is correct then, thecorrect method for fillingthe bubble is
A B C D
For example if only'B & D' choicesare correct then, thecorrect method for fillingthe bubbles is
A B C D
The answer of the question in anyother manner (such as putting , cross , or partial shading etc.)
will be treated as wrong.
USEFUL DATA
Atomic Mass:Al = 27, Mg = 24, Cu = 63.5, Mn = 55, Cl = 35.5, O = 16, H = 1, P = 31, Ag = 108, N = 14,
Li = 7, I = 127, Cr = 52, K=39, S = 32, Na = 23, C = 12, Br = 80, Fe = 56, Ca = 40, Zn = 65.4, Ba = 137,
Co = 59, Hg = 200, Pb = 207, He = 4, F=19.
Radius of nucleus =10–14 m; h = 6.626 ×10–34 Js; me = 9.1 ×10–31 kg, R = 109637 cm–1.

2. XII & XIII MATHEMATICS (DATE: 11-03-2007) REVIEWTEST-12 / 9
Select the correct alternative(s). (Only One is correct) [15 × 5 = 75]
There is NEGATIVE marking. 1 mark will be deducted for eachwrong answer.
Q.1func Let f (x) = ax7 + bx3 + cx – 5, where a, b and c are constants. If f (–7) = 7, then f (7) equals
(A*) –17 (B) –7 (C) 14 (D) 21
[Hint: f (–x) = –ax7 – bx3 – cx – 5
 f (x) + f (–x) = – 10; Put x = 7
f (7) = – 10 – f (–x) = – 17 ]
Q.2mod For f (x) = sin 










 

2
x
2
cos , then the f ' 




 

4
(A) – cos 1 (B) 1 (C) – 2 (D*) 0
[Hint: f ' (x) = – cos 










 

2
x
2
cos sin 




 

2
x
2 · 2
f ' 




 

4
= – cos(cos 0) · 2 · 0 = 0 Ans. ]
Q.3trig If x and yare real numbers such that x2 + y2 = 8, The maximum possible value of x – y, is
(A) 2 (B*) 4 (C) 2 (D) 2
2
[Sol. x2 + y2 = 8
x = 2
2 cos ; y= 2
2 sin 
 x – y = 2
2 (cos  – sin ) = 4 cos( + /4)
 (x – y)max = 4 Ans. ]
Q.43d A plane passes through the point P(4, 0, 0) and Q(0, 0, 4) and is parallel to the y-axis. The distance of
theplanefromtheoriginis
(A) 2 (B) 4 (C) 2 (D*) 2
2
[Sol. x and z intercept of the plane is 4 and it is parallel to y-axis, hence equation of the plane is x + z = 4.
Its distance from (0, 0, 0) is 2
2 Ans. ]
Q.5flcd Let f (x)becontinuous and differentiable function for all reals.
f (x + y) = f (x) – 3xy + f (y). If
h
)
h
(
Lim
0
h
f

= 7, then the value of f ' (x) is
(A) – 3x (B) 7 (C*) – 3x + 7 (D) 2 f (x) + 7
Q.6auc For b> a >1, the area enclosed bythecurve y= ln x, yaxis and the straight lines y = ln a and y= ln b is
(A*) b – a (B) b(ln b – 1) – a(ln a – 1)
(C) (ln a)(b – a) (D) (ln b)(ln a)
[Sol. y = ln x x = ey
A = 
b
n
a
n
dy
x
l
l
= 
b
n
a
n
y
dy
e
l
l
= (eln b) – (eln a) = b – a Ans. ]

3. Q.7complex The set of positive values of n for which
n
2
30
cos
1
i
2
30
cos
1







 




is real, is
(A) 4, 8, 12,  (B) 6, 12, 18,  (C) 8, 16, 24,  (D*) 12, 24, 36, 
[Sol.
n
12
sin
i
12
cos 




 


=
12
n
sin
i
12
n
cos



for this to be real
12
n
= m
 n = 12 m, where m = 1, 2, 3, 
 12, 24, 36,   C ]
Q.8para PointsAandBareselectedonthegraph ofx2 +2y= 0so that thetriangleABOis equilateral.Thelength
of the sideof thetriangle is ('O'is the origin)
(A*) 3
4 (B)
3
4
(C)
3
4
(D) 3
2
[Sol. y = –
2
x2
tan 30° = 2
t
t
2

3
1
=
t
2
 t = 3
2
l(OP) = 3
4 Ans. ]
Q.9flcd If f (x) is continuous and differentiable over [–2, 5] and – 4  f ' (x)  3 for all x in (–2, 5) then the
greatest possible value of f (5) – f (–2) is
(A) 7 (B) 9 (C) 15 (D*) 21
[Hint: UsingLMVTin[–2, 5]
– 4 
7
)
2
(
f
)
5
(
f 

 3
– 28  f (5) – f (–2)  21 ]
Q.10aod The graph of f (x) =
20
x5
–
12
x4
+ 5 has
(A)norelativeextrema, onepointof inflection.
(B)tworelativemaxima,onerelativeminimum,twopointsofinflection.
(C*)onerelativemaximum,onerelativeminimum,onepointofinflection.
(D)onerelativemaximum,onerelativeminimum,twopointofinflection.
[Sol. f ' (x) =
3
x
4
x 3
4
 = )
4
x
3
(
12
x3

f '' (x) = x3 – x2 = x2(x – 1)
for x < 0, f ' (x) > 0 and
x > 0, f ' (x) < 0 hence at x = 0, we have maxima
Also f '' (0) = 0 when x = 1 and x = 0 but at x = 1 only we have an inflection point
 (C) ]

4. Q.11trig Minimum value of 






2
A
tan
3
2
A
cot
2
1
whereA (0, 180°) occurs whenAequals
(A*) 60° (B) 90° (C) 120° (D) 150°
[Sol. UsingAM  GM







2
A
tan
3
2
A
cot
2
1
 3
Henceminimumvalueis 3
now cot
2
A
+ 3 tan
2
A
= 3
2 ; 3tan2
2
A
– 3
2 tan
2
A
+ 1 = 0
tan
2
A
=
6
12
12
3
2 

=
3
1
;
2
A
= 30°  A= 60°Ans. ]
Q.12qe Let k be an integer and p is a prime such that the quadratic equation x2 + kx + p = 0 has two distinct
positive integer solutions. Thevalue of (k+ p)equals
(A*) – 1 (B) 0 (C) 3 (D) 6
[Sol. Let r1 and r2 aretwo integral solutions
r1 + r2 = – k and r1r2 = p
since p is prime hence either r1 = 1 or r2 = 1
let r1 = 1; r2 = p
1 + p = – k  k + p = – 1 ]
Q.13elli Consider the particle travellingclockwise on the elliptical path
25
y
100
x 2
2
 =1. The particleleaves the
orbitat thepoint (–8, 3)andtravels inastraightlinetangent to theellipse.Atwhat pointwill theparticle
cross the y-axis?
(A*) 





3
25
,
0 (B) 






3
25
,
0 (C) (0, 9) (D) 





3
7
,
0
Q.14de Thesolutionof thedifferentialequation,ex(x +1)dx+(yey –xex)dy=0 withinitialconditionsf (0) = 0,
is
(A) xex + 2y2ey = 0 (B*) 2xex + y2ey = 0 (C) xex – 2y2ey = 0 (D) 2xex – y2ey = 0
[Sol. put xex = t
(ex + xex)dx = dt

dy
dt
+ (yey – t) = 0 
dy
dt
– t + yey = 0
I.F. 
 dy
e = e–y
t · e–y = – 

dy
e
ye y
y
x ex e–y = –
2
y2
+ C
f (0) = 0  C = 0; 2x ex e–y + y2 = 0 Ans.]

5. Q.15trig The value of the expression, cos 35° + cos 125° +2 sin 185°(sin 130° + sin140°) when simplified is
(A) positive andgreater than(1/2) (B) negative and less than (– 1/2)
(C) 1/2 (D*) zero
[Sol. 2 cos 80° cos 45° – 2 sin 5°(2 sin 135° cos 5°)
2 sin 10° cos 45° – 2 sin 10° sin 45° = 0 Ans. ]
Q.16func f (x) = 1
x
4 2
5 
. How manyof the following are true?
I The domain of f is | x | 
2
1
II The range of f (x) is y  1
III Thegraphoff (x)is symmetricabout the y-axis.
IV f(x)hasnocritical points.
V The graph of f (x) never decreases.
(A) one (B) two (C*) three (D)four
[Hint: only(i), (ii)and (iii) are correct. f(x) has critical points at x = ±
2
1
as
dx
dy
  and f (x) is decreasing
for x < – 1/2]
Q.17mat IfA= 





 1
x
tan
x
tan
1
then let us define a function f (x) = det. (ATA–1) then which of the following
can not be the value of  
 
 


 


 

times
n
)
x
(
f
.
..........
f
f
f
f is (n  2)
(A) f n(x) (B) 1 (C) f n – 1(x) (D*) n f (x)
[Sol. ATA–1 = 




 
x
2
cos
x
2
sin
x
2
sin
x
2
cos
so det. (ATA–1) = 1
so f (x) = 1 is a constant function.]
Q.18st.line The equations of L1 and L2 are y= mx and y= nx, respectively. Suppose L1 makes twice as large of
ananglewiththehorizontal(measuredcounterclockwisefromthepositivex-axis)asdoesL2 andthatL1
has 4 times the slope of L2. If L1 is not horizontal, then the value of the product (mn) equals
(A)
2
2
(B) –
2
2
(C*) 2 (D) –2
[Sol. Let m = tan 2 and n = tan 
and m = 4n & m  0,  tan 2  0;    0
tan 2 = 4 tan 



2
tan
1
tan
2
= 4 tan 
 1 = 2 – 2 tan2
2 tan2 = 1  tan2 = 1/2
 mn = tan 2· tan =



2
2
tan
1
tan
2
= )
2
1
(
1
1
 = 2 Ans. ]

6. Q.19prob Three balls marked 1, 2 and3 are placed inan urn. One ball is drawn, its number is recorded, then the
ballisreturnedtotheurn.This process isrepeated andthenrepeated oncemore,and each ballis equally
likelytobedrawnon eachoccasion. Ifthesumofthenumbersrecordedis 6,theprobabilitythat theball
numbered 2 was drawn at all the three occassions, is
(A)
27
1
(B*)
7
1
(C)
6
1
(D)
3
1
[Sol. He picked 3times, the followingoutcome add up to 6
1, 2, 3









1, 3, 2
2, 3, 1
2, 1, 3 all theseevents are equallylikely,ME and exhaustive. Out of7 cases onlyonefavours
2, 2, 2
3, 1, 2
3, 2, 1
Therefore, there is 1/7 chance. ]
Q.20log Number of integral values ofx the inequality 







1
x
2007
x
2
log10  0 holds true, is
(A) 1004 (B*) 1005 (C) 2007 (D) 2008
[Sol. 0 <
1
x
2007
x
2


 1
0 < 2x – 2007  x + 1
0 < x  2008 ....(1)
also 2x – 2007 > 0
x > 1003.5 or x  1004 ....(2)
from (1) and (2)
]
Q.21def Let f bea one-to-one continuous function such that f (2) =3 and f (5) =7. Given 
5
2
dx
)
x
(
f =17, then
thevalueofthedefiniteintegral 

7
3
1
dx
)
x
(
f equals
(A) 10 (B) 11 (C*) 12 (D) 13
[Sol. y= f (x)  x = f–1(y) and dy = f ' (x) dx
now I = 

7
3
1
dx
)
x
(
f = 

7
3
1
dy
)
y
(
f = 
5
2
dx
)
x
(
'
f
x ; (when y = 3 then x = 2 and y = 7 then x = 5)
hence I= 
5
2
dx
)
x
(
'
f
x . Integrating byparts gives,
I =
5
2
)
x
(
x f – 
5
2
dx
)
x
(
f
I = 5 · 7 – 2 · 3 – 17 = 35 – 6 – 17 = 12 Ans. ]

7. Q.22trig Tworays withcommon endpoint 'O'forms a30° angle. PointAlies on oneray,point B on theother ray
andAB =1.Themaximum possible lengthof OB is
(A) 3 (B)
3
4
(C*) 2 (D)
2
1
3 
[Sol. Usingcosinerule
1 = x2 + y2 – 2xy cos30°
 y2 – xy 3 + x2 – 1 = 0
as y is real, D  0
3x2 – 4(x2 – 1)  0
x2  4  x  2
Maximumvalueofxis2
 C ]
Q.23 Suppose that thequadraticfunction f (x) =ax2 +bx + c is non-negativeontheinterval [–1, 1]. Then the
area under the graph of f over the interval [–1, 1] and the x-axis is given bythe formula
(A) A = f (–1) + f (1) (B) A= 













2
1
2
1
f
f
(C) A = )]
1
(
)
0
(
2
)
1
(
[
2
1
f
f
f 

 (D*) A= )]
1
(
)
0
(
4
)
1
(
[
3
1
f
f
f 


[Sol. A = 



1
1
2
dx
)
c
bx
ax
( =  
1
0
2
dx
)
c
ax
(
2
= 





 c
3
a
2 =
3
1
[2a + 6c]
 A = )]
1
(
)
0
(
4
)
1
(
[
3
1
f
f
f 

 Ans. ]
Q.24hyp Let F1, F2 arethe foci of thehyperbola
9
y
16
x 2
2
 =1 and F3, F4 arethe foci of its conjugate hyperbola.
If eH and eC aretheir eccentricities respectivelythenthe statement which holdstrue is
(A) Theirequations oftheasymptotes are different.
(B) eH > eC
(C*)Area of the quadrilateral formed bytheir foci is 50 sq. units.
(D)Theirauxillarycircleswillhavethesameequation.
[Hint: eH = 5/4; eC = 5/3
area =
2
d
d 2
1
=
2
100
= 50
AC: x2 + y2 = 16; AH = x2 + y2 = 9 ]

8. Q.25qe Consider the two functions f (x) = x2 + 2bx + 1 and g(x) = 2a(x + b), where the variable x and the
constants aandbarereal numbers.Each suchpair oftheconstants aandb maybeconsideredas apoint
(a, b) in an ab – plane. Let S be the set of such points (a, b) for which the graphs of y = f (x) and
y = g (x) do not intersect (in the xy– plane.). The area of S is
(A) 1 (B*)  (C) 4 (D) 4
[Sol. We need x2 +2bx + 1 =2ax + 2ab not to have anyreal solutions, implying that thediscriminant is less
than or equal to zero.Actuallycalculating the discriminant and simplifying, we get a2 + b2 <1, which
describes a circle of area  in the a-b plane]
Q.26det Let 0 =
33
32
32
23
22
21
13
12
11
a
a
a
a
a
a
a
a
a
and let 1 denote the determinant formed bythe cofactors of elements of
0 and 2 denote the determinant formed by the cofactor at 1 and so on n denotes the determinant
formed bythe cofactors at n – 1 then the determinant value of n is
(A) n
2
0
 (B*)
n
2
0
 (C)
2
n
0
 (D) 2
0

[Sol. 1 = 2
0

2 = 2
1
 = 4


3 = 2
2
 = 8
0

4 = 2
3
 = 16
0

and so on n =
n
2
0
 Ans. ]
Q.27 The polynomial P(x) = x3 + ax2 + bx + c has the propertythat the mean of its zeroes, the product of its
zeroes,andthesumofitscoefficientsareallequal.Ifthey-interceptofthegraph ofy=P(x)is 2,thenthe
value of b is
(A*) – 11 (B) – 9 (C) 1 (D) 5
[Sol. The y-intercept is at x = 0, so we have c= 2, meaning that the product of theroots is – 2.We know that
a is the sum of the roots. The average of theroots is equal to the product, so the sum of the roots is – 6,
and a = 6. Finally, 1 + a + b + c = 2 as well, so we have 1 + 6 + b + 2 = – 2  b = – 11 Ans.]
Q.28bin If 

2006
10
k
10
k
C simplifies to nCp where p is prime then (n + p) has the value equal to
(A) 2017 (B*) 2018 (C) 2019 (D) 2020
[Sol. S = 10C10 + 11C10 + 12C10 + 13C10 + ........... + 2006C10
=
 
 
 1
11
0
11
C
C  + 12C2 + 13C3 + ........... + 2006C1996
= ........
C
C 2
12
1
12


 
 

13C2 and so on
 S = 2007C1996 = 2007C11 = nCp
 n + p = 2018 Ans. ]

9. Q.29p&c The number of three elements sets of positive integers {a, b, c} such that a × b × c = 2310, is
(A*) 40 (B) 30 (C) 25 (D) 15
[Sol. N = 2310 = 2 · 3 · 5 · 7 · 11 [T/S, Q.26, Ex-2, P&C]
Wewanttoeitherpartitionthese5primenumbers ingroupsof2 withtheother third number1 orgroups
of 3.
Now
5 = 1 + 4 = 2 + 3 and 5 = 1 + 1 + 3 = 1 + 2 + 2
Hence, numberof ways
!
4
!
1
!
5
+ !
3
!
2
!
5
+ !
3
!
2
!
1
!
1
!
5
+ !
2
!
2
!
2
!
1
!
5
= 5 + 10 + 10 + 15 = 40 Ans. ]
Q.30def(prop) Let u = dx
x
sin
3
2
cos
2
/
0
2







 
and v = dx
x
sin
3
cos
2
/
0







 
, then the relation between u and v is
(A*) 2u = v (B) 2u = 3v (C) u = v (D) u = 2v
[Sol. u = dx
x
sin
3
2
cos
2
/
0
2







 
u = dx
x
cos
3
2
cos
2
/
0
2







 
Onadding
2u = dx
x
2
cos
3
cos
·
3
cos
2
2
/
0







 

(using cos C + cos D)
= dt
t
cos
3
cos
2
1
0







 
[Put 2x = t]
= dt
t
cos
3
cos
2
/
0







 
= dt
t
sin
3
cos
2
/
0







 
= v  (A) ]
Q.31cir In the figure given,two circles with centres C1 and C2 are35units apart, i.e. C1C2 =35. Theradii of the
circles with centres C1 and C2 are 12 and 9 respectively. If Pis theintersection of C1C2 and a common
internaltangent to the circles, then l(C1P)equals
(A) 18 (B*) 20
(C) 12 (D) 15
[Sol.
Let C1P = a
PC2 = 35 – a
Hence,
a
35
9

=
a
12
 3a = 4(35 – a)
 7a = 4 · 35  a = 20 Ans. ]

10. Q.32complex Let Z = 

2007
1
k
k
ki , then Re(Z) + Im(Z) equals (where i = 1
 )
(A) 0 (B) 2007 (C*) – 2008 (D) 3012
[Hint: Z = i + 2i2 + 3i3 +  + 2007i2007
Use method of difference to get Z = –1004 – 1004i
 Re(Z) + Im(Z) = –2008 ]
Q.33 Position vectorsofthefour angularpoints ofatetrahedronABCDareA(3, –2, 1); B(3, 1,5); C(4, 0, 3)
and D(1, 0, 0).Acute angle between the plane facesADC andABC is
(A*) tan–1 





2
5
(B) cos–1 





5
2
(C) cosec–1 





2
5
(D) cot–1 





2
3
[Sol. k̂
4
ĵ
3
î
0
V1 



k̂
2
ĵ
2
î
V2 



k̂
ĵ
2
î
2
V3 




2
2
1
4
3
0
k̂
ĵ
î
V
V
n 2
1
1 





k̂
)
3
0
(
ĵ
)
4
0
(
î
)
2
(
n1 






k̂
3
ĵ
4
î
2
n1 




1
2
2
2
2
1
k̂
ĵ
î
V
V
n 3
2
2








k̂
)
4
2
(
ĵ
)
4
1
(
î
)
4
2
(
n2 








k̂
6
ĵ
3
î
6
n2 




= )
k̂
2
j
ˆ
î
2
(
3 


cos  = |
n
|
|
n
|
n
·
n
2
1
2
1




=
3
·
29
6
4
4 


=
29
2
hence tan  =
2
5
  = tan–1 





2
5
]