1. REVIEW TEST-3
PAPER CODE : A
P A P E R - 1
Class : XIII (XYZ)
Time : 3 hour Max. Marks : 216
INSTRUCTIONS
1. The question paper contains 72 questions and 16 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 correctanswer for each question byfillingappropriate bubble inyour answer sheet.
3. Use onlyHB pencil fordarkeningthe bubble.
4. Use ofCalculator, LogTable, SlideRule and Mobile is not allowed.
5. TheanswerofthequestionsmustbemarkedbyshadingthecirclesagainstthequestionbydarkHBpencilonly.
6. The answer(s) ofthe questions must bemarked byshading thecircles against the questionbydark HB
pencilonly.
For exampleifonly'B'choice is correct then, thecorrect method for fillingthe bubble is
A B C D
thewrongmethodfor fillingthebubbleare
(i) A B C D
(ii) A B C D
(iii) A B C D
The answer of the questions inwrong or anyothermanner will be treated as wrong.
USEFUL DATA
Atomic weights: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,
Radius of nucleus =10–14 m; h = 6.626 ×10–34 Js; me = 9.1 ×10–31 kg, R = 109637 cm–1

2. XIII (XYZ) MATHS (24-9-2006) REVIEW TEST-3
PART-A
Select the correct alternative. (Only one is correct) [24 × 3 = 72]
There is NEGATIVE marking. 1 mark will be deducted for eachwrong answer.
Q.1cir The area of the region of the plane bounded above bythe graph of x2 + y2 + 6x + 8 = 0 and below by
the graph of y = | x + 3 |, is
(A*)
4

(B)
4
2

(C)
2

(D) 
[Sol. Completing the square, the top curve is the circle (x + 3)2 + y2 = 1
and the lowercurvewhichis aright angle"vee-shape",
cuts the circle (of radius 1)into quarters.
The area of the region is
4
r2

=
4

Ans.]
Q.2st.line Consider straight line ax + by = c where a, b, c  R+ and a, b, c are distinct. This line meets the
coordinate axes at Pand Q respectively. If area of OPQ, 'O' being origin does not depend upon a, b
and c, then
(A) a, b, c are in G.P. (B*) a, c, b are in G.P. (C) a, b, c are in A.P. (D) a, c, b are inA.P.
[Hint: A =
2
1
1
0
a
c
1
b
c
0
1
0
0
=
2
1
ab
c
0
2

IfAis independent of a, b and c then c2 = ab
 a, c, b are in G.P. ]
Q.3ph-1 If x and y are real numbers and x2 + y2 = 1, then the maximum value of (x + y)2 is
(A) 3 (B*) 2 (C) 3/2 (D) 5
[Sol. x2 + y2 = 1
let x = cos and y= sin so (x + y)2 can be written in the form of
(cos + sin)2 = 2 








4
sin2
maximum value=2 Ans. ]
Q.4def The value of thedefinite integral 



0
2
a
)
x
1
)(
x
1
(
dx
(a >0) is
(A*)
4

(B)
2

(C)  (D) somefunction of a.
[Hint: put x = tan 
I = 




2
0
a
)
(tan
1
d
= 






2
0
a
a
a
d
)
(cos
)
(sin
)
(cos
 I =
4

Ans. ]

3. Q.5mod Let a, b, c are non zero constant number then
r
c
sin
r
b
sin
r
c
cos
r
b
cos
r
a
cos
Lim
r



equals
(A)
bc
2
c
b
a 2
2
2


(B)
bc
2
b
a
c 2
2
2


(C*)
bc
2
a
c
b 2
2
2


(D) independent of a, b and c
[Sol. Let
r
1
= x so that as r  , x  0
2
0
x
x
·
bc
·
cx
cx
sin
·
bx
bx
sin
cx
cos
·
bx
cos
ax
cos
Lim


= 2
0
x x
cx
cos
·
bx
cos
ax
cos
Lim
bc
1 

use L'Hospital's to get (C) ]
Q.6de A curve y = f (x) such that f ''(x) = 4x at each point (x, y) on it and crosses the x-axis at (–2, 0) at an
angle of 45°. The value of f (1), is
(A) – 5 (B*) – 15 (C) –
3
55
(D) –
3
35
[Sol. f ' (x) = 2x2 + c
f ' (–2) = 1 = 8 + c  c = – 7
now f ' (x) = 2x2 – 7
f (x) = d
x
7
x
3
2



f (–2) = 0  –
3
16
+ 14 + d = 0  d = –
3
26
 f (x) =
3
1
[2x3 – 21x – 26]
f (1) = –
3
45
= – 15 Ans. ]
Q.7ph-1 Theminimumvalueofthefunctionf(x)=
x
cos
1
x
sin
2

+
x
sin
1
x
cos
2

+
1
x
sec
x
tan
2

+
1
x
cosec
x
cot
2

as x varies over all numbers inthe largest possible domain of f (x) is
(A) 4 (B*) – 2 (C) 0 (D) 2
[Hint:
2
1
1
1
1
y
then
quadrant
4
x
if
0
1
1
1
1
y
then
quadrant
3
x
if
2
1
1
1
1
y
then
quadrant
2
x
if
4
1
1
1
1
y
then
quadrant
1
x
if
th
rd
nd
st




























2
ymin 







Q.8 A non zero polynomial with real coefficients has the property that f (x) = f ' (x) · f ''(x). The leading
coefficient of f(x) is
(A)
6
1
(B)
9
1
(C)
12
1
(D*)
18
1

4. [Sol. Degree of f (x) = n; degree of f ' (x) = n – 1
degree of f ''(x) = (n – 2)
hence n = (n – 1) + (n – 2) = 2n – 3
 n = 3
hence f (x) = ax3 + bx2 + cx + d, (a  0)
f ' (x) = 3ax2 + 2bx + c
f '' (x) = 6ax + 2b
 ax3 + bx2 + cx + d = (3ax2 + 2bx + c)(6ax + 2b)
 18a2 = a  a =
18
1
Ans. ]
Q.9def Let Cn = 



n
1
1
n
1
1
1
dx
)
nx
(
sin
)
nx
(
tan
then n
2
n
C
·
n
Lim


equals
(A) 1 (B) 0 (C) – 1 (D*)
2
1
[Sol. Cn = 



n
1
1
n
1
1
1
dx
)
nx
(
sin
)
nx
(
tan
(put nx = t)  Cn = 



1
1
n
n
1
1
dt
)
t
(
sin
)
t
(
tan
n
1
L = 








1
1
n
n
1
1
n
n
2
n
dt
t
sin
t
tan
·
n
Lim
C
·
n
Lim ( × 0); L = =
n
1
dt
t
sin
t
tan
1
1
n
n
1
1




applyingLeibnitzrule
L =
2
2
1
1
n
n
1
)
1
n
(
1
1
n
n
sin
1
n
n
tan
Lim

















=

 2
·
4
=
2
1
Ans. ]
Q.10complex Let z1, z2, z3 be complex numbers such that z1 + z2 + z3 = 0 and | z1 | = | z2 | = | z3 | = 1 then
2
3
2
2
2
1 z
z
z 
 , is
(A) greater than zero (B) equal to 3 (C*) equal to zero (D) equal to 1
[Sol. 1
1z
z = 1 ; 2
2z
z = 1 ; 3
3z
z = 1
given, z1 + z2 + z3 = 0  3
2
1 z
z
z 
 = 0
hence (z1 + z2 + z3)2 = 0
 2
1
z + 2z1z2z3 









3
2
1 z
1
z
1
z
1
= 0 ; hence  2
1
z + 2z1z2z3 [ 3
2
1 z
z
z 
 ] = 0
  2
1
z = 0 Ans. ]

5. Q.11p/c Numberof rectangleswith sides parallel to the coordinateaxes whose verticesare all of theform (a, b)
with a and b integers such that 0  a, b  n, is (n  N)
(A*)
4
)
1
n
(
n 2
2

(B)
4
n
)
1
n
( 2
2

(C)
4
)
1
n
( 2

(D) n2
[Hint: number of rectangles = n+1C2 · n+1C2 =
4
)
1
n
(
n 2
2

]
Q.12aod Number of roots of the function f (x) = 3
)
1
x
(
1

– 3x + sin x is
(A) 0 (B) 1 (C*) 2 (D) more than 2
[Sol. f ' (x) = – 4
)
1
x
(
3

– 3 + cos x < 0
hence f (x) is always decreasing,Also as x  , f (x)  –  and as x  – , f (x)  + 
hence one positive and onenegative root
Graph is as shown
]
Q.13QE If p (x) = ax2 + bx + c leaves a remainder of 4 when divided by x, a remainder of 3 when divided by
x + 1, and a remainder of 1 when divided byx – 1 then p(2) is
(A) 3 (B) 6 (C) – 3 (D*) – 6
[Hint: p(x) = ax2 + bx + c
p(x) = Q1x + 4
= Q2(x + 1) + 3
= Q3(x – 1) + 1
 p(0) = c = 4 ....(1)
p(–1) = a – b + 4 = 3  b – a = 1
p(1) = a + b + 4 = 1  b + a = – 3
 b = – 1; a = – 2
 p(x) = – 2x2 – x + 4
 p(2) = – 8 – 2 + 4 = – 6 Ans. ]
Q.14aod Let f (x) bea function that has a continuous derivative on [a, b], f (a) and f (b)have opposite signs, and
f ' (x)  0 for all numbers x between a and b, (a < x < b). Number of solutions does the equation
f (x) = 0 have (a < x < b).
(A*) 1 (B) 0 (C) 2 (D) cannot be determined
[Sol. As f (x) has contained derivative  x [a, b]  f (x) is continuous and as f (a) & f (b) have
opposite signs at least one value a < x = c < b such that f (c) = 0 also since f '(x)  0  f ' (x) > 0
or f '(x) < 0 throughout  function is increasing ordecreasing  will have onlyonepossible root. ]

6. Q.15def Whichofthefollowingdefiniteintegralhasapositivevalue?
(A) 



3
2
0
dx
)
x
3
sin( (B) 


0
3
2
dx
)
x
3
sin( (C*) 



0
2
3
dx
)
x
3
sin( (D) 




2
3
0
dx
)
x
3
sin(
[Sol. put 3x +  = t
(I) 


3
dt
3
t
sin
= 0; (II) 


3
dt
3
t
sin
= 0;
(III) 


 2
7
dt
3
t
sin
= + ve (IV) 



2
7
dt
3
t
sin
= – ve  (C) ]
Q.16p&c Let setAconsists of 5 elements and set B consists of 3 elements. Number of functions that can be
defined fromAtoBwhichareneitherinjectivenorsurjective,is
(A) 99 (B*) 93 (C) 123 (D) none
[Sol. numberofmapping
= 3C1 (25 – 2) + 3C2
= 90 + 3 = 93 Ans. ]
Q.17cir Acircle with centerAandradius 7 is tangent to thesides ofan angle of
60°.AlargercirclewithcenterBis tangent tothesides oftheangleand
to the first circle. The radius of thelarger circle is
(A) 3
30 (B*) 21
(C) 3
20 (D) 30
[Sol. r = 7
sin 30° =
r
R
r
R


=
2
1
2R – 2r = R + r
R = 3r = 21 Ans. ]
Q.18vector The value of the scalar    
s
r
·
q
p





 can be expressed in the determinant form as
(A)
s
·
p
r
·
p
s
·
q
r
·
q








(B)
r
·
q
s
·
q
s
·
p
r
·
p








(C)
s
·
p
r
·
q
s
·
q
r
·
p








(D*) s
·
q
r
·
q
s
·
p
r
·
p 







[Sol.   s
·
r
)
q
p
(





 =   s
·
p
)
q
·
r
(
q
)
p
·
r
(







 = )
s
·
p
)(
q
·
r
(
)
s
·
q
)(
p
·
r
(








 ]
Q.19limit If
x
1
0
1
x
1
0
1
x
n
·
x
Lim
x





l = – 5, where , ,  are finite real numbers then
(A)  = 2,  = 1,   R (B)  = 2,  = 2,  = 5
(C)   R,  = 1,   R (D*)   R,  = 1,  = 5
[Hint: Let, L= 




 





 x
x
n
·
x
Lim 3
x
l ; for limit to exist, 1


now L = 




 




 x
x
1
n
x
Lim 3
x
l = 




 



 x
x
x
Lim 3
x
= 









 2
x x
Lim = –  = – 5
 5

 and R

 ]

7. Q.20ITF If f (x, y) = sin–1( | x | + | y | ), then the area of the domain of f, is
(A*) 2 (B) 2
2 (C) 4 (D) 1
[Sol. | x | + | y |  1  interior of a square with vertices (1, 0), (0, 1), (–1, 0), (0, – 1)  area = 2Ans.]
Q.21s&p A,B and C aredistinct positive integers, less than or equal to10.The arithmetic meanofAand Bis 9.
The geometric mean ofAand C is 2
6 . The harmonic mean of Band C is
(A*)
19
9
9 (B)
9
8
8 (C)
19
7
2 (D)
17
8
2
[Sol. A + B = 18 .....(1)
AC = 72 .....(2)
Thereareonlytwopossibilities
A = 10 and B = 8
If A= 10 then from (2) C is not an integer.
Hence A = 8 and B = 10; C = 9
 H.M. between B and C =
9
10
9
·
10
·
2

=
19
180
=
19
9
9 Ans. ]
Q.22 If x is real and 4y2 + 4xy+ x + 6 = 0, then the complete set of values of x for which yis real, is
(A) x  2 or x  3 (B*) x  – 2 or x  3
(C) – 3  x  2 (D) x  – 3 or x  2
[Hint: D  0  16x2 – 16(x + 6)  0  (x – 3)(x + 2)  0]
Q.23prob Ialternativelytoss afaircoinandthrowafairdieuntil I, eithertossahead orthrowa2. IfItoss thecoin
first, the probabilitythat Ithrow a 2 before I toss a head, is
(A*) 1/7 (B) 7/12 (C) 5/12 (D) 5/7
[Sol. H: tossing a Head, P(H) =
2
1
; A : event of tossing a 2 with die, P(A) =
6
1
E: tossinga 2 beforetossinga head
P(E) = P  
 
.......
or
)
A
H
(
and
)
A
H
(
or
A
H 


= 





6
1
·
2
1
+ 





6
5
·
2
1
· 





6
1
·
2
1
+ .......
=
12
1
+
12
5
·
12
1
+ ....... 
P(E) =
12
5
1
12
1

=
7
1
Ans. ]
Q.24mat LetA, B, C, D be (not necessarilysquare) real matrices such that
AT = BCD; BT = CDA; CT = DAB and DT = ABC
for the matrix S =ABCD, consider the two statements.
I S3 = S
II S2 = S4
(A) II is true but not I (B) I is true but not II
(C*) both I and II are true (D) both I and II are false.

8. [Sol. S = ABCD = A(BCD) = AAT ....(1)
S3 = (ABCD)(ABCD)(ABCD)
= (ABC)(DAB)(CDA)(BCD)
= DTCTBTAT
= (BCD)TAT = AAT ....(2)
from (1) and (2) 3
S
S   I is correct
multiplybothsidesbyS
4
2
S
S   II is correct  Both I and II are true Ans. ]