1. REVIEW TEST-5
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 : XIII (XYZ)
Time : 3 hour Max. Marks : 243
INSTRUCTIONS
1. The question paper contain 24 pages and 4-parts. Part-A contains 9 objective question, Part-B contains
6 questions of "Match the Column" type and Part-C contains 15 and Part-D contains 2 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 eachwrong match. Marks will be awarded onlyifall the correct
alternatives are selected.
PART-C
(iv) Q.1 to Q.15 are "Subjective" questions. There is NO NEGATIVE marking. Marks will be awarded only if all the correct
bubbles are filled in your OMR sheet.
PART-D
(iv) Q.1 to Q.2 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 ]
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] ]
Class - XIII Mathematics Paper - 2

3. 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]

4. PART-C
SUBJECTIVE:
Q.11vector 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. ]
Q.12217/3 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]
[Sol. Considering tan–1x as 1st and
x
1
1

as 2nd and applying IBP [Ans. k = 8]
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. ]

5. Q.13249/3 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. ]
Q.14 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.
Let C =   A = 2
also a = n + 2, b = n + 1, c = n

6. 

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)
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. ]
Q.1533/3 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 ]