1. REVIEW TEST-8
PAPER CODE : A
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 will be
treated as incorrect. Insert
leadingzero(s)ifrequiredafter
rounding the result to 2
decimal places.
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 the bubble
is
A B C D
For example if only 'B & D'
choices are correct then, the
correct method for filling the
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.
PART-D
Ensure that all columns
{1 before decimal and 2 after
decimal with proper sign (+)
or (–)} arefilledand columns
after 'E'usedfor fillingpower
of 10 with proper sign (+) or
(–). Answer having blank
column will be treated as
incorrect.
e.g. – 4.19 × 1027 should be
filled as – 4.19 E + 27
P A P E R - 2
Class : XII (ABCD)
Time : 3 hour Max. Marks : 243
INSTRUCTIONS
1. The question paper contain 00 pages and 4-parts. Part-A contains 9 objective question, Part-B contains
6 questions of "Match the Column" type and Part-C contains 14 and Part-D contains 3 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.9 have One or More than one is / are correct alternative(s) and carry 5 marks each.
There is NEGATIVE marking. 1 mark will be deducted for each wrong answer.
PART-B
(iii) Q.1 to Q.6 are "Match the Column" type which may have one or more than one matching options and carry 8 marks
for each question. 2 marks will be awarded for each correct match within a question.
There is NEGATIVE marking. 0.5 Marks will be deducted for each wrong match. Marks will be awarded onlyif all the
correct alternatives are selected.
PART-C
(iv) Q.1 to Q.14 are "Subjective" questions. There is NO NEGATIVE marking. Marks will be awarded onlyifall the correct
bubbles are filled in your OMR sheet.
PART-D
(iv) Q.1 to Q.3 are "Subjective" questions. There is NO NEGATIVE marking. Marks will be awarded only if all the correct
bubbles are filled in your OMR sheet.
2. Indicate the correct answer for each question by filling appropriate bubble(s) in your answer sheet.
3. Use only HB pencil for darkening the bubble(s).
4. Use of Calculator, Log Table, Slide Rule and Mobile is not allowed.
5. The answer(s) of the questions must be marked by shading the circles against the question by dark HB pencil only.

2. Select the correct alternative. (More than one are correct) [3 × 5 = 15]
Q.7 Let a1, a2, a3 ....... and b1, b2, b3 ...... be arithmetic progressions such that a1 = 25, b1 = 75 and
a100 + b100 = 100. Then
(A*)thedifferencebetweensuccessivetermsinprogression'a'isoppositeofthedifferenceinprogression'b'.
(B*) an + bn = 100 for any n.
(C*) (a1 + b1), (a2 + b2), (a3 + b3), ....... are in A.P.
(D*) 


100
1
r
r
r )
b
a
( = 10000
[Sol. a1 + (a1 + d), (a1 + 2d), .........
b1 + (b1 + d1), (b1 + 2d1), .........
hence a100 = a1 + 99d
b100 = b1 + 99d1
add ———————
a100 + b100 = 100 + 99(d + d1)
hence d + d1 = 0  d = – d1  (A)
(B) and (C) areobviouslytrue.



100
1
r
r
r )
b
a
( =
2
100
[(a1 + b1) + (a100 + b100)] =
2
200
100
= 104  (D) (using Sn = )
d
a
(
2
n
 ) ]
Q.8 Let f (x)=



0
x
if
0
0
x
if
]
x
[
x
x
1
x









where [x] denotes the greatest integer function, then thecorrect
statements are
(A*) Limit exists for x = – 1.
(B*) f (x) has a removable discontinuityat x = 1.
(C*) f (x) has a non removable discontinuityat x = 2.
(D*) f (x)is discontinuous at allpositiveintegers.
[Hint: f (1+) = 1 = f (1–) but f (1) = 2; f (–1+) = 3 = f(–1–)
f (2+) = 4; f (2–) = 2; f (2) = 4 ]
Only for12th (Kota andAjmer)
Q.9 Let P andQ are two points denoting thecomplex numbers and  respectivelyon the complex plane.
Which of thefollowingequations can representthe equationofthecirclepassingthrough Pand Qwith
least possible area?
(A)Arg 











z
z
=
2

(B*) Re (z – )( 

z ) = 0
(C*) 2
2
2
|
|
|
Z
|
|
Z
| 







 (D) 










 










 


 z
2
z
2
z
z = 0

3. [Sol. (A) is not correct as it should be ±
2

(B)equationofcircle is,




z
z
ispurelyimaginary
 Re 











z
z
= 0
or Re (z – )( 

z ) = 0  (B) is correct ]
(C) is obviouslycorrect.
(D) can not be correct as centre of required circle is 




 


2
but from this given equation
centre of the circle is – coefficient of z = – 




 


2
.
Only for 12th (Jaipur)
Q.3 Takingyas thedependentandxastheindependentvariable,whichofthefollowingordinarydifferential
equations arenot linear?
(A*) y'y'' + y2 = x2 (B) x2y'' - xy' + 6y = lnx
(C*) [1 + (y')2]1/2 = 5y (D*) y' = y
x 
Only for13th (Kota, Jaipur &Ajmer)
Q.9 f ''(x) > 0 for all x  [–3, 4], then which of the following are always true?
(A) f (x)has a relative minimum on(–3, 4)
(B*) f (x) has a minimum on [–3, 4]
(C*) f (x) is concave upwards on [–3, 4]
(D*) if f (3) = f (4) then f (x) has a critical point on [–3, 4]
[Hint: (A) f(x)has norelative minimumon (–3, 4)
(B) f(x)is continuous function on[–3, 4]
 f (x) has min. and max. on [–3, 4] by IVT
(C) f '' (x) > 0  f (x) is concave upwards on [–3, 4]
(D) f (3) = f (4)
ByRolle'stheorem
 c  (3, 4), where f ' (c) = 0
  critical point on [– 3, 4] ]

4. PART-B
MATCH THE COLUMN [2 × 8 = 16]
INSTRUCTIONS:
Column-Iand column-IIcontainsfour entries each. Entries ofcolumn-Iareto bematchedwith some
entriesofcolumn-II.Oneormorethanoneentriesofcolumn-Imayhavethematchingwiththesameentries
ofcolumn-IIandoneentryofcolumn-Imayhaveoneormorethanonematchingwithentriesofcolumn-II.
Q.52 Column–I Column–II
(A) Let Z = (cos 12° + i sin 12° + cos 48° + i sin 48°)6 (P) – 1
then Im(z) is equal to
(B)  
)
ex
x
(
n
x
sin
n
Lim 3
4
3
0
x




l
l equals (Q) 0
(C) If twice thesquare on the diameter of a circle is equal to the (R) 1
sum of the squares on thesides of the inscribed triangleABC
then sin2A + sin2B + sin2C equals
(D) Let g (x) be a function such that g(a + b) = g(a) · g(b) (S) 2
for all real numbers a and b. If 0 is not an element of the
range ofgthen g(x) · g(–x) equals [Ans. (A) Q; (B) P; (C) S; (D) R]
[Hint: (A) Z = cos 12° + cos 48° + i (sin 48° + sin 12°)
= 2 cos 30° · cos 18° + 2i sin 30° · cos 18°
= 2 cos 18° (cos 30° + i sin 30°)
= 2 cos 18° [cos(/6) + i sin(/6)]
Z6 = 26 · cos618°(cos  + i sin )
= – 26 (cos618°)6 + 0i
Im Z6 = 0 Ans.
(B)  
)
ex
x
(
n
x
sin
n
Lim 3
4
3
0
x




l
l = 








 )
e
x
(
x
x
sin
n
Lim 3
3
0
x
l = ln 





e
1
= – 1 Ans.
(C) Given 2d2 = a2 + b2 + c2; hence 8R2 = 4R2
 A
sin2
  A
sin2
= 2
(D) g(0) = g2(0)  g(0) = 1 (as g (0)  0)
again put a = x, b = – x
g (0) = g (x) · g(–x) = 1 Ans. ]
Q.62 ColumnI ColumnII
(A) ABC is a triangle. If P is a point inside the ABC such that (P) centroid
areas of the triangles PBC, PCAand PAB are equal, then
w.r.t the ABC, Pis its (Q) orthocentre
(B) If c
,
b
,
a



are the position vectors of thethree non collinear
pointsA,B and C respectivelysuch that the vector (R) Incentre
C
P
B
P
A
P
V 



is anull vector then w.r.t.
the ABC, P is its
(C) If Pis a point inside the ABC such that the vector (S) circumcentre
)
C
P
)(
AB
(
)
B
P
)(
CA
(
)
A
P
)(
BC
(
R 



is anull vectorthen
w.r.t. the ABC, Pis its
(D) If Pis a point in theplane of the triangleABC such that the
scalar product B
C
·
A
P and C
A
·
B
P , vanishes then w.r.t.
the ABC, P is its [Ans. (A) P; (B) P; (C) R; (D) Q]

5. PART-C
SUBJECTIVE:
Q.10vector ABCD is a tetrahedron with pv's of its angular points as A(–5, 22, 5); B(1, 2, 3); C(4, 3, 2) and
D(–1,2,–3).IftheareaofthetriangleAEFwherethequadrilateralsABDEandABCFareparallelograms
is S thenfindthe valueof S. [10]
[Ans. 110]
[Sol. pv of M =
2
d
a



= k̂
j
ˆ
12
î
3 


|||ly pv of N =
2
c
a



= k̂
2
7
j
ˆ
2
25
î
2
1



Now the AEF is as shown
5
1
5
6
0
2
k̂
ĵ
î
2
1
S 

k̂
ĵ
10
î
3
S 




= 110
 S = 110 Ans. ]
For Both 12th & 13th (Kota / Jaipur /Ajmer)
Q.11217/3 If the value of the definite integral  

1
0
1
dx
x
1
x
tan
is equal to
k
2
n
l

then find the value of k. [10]
[Ans. k = 8]
[Sol. Considering tan–1x as 1st and
x
1
1

as 2nd and applying IBP
I =
1
0
1
)
x
1
(
n
·
x
tan 

l –  

1
0
2
dx
x
1
)
x
1
(
n
l
I =
4

ln 2 –

 

 

1
I
1
0
2
dx
x
1
)
x
1
(
n
 

l
....(1); put x = tan  in I1
I1 = 




4
0
d
)
tan
1
(
n
l (applying King) ....(2); I1 = 













4
0
d
tan
1
tan
1
1
n
l = 










4
0
d
tan
1
2
n
l
I1 = 


4
0
d
2
n
l – 




4
0
d
)
tan
1
(
n
l ....(3)
2I1 = 


4
0
d
2
n
l =
4

ln 2  I1 =
8

ln 2
 I =
4

ln 2 –
8

ln 2  I =
8

ln 2  k = 8 Ans. ]

6. For Both 12th & 13th (Kota / Jaipur /Ajmer)
Q.12249/3 Suppose f : R  R+ be a differentiable function and satisfies 3 f (x + y) = f (x) · f (y) for all x, yR
with f (1) = 6. If U = 

















)
1
(
n
1
1
n
Lim
n
f
f and V = 
3
0
dx
)
x
(
f . Find the value of (U·V). [10]
[Ans. 126]
[Sol. f ' (x) =
h
)
x
(
f
)
h
x
(
f
Lim
0
h



=
h
)
x
(
f
3
)
h
(
f
·
)
x
(
f
Lim
0
h


(using f rule)
=
h
3
)
x
(
f
3
)
h
(
f
·
)
x
(
f
Lim
0
h


f ' (x) =
h
]
3
)
h
(
f
[
·
3
)
x
(
f
Lim
0
h


=
h
)
0
(
f
)
h
(
f
Lim
3
)
x
(
f
0
h


(from f rule x = 0; y = 0; 3 f (0) = f 2(0);  f (0)  0 hence f (0) = 3)
f ' (x) =
3
)
x
(
f
· f ' (0);  3 ·
)
x
(
f
)
x
(
'
f
= f ' (0) = k (say)
integrating
3 ln 
)
x
(
f = kx + C; put x = 0; f (0) = 3  3 ln 3 = C
hence, 3 ln 
)
x
(
f = kx + 3 ln 3
3 ln 





3
)
x
(
f
= kx; put x = 1; f (1) = 6  3 ln 2 = k
hence 3 ln 





3
)
x
(
f
= (3 ln 2) x  ln 





3
)
x
(
f
= x ln 2






3
)
x
(
f
= e(ln 2)x = 2x  f (x) = 3 · 2x .....(1)
now U =
h
)
1
(
f
)
h
1
(
f
Lim
0
h



where n =
h
1
U = f ' (1) =
1
x
x
2
n
2
·
3

l ; hence U = 6 ln 2
now, V = 
3
0
x
dx
2
3 =  3
0
x
2
2
n
3
l
=
2
n
21
l
, hence V =
2
n
21
l
 U · V = (6 ln 2) 







2
n
21
l
= 126 Ans. ]

7. Only for 12th (Kota / Jaipur /Ajmer)
Q.1318/5 Withrespecttoaparticularquestiononamultiplechoicetest(having4 alternatives withonly1correct)
a student knows the answer and therefore can eliminate 3 of the 4 choices from consideration with
probability2/3,caneliminate2ofthe4choicesfromconsideration withprobability1/6,caneliminate1
choice from consideration with probability1/9, and can eliminate none with probability 1/18. If the
student knows the answer, he answers correctly, otherwise he guesses from among the choices not
eliminated.
If theanswergivenbythestudent was found correct,then the probabilitythathe knewthe answeris
b
a
where a and b are relativelyprime. Find the value of (a + b). [10]
[Ans. p = 144 + 173 = 317]
[Sol. A: his answer was correct
B1 : hecan eliminate3 out of 4 alternatives
B2 : hecan eliminate2 out of 4 alternatives
B3 : hecan eliminate1 out of 4 alternatives
B4 : hecan eliminatenoneout of4alternatives
P(B1) =
3
2
; P(B2) =
6
1
; P(B3) =
9
1
; P(B4) =
18
1
P(A/B1) = 1; P(A/B2) =
2
1
; P(A/B3) =
3
1
; P(A/B4) =
4
1
P(B1 / A) =
)
A
(
P
)
A
B
(
P 1 
= )
B
A
(
P
)
B
A
(
P
)
B
A
(
P
)
B
A
(
P
)
A
B
(
P
4
3
2
1
1








=


4
1
i
i
i
1
1
)
B
/
A
(
P
·
)
B
(
P
)
B
/
A
(
P
·
)
B
(
P
=
4
1
·
18
1
9
1
·
3
1
2
1
·
6
1
1
·
3
2
1
·
3
2



=
173
216
·
3
2
=
173
144
=
b
a
 a + b = 144 + 173 = 317 Ans. ]
Only for 13th (Kota / Jaipur /Ajmer)
Q.13 The sides of a triangle are consecutive integers n, n + 1 and n + 2 and the largest angle is twice the
smallestangle.Findn. [10]
[Ans. 4]
[Sol. Let A> B > C i.e. Ais the largest and C is the smallest. [12th & 13th 03-12-2006]
Let C =   A = 2 [T/S, Q.25, Ex-2, Ph-3]
also a = n + 2, b = n + 1, c = n


2
sin
a
=

sin
c




cos
sin
2
2
n
=

sin
n
 2 cos  =
n
2
n 
 cos C =
n
2
2
n 
....(1)
also cos C =
ab
2
c
b
a 2
2
2


=
)
1
n
)(
2
n
(
2
n
)
1
n
(
)
2
n
( 2
2
2






cos C =
)
2
n
3
n
(
2
5
n
6
n
2
2




.....(2)

8. from (1) and (2)
)
2
n
3
n
(
2
5
n
6
n
2
2




=
n
2
2
n 
=
2
1
+
n
1

2
n
3
n
5
n
6
n
2
2




– 1 =
n
2

2
n
3
n
3
n
3
2



=
n
2
 2n2 + 6n + 4 = 3n2 + 3n
 n2 – 3n – 4 = 0  (n – 4)(n + 1) = 0  n = 4 Ans. ]
(Only in 12th and only forKota andAjmer)
Q.14 If the expression z5 – 32 canbe factorised intolinear and quadratic factors over real coefficients 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 [12th 3-12-2006]
zi = 2 




 


5
m
2
sin
5
m
2
cos i (usingDemoivre'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.]

9. (Only in 13th and only for Kota andAjmer) and (Both in 12th & 13th ForJAIPUR)
Q.1433/3 f : R  R, f(x) =
1
x
n
mx
x
3
2
2



. Iftherangeof this function is[– 4, 3)then find thevalueof (m2 +n2).
[10]
[Ans. m = 0; n = – 4 and (m2 + n2) = 16]
[Sol. f (x) = 2
2
x
1
3
n
mx
)
1
x
(
3





; f (x) = 3 + 2
x
1
3
n
mx



y = 3 + 2
x
1
3
n
mx



for y to lie in [– 4, 3)
mx + n – 3 < 0  x  R
this ispossible onlyif m = 0
when, m = 0 then y = 3 + 2
x
1
3
n


note that n – 3 < 0 (think !)
n < 3
if x  , ymax  3–
now ymin occurs at x = 0 (as1 +x2 is minimum)
ymin = 3 + n – 3 = n  n = – 4 ]