1. Class : XIII (XYZ)
Time : 3 hour Max. Marks : 276
INSTRUCTIONS
1. The question paper contain 24 pages and 3-parts. Part-A contains 9 objective question, Part-B contains
9 questions of "Match the Column" type and Part-C contains 15 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 More than one are correct alternative and carry 6 marks each.
There is NEGATIVE marking. 1 mark will be deducted for each wrong answer.
PART-B
(iii) Q.1 to Q.9 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.15 are "Subjective" questions and carry 10 marks for each. There is NO NEGATIVE marking. Marks will be
awarded only if all the correct bubble are filled in you answer sheet.
2. Indicate the correct answer for each question by filling appropriate bubble in your answer sheet.
3. Use only HB pencil for darkening the bubble.
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.
REVIEW TEST-3
PAPER CODE : A
P A P E R - 2
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
(3 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
bubble is
A B C D
the wrong method for filling
the bubble are
The answer of the questions
inwrong or anyother manner
will be treated as wrong.
PART-D
Ensure that all columns
(3 before decimal and 2 after
decimal) are filled. Answer
having blank column will be
treated as incorrect. Insert
leading zero(s) if required
after rounding the result to 2
decimal places.
e.g. – 4.19 × 1027 should be
filled as – 4.19 E + 27

2. XIII (XYZ) REVIEW TEST-3
Select the correct alternative. (More than one is/are correct) [3 × 6 = 18]
There is NEGATIVE marking. 1 mark will be deducted for eachwrong answer.
Q.1 The function f (x) is defined for x  0 and has its inverse g (x) which is differentiable. If f (x) satisfies

)
x
(
g
0
dt
)
t
(
f = x2 and g (0) = 0 then
(A*) f (x) is an odd linear polynomial (B) f (x)issomequadraticpolynomial
(C*) f (2) = 1 (D*) g (2) = 4
[Sol. differentiating 
)
x
(
g
0
dt
)
t
(
f =x2
  
)
x
(
g
f g ' (x) = 2x ....(1)
since f and g inverse of each other hence  
)
x
(
g
f = x.
Hence (1) becomes x g'(x) = 2x or g'(x) = 2
integrating g (x) = 2x + C
but theinitial conditionshows C =0
hence , g (x) = 2x
its inverse f (x) =
2
x
 A, C, D ]
Q.2 Consider a triangleABC in xyplanewith D,E and F as the middle points of the sides BC, CAandAB
respectively. If thecoordinates of the points D, E and Fare (3/2, 3/2); (7/2, 0) and (0, –1/2) then which
ofthefollowingarecorrect?
(A*)circumcentreof thetriangleABCdoes not lie insidethetriangle.
(B) orthocentre, centroid, circumcentreand incentre of triangleDEF are collinear but of triangleABC
arenon collinear.
(C*) Equation ofalinepasses through theorthocentreoftriangleABC and perpendiculartoits planeis
k̂
)
j
ˆ
î
(
2
r 




(D*) distancebetween centroid and orthocentreof the triangleABC is
3
2
5
.
[Hint: Note that D is 90°
hence  A = 9°
ABC and  DEF are both right isosceles triangle.
 Nowverifyallalternatives ]
Q.3 Ifacontinuousfunctionf (x)satisfies therelation,  
x
0
dt
)
t
x
(
t f = 
x
0
dt
)
t
(
f +sinx +cosx –x –1, for
all real numbersx, thenwhichof thefollowingdoesnot hold good?
(A*) f (0) = 1 (B*) f ' (0) = 0 (C) f '' (0) = 2 (D*) 

0
dx
)
x
(
f = e
[Sol.  
x
0
dt
)
t
x
(
t f = 
x
0
dt
)
t
(
f + sin x + cos x – x – 1 [T/S, Q.10, Ex-IV, DE]

 

 

King
using
x
0
dt
)
t
(
f
)
t
x
(
  = 
x
0
dt
)
t
(
f + sin x + cos x – x – 1

3. x 
x
0
dt
)
t
(
f – 
x
0
dt
)
t
(
t f = 
x
0
dt
)
t
(
f + sin x + cos x – x – 1
differentiating
x · f (x) + 
x
0
dt
)
t
(
f – x · f (x) = f (x) + cos x – sin x – 1 ....(1)
againdifferentiation
f (x) = f '(x) – sin x – cos x ....(2)
which isalinear differential equation of 1st order
let f (x) = y

dx
dy
– y = sin x + cos x
I.F. = e–x
 y · e–x =  

dx
)
x
cos
x
(sin
e x
; put x = – t
=   dt
)
t
cos
t
(sin
et
y · e–x = – et cos t + C
y · e–x = – e– x cos x + C
f (x) = y = C ex – cos x
if x = 0, f (0) = 0 from (1)
 C = 1
f (x) = ex – cos x Ans. ]
PART-B
MATCH THE COLUMN [3 × 8 = 24]
There is NEGATIVE marking.0.5 mark will be deductedfor each wrong match within a question.
INSTRUCTIONS:
Column-Iand column-IIcontainsfour entries each. Entries ofcolumn-Iareto bematchedwith some
entriesofcolumn-II.Oneormorethanoneentriesofcolumn-Imayhavethematchingwiththesameentries
ofcolumn-IIandoneentryofcolumn-Imayhaveoneormorethanonematchingwithentriesofcolumn-II.
Q.1 ColumnI ColumnII
(A) 


x
e
x 3 t
n
dt
x
x
n
Lim
l
l
is (P) 0
(B) 





 



1
x
1
x
x
2
4
e
e
Lim is (Q)
2
1
(C)
n
4
)
1
n
(
sin
1
n
5
.
0
n
sin
)
1
(
Lim 2
n
n







 





is where n  N (R) 1
(D) The valueof the integral 







 










1
0
2
1
1
dx
2
x
2
x
2
1
tan
1
x
x
tan
is (S) nonexistent
[(A) R; (B) S; (C) Q; (D) Q]
[Hint: (A)
x
t
n
dt
·
x
n
Lim
x
e
x
3



l
l








form

4. usingL'Hospital'srule
ln x ·
x
n
1
l
+
x
1
·
t
n
dt
x
e3








l
(1 + 0) = 1 Ans.
(B) 






 





1
e
e
Lim )
1
x
(
1
x
1
x
x
2
4
2
l
now as )
1
x
(
1
x
Lim 2
4
x





=
)
1
x
(
1
x
)
x
2
1
x
(
1
x
Lim
2
4
2
4
4
x









=
  




 





)
x
1
(
1
x
1
1
x
x
2
Lim
2
4
2
2
x
= – 1
now as x 
)
1
x
(
1
x 2
2


  – 1
 × 





1
e
1
 –  and hence limit does not exist Ans.
(C) 













 


1
2
n
n
n
sin
)
1
(
)
1
(
Lim 2
1
n
n
n
=








































 


1
2
n
n
n
1
2
n
n
n
1
2
n
n
n
sin
)
1
(
Lim
2
2
2
1
n
2
n
=































 


2
2
2
1
n
2
n
n
1
n
2
1
1
1
n
1
2
n
n
n
sin
)
1
(
Lim =




















2
n
2
n
n
1
n
2
1
1
1
n
2
1
2
1
sin
)
1
(
Lim
as n  
sin
4

=
2
1
also as n  , sin
2
1
n
4
)
1
n
(



 final answeris
2
1
Ans.
(D) Use King and add to get f (x) + f (1 – x) = 1
 2I = 1  I = 1/2 Ans. ]

5. Q.2 Consider the matricesA= 







1
1
4
3
andB= 





1
0
b
a
andlet Pbeanyorthogonal matrix and Q= PAP
APT
and R = PTQKP also S = PBPT and T = PTSKP
ColumnI ColumnII
(A) If we varyK from 1 to n then the first row (P) G.P. with common ratio a
firstcolumnelementsatRwillform
(B) If we varyK from 1 to n then the 2nd row 2nd (Q) A.P. with common difference 2
columnelementsatRwillform
(C) If we varyK from 1 to n then the first row first (R) G.P. with common ratio b
columnelementsofTwillform
(D) If we varyK from 3 to n then the first row 2nd column (S) A.P. with common difference– 2.
elements ofT willrepresent thesum of
[Ans. (A) Q; (B) S; (C) P; (D) P]
[Sol. R = PTQKP
= PT(PAPT)KP
= P
PAP
....
..........
PAP
PAP
P
times
K
T
T
T
T



 



 

= AK as PPT = I as P is orthogonal
= AK = 








K
2
1
K
K
4
K
2
1
|||ly T = PTSKP = BK
BK =










1
0
1
a
)
1
a
(
b
a
K
K
]
Q.3 ColumnI ColumnII
(A) Given two vectors b
and
a


such that |
b
|
|
a
|


 = |
b
a
|


 = 1 (P) 30°
The angle between the vectors b
a
2


 and a

is
(B) In a scalene triangleABC, if a cosA= b cos B (Q) 45°
then  C equals
(C) In a triangleABC, BC = 1 andAC = 2.The maximum possible (R) 60°
value which the Acanhave is
(D) In a ABC  B = 75° and BC = 2AD whereAD is the (S) 90°
altitude fromA, then C equals
[Ans. (A) P; (B) S; (C) P; (D) P ]
[Sol. (C) Usingcosinerule
cos A =
x
4
1
x
4 2


where AC = x
=
x
4
3
x2

= 






x
3
x
4
1
=

















 3
2
x
3
x
4
1
2

6. hence minimum value of cosA=
2
3
i.e. cosA
2
3
 Amax = 30°
(D) BD = x cot 75°
and DC = x cot 
add ——————
2x = x(cot 75 + cot )
 cot  = 2 – (2 – 3 ) = 3   = 30°]
PART-C
SUBJECTIVE: [5 × 10 = 50]
Q.1 Suppose V = 


2
0
2
dx
2
1
x
sin
x , find the value of

V
96
. [Ans. 12]
[Sol. V = 


2
0
2
dx
2
1
x
sin
x = 


2
0
2
dx
1
x
sin
2
x
2
1
= 
 2
0
dx
x
2
cos
x
2
1
put 2x = t  dx =
2
dt
V = 

0
dt
t
cos
t
8
1
= 



0
dt
t
cos
)
t
(
8
1
2V = 


0
dt
t
cos
8
=
8
2
; V =
8


8
·
96 

= 12 Ans. ]
Q.2134/1 One of the roots of the equation 2000x6 + 100x5 + 10x3 + x – 2 = 0 is of the form
r
n
m 
, where
m is non zero integer and n and r are relativelyprime natural numbers. Find the value of m + n + r.
[Ans. 200]
[Sol. 2
x
x
10
x
100
x
2000
.
P
.
G
a
3
5
6



 
 

 
 = 0
2000x6 +
 
1
x
10
1
)
x
10
(
x
2
3
2


– 2 = 0
1
x
10
)
1
x
1000
(
x
2
6


= – 2(1000x6 – 1)
 1000x6 – 1 = 0 or
1
x
10
x
2

= – 2  x = – (10x2 – 1)
x2 =
10
1
which is not possible

7. 20x2 + x – 2 = 0
x =
40
160
1
1 


=
40
161
1

or
40
161
1

 m = – 1; n = 161; r = 40
m + n + r = 200 Ans. ]
Q.3225/2 Acircle C is tangent to the x and yaxis in the first quadrant at the points Pand Q respectively. BC and
AD are parallel tangents to thecircle with slope – 1. If the pointsAand B areon the y-axis while C and
Dareonthex-axisandtheareaofthefigureABCDis 900 2 sq.units thenfindtheradius ofthecircle.
[Ans. r = 15]
[Sol. Note that ABCD is a trapezium and its area
Area =
2
1
(a + b) · h
where
2
b
a 
= EF (median)
and h = 2R
A = (2r)(EF)
now equation of EF is y= – x + c ....(1)
(1) passes through (r, r)  c = 2r
x + y = 2r, hence E = (2r, 0) and F = (0, 2r)
 EF = 2
2
r
4
r
4  = 2
2 r
 AreaABCD = (2r)( 2
2 r) = 2
4 r2
 2
4 r2 = 2
900  r2 = 225  r = 15 Ans. ]
Q.4 Let f (x) = ax2 –4ax + b(a>0) be definedin 1 x  5. Supposethe average ofthemaximum value and
theminimumvalueofthefunctionis14,andthedifferencebetweenthemaximumvalueand minimum
value is 18. Find the value of a2 + b2. [Ans. a2 + b2 = 173]
[Sol. f (x) = ax2 – 4ax + b (a > 0)
f ' (x) = 2ax – 4a = 0
x = 2
also, f ' (x) = 2a(x – 2)  for x  (1, 2) f is 
Henceminimumoccursas
x = 2
f (2) = 4a – 8a + b
f (2) = b – 4a
maximumwilloccuratf(5)and
f (5) = 25a – 20a + b
= b + 5a
 M = b + 5a
m = b – 4a
—————
M – m = 9a = 18  2
a 
also
2
m
M 
= 14  M + m = 28 = 2b + a  13
b 
Hence a = 2 and b = 13
 a2 + b2 = 4 + 169 = 173 Ans. ]

8. Q.513/3 If the 









 bx
1
ax
1
x
1
1
x
1
Lim 3
0
x
existsand hasthevalueequal to l, thenfind thevalueof
b
3
2
a
1


l
.
[Sol: 









 bx
ax
x
x
Lt
x 1
1
1
1
1
3
0
3
0
1
)
1
(
1
x
x
ax
bx
Lt
x




 

 


 

1
0 )
1
(
1
1

 
 bx
x
Lt
x
  
3
2
/
1
0
1
1
1
x
x
ax
bx
Lt
x





UseBinomialExpansion
Toget relations













0
2
8
1
0
2
1
a
a
b
a =
4
1
b =
4
3
l =
32
1
 ]